1. No category

PART A TOPOLOGY COURSE: HT 2011
Contents
1. What is Topology about ?
2. Metric spaces
2.1. Definition and examples
2.2. Balls, open, closed and bounded sets
2.3. Convergent sequences and complete metric spaces
2.4. Continuous maps of metric spaces
2.5. The space of continuous functions
2.6. Various strong versions of continuity
3. Topological spaces
3.1. Definitions and examples
3.2. Closure of a set
3.3. Interior of a set
3.4. Boundary of a set
3.5. Completion of a metric space
3.6. Separation axioms
3.7. Subspace of a topological space
3.8. Basis for a topology
3.9. Product of topological spaces
3.10. Product of topological spaces and continuous maps
4. Connected spaces
4.1. Definition and properties of connected spaces
4.2. Path-connected spaces
5. Compact spaces
5.1. Definition and properties of compact sets
5.2. Compact spaces and continuous maps
5.3. Sequential compactness
6. Quotient spaces, quotient topology
6.1. Definition and examples
6.2. Separation axioms
6.3. Quotient spaces and continuous maps
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Recommended books in Topology:
(1) J. Dugundji, Topology;
(2) J.R. Munkres, Topology, A First Course;
(3) W. Sutherland, Introduction to Metric and Topological Spaces (particularly recommended);
(4) O. Viro, O. Ivanov, V. Kharlamov, N. Netsvetaev, Elementary Topology,
http://www.math.uu.se/ ∼ oleg/educ-texts.html
For further reading see the list at the end of the synopsis.
These notes were created by Cornelia Drutu, and were developed out of the set of slides written by
Wilson Sutherland for the course as given in 2007. I wish to thank them both for having allowed me
to use them.
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PART A TOPOLOGY COURSE: HT 2011
1. What is Topology about ?
You have already seen and used some notions of Topology up to now:
(1) In Analysis in MT. Several notions in Analysis, like:
• convergent sequence, limit of a function in a point;
• bounded sequence, bounded function;
• continuous function;
need a way of defining nearby points as well as a way of defining points not too far apart.
This can be done via a distance (or metric).
In order to differentiate, one needs to consider functions defined on open sets.
(2) In Linear Algebra MT. There you have seen the notion of inner product space. In such a space
one can define
• a notion of length of a vector (hence of distance between two points);
• and a notion of angle between vectors: in particular one may speak of orthogonality, and
Trigonometry.
In what follows we weaken that structure, and we only require to have a notion of distance between
two points. In particular we give up the notion of angle.
For the distance we require three simple and easy to admit properties.
PART A TOPOLOGY COURSE: HT 2011
3
2. Metric spaces
2.1. Definition and examples.
Definition 2.1. A metric space is a non-empty set X endowed with a function d : X × X → R with
the following properties:
(M1) d(x, y) > 0 for all x, y ∈ X ; d(x, y) = 0 if and only if x = y;
(M2) (Symmetry) for all x, y ∈ X, d(y, x) = d(x, y);
(M3) (Triangle inequality) for all x, y, z ∈ X, d(x, z) 6 d(x, y) + d(y, z).
The function d is called metric or distance.
Y
HH
x AH
H H
A HH HH
HH d(x, z)
A
HH H
A
HHH H
A
H HHH
A
j
H
HH
A
y
z
A first example of metric space is provided by the following familiar structure.
Definition 2.2. A normed vector space (NVS) is a vector space V over R or C together with a
function V → R called a norm and written v 7→ ||v|| such that
(N1) for all v ∈ V, ||v|| > 0; ||v|| = 0 if and only if v = 0;
(N2) for all v ∈ V and any scalar α, ||αv|| = |α| ||v||;
(N3) (subadditivity) for all u, v ∈ V, ||u + v|| 6 ||u|| + ||v||.
Proposition 2.3. A norm || || defines a metric on V , d(x, y) = ||x − y|| , for every x, y ∈ V .
Proof. Property (M1) follows from (N1), property (M2) follows from (N2) applied to α = −1 and to
v = x − y.
Property (M3) follows from (N3) applied to u = x − y and v = y − z .
Remarks 2.4.
(1) A vector subspace U of a normed vector space (V, k k) is naturally endowed
with a norm: the restriction of k k to U.
(2) A Cartesian product V × W of two normed vector spaces (V, k kV ) p
and (W, k kW ) can be
endowed with several norms: k kV + k kW , max (k kV , k kW ) and k k2V + k k2W .
Example 2.5. On Rn several norms can be defined. For x = (x1 , x2 , . . . , xn ), the l2 , l1 and l∞
norms are
v
u n
n
X
uX
||x||2 = t
|xi | ,
||x||∞ = max |xi | .
x2i ,
||x||1 =
i=1
See Sheet 1, Exercise 4.
i=1
1≤i≤n
4
PART A TOPOLOGY COURSE: HT 2011
n
Remark 2.6.
p The norm || ||2 is defined by the standard inner product (or dot product) on R , that
is ||x||2 = hx, xi , where
hx, yi = x1 y1 + x2 y2 + · · · + xn yn .
n
The space R endowed with the above inner product is sometimes called Euclidean space, therefore
|| ||2 is called the Euclidean norm and the metric it defines the Euclidean metric on Rn .
The other two norms || ||1 and || ||∞ are not defined by any inner product on Rn .
A more complicated example is the following.
Example 2.7. (Function spaces) Let S be any non-empty set and let X be the set B(S, R) of all
bounded functions from S to R . Then X is a real vector space, where for f, g ∈ X and α ∈ R we
define f + g and αf by:
(f + g)(s) = f (s) + g(s),
(αf )(s) = αf (s) .
Moreover ||f ||∞ = sups∈S |f (s)| is a norm on B(S, R) .
Proof. Property (N1) is immediate.
For any scalar α and function f ∈ X , ||αf ||∞ = sups∈S |αf (s)| = |α| sups∈S |f (s)| = |α|||f ||∞ .
For any two functions f, g ∈ X and any s ∈ S , |f (s) + g(s)| ≤ |f (s)| + |g(s)| ≤ ||f ||∞ + ||g||∞ .
Whence ||f + g||∞ ≤ ||f ||∞ + ||g||∞ .
There are many metric space with metric not defined by a norm, as in the following example. This
is also a rather counter-intuitive example: a metric space in which all distinct points are at the same
distance.
Example 2.8 (Discrete metric spaces). The discrete metric on any non-empty set X is defined by
0 if x = y,
d(x, y) =
1 if x 6= y.
This space (X, d) is called a discrete metric space.
Proof. Properties (M1) and (M2) are clearly satisfied.
In (M3) it is enough to study the case when x 6= z . Then either x 6= y or z 6= y . Thus d(x, z) = 1
and d(x, y) + d(y, z) ≥ 1.
Examples 2.9.
(1) A non-empty subset A of a metric space (X, d) is naturally endowed with a
metric: the restriction d|A×A of d to A × A.
(2) A Cartesian product X1 × X2 of two metric spaces can be endowed with a metric (see Sheet
1, Exercise 6).
2.2. Balls, open, closed and bounded sets.
Definition 2.10. Let (X, d) be a metric space, x0 a point in X and r > 0.
The open ball in X of radius r centred at x0 is the set B(x0 , r) = {x ∈ X : d(x, x0 ) < r}.
The closed ball in X of radius r centred at x0 is the set B ′ (x0 , r) = {x ∈ X : d(x, x0 ) ≤ r}.
The sphere in X of radius r centred at x0 is the set S(x0 , r) = {x ∈ X : d(x, x0 ) = r}.
If more than one metric is defined on X then we write Bd (x0 , r), Bd′ (x0 , r), Sd (x0 , r) .
Examples 2.11.
(a) X = R2 , d = d2 : Bd2 (x0 , r) is the open disc of radius r centred at x0 .
(b) X = R2 , d = d1 : Bd1 (x0 , r) is the inside of the square centred on x0 with diagonals of
length 2r parallel to the axes, as in the diagram below.
PART A TOPOLOGY COURSE: HT 2011
2
(c) X = R , d the discrete metric: Bd (x0 , r) =
5
{x0 } if r 6 1
R2 if r > 1.
(d) X the set B([0, 1], R) of all bounded real-valued functions on [0, 1] , d = d∞ the sup metric:
for f0 ∈ X and r > 0, Bd∞ (f0 , r) is the set of all f ∈ X whose graphs lie inside a ribbon of
vertical width 2r centred on the graph of f0 .
@
@
@
@
@
@
(r, 0)
@
@
Definition 2.12. A subset A of a metric space X is bounded if A ⊆ B(x0 , r) for some x0 ∈ X and
r > 0.
Definition 2.13. Let (X, d) be a metric space and U ⊆ X. We say U is open in X if for every
x ∈ U , there exists εx > 0 such that B(x, εx ) ⊆ U.
Examples 2.14.
(a) X = R2 , d = d2 : open sets are as in MT Analysis course.
(b) X = R2 , d = d1 or d∞ : open sets exactly as in (a) above (see Sheet 1, Exercise 5).
(c) X endowed with the discrete metric: any subset U ⊆ X is open.
(d) Any open ball B(x0 , r) in a metric space (X, d) is open in X .
Let us explain why (d) is true: take an arbitrary point x ∈ B(x0 , r) and εx = r − d(x, x0 ) . Then
for every y ∈ B(x, εx ) , d(x0 , y) ≤ d(x0 , x) + d(x, y) < r .
Proposition 2.15. A subset in a metric space is open if and only if it is a union of open balls.
S
Proof. Consider a union of open balls A = i∈I B(xi , ri ) . For every x ∈ A there exists j ∈ I such
that x ∈ B(xj , rj ) . According to the argument justifying Example (d) above, for εx = rj − d(x, xj )
the ball B(x, εx ) is contained in B(xj , rj ) , hence in A .
Conversely, let U be an open subset in a metric space (X, d) . For every x ∈ U , there exists εx > 0
such thatSB(x, εx ) ⊆ U.
S
Then x∈U B(x, εx ) ⊆ U . Since x ∈ B(x, εx ) , it follows that U ⊆ x∈U B(x, εx ) . Whence the
equality.
Corollary 2.16. A subset in (R, | |) is open if and only if it is a union of open intervals.
Proposition 2.17. Suppose that X is a metric space. Then
(a) X and ∅ are open in X ;
(b) if U and V are open in X , then so is U ∩ V ;
[
(c) if Ui is open in X for all i ∈ I , then so is
Ui .
i∈I
Remark 2.18. Note that in Proposition 2.17, (c), I might be infinite uncountable.
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PART A TOPOLOGY COURSE: HT 2011
Proof. (b) Consider an arbitrary point x ∈ U ∩ V .
As U is open there exists αx > 0 such that B(x, αx ) ⊆ U . Likewise, as x ∈ V and V is open
there exists βx such that B(x, βx ) ⊆ V .
Then for δx = min(αx , βx ) , B(x, δx ) ⊆ B(x, αx ) ⊆ U and B(x, δx ) ⊆ B(x, βx ) ⊆ V . Therefore
B(x, δx ) ⊆ U ∩ V .
S
(c) For every x ∈ i∈I Ui thereSexists k ∈ I such that x ∈ Uk . Since Uk is open there exists
εx > 0 such that B(x, εx ) ⊆ Uk ⊆ i∈I Ui .
Remark 2.19. Property (b) in Proposition 2.17 is not true for infinite intersections:
\
n∈N,n≥1
1 1
− ,
n n
= {0} .
Definition 2.20. A subset V of a metric space X is closed in X if X \ V is open in X .
Examples 2.21.
(a) In Euclidean spaces, the closed sets are exactly as you saw last term.
(b) In a discrete metric space X , any subset of X is closed in X (since any subset is open in X ).
(c) Any closed ball Bd′ (x0 , r) in a metric space (X, d) is closed in X .
Remark 2.22.
(1) In a metric space there may exist subsets that are neither open nor closed. For
instance intervals [a, b) in (R, | |) .
(2) On the other hand in a discrete metric space X all subsets are both open and closed.
Proposition 2.23. Suppose that X is a metric space. Then
(a) X and ∅ are closed in X ;
(b) if A and B are closed in X , then so is A ∪ B ;
\
(c) if Ai is closed in X for all i ∈ I , then so is
Ai .
i∈I
Proof. It follows from the fact that A is closed if and only if X \ A is open, from Proposition 2.17,
and from the De Morgan laws (see Appendix A).
Remark 2.24. Property (b) in Proposition 2.23 is not true for infinite unions:
[ 1
1
= (0, 1) .
,1 −
n
n
n∈N,n≥1
Proposition 2.25. Let (X, d) be a metric space and let A ⊂ X be endowed with the restricted
metric dA .
Then a subset Y ⊂ A is open (closed) in (A, dA ) if and only if Y = A ∩ U , where U is open
(respectively closed) in X .
PART A TOPOLOGY COURSE: HT 2011
7
Proof. We begin by noting that if a ∈ A and r > 0 then BdA (a, r) = B(a, r) ∩ A .
Assume that Y ⊂ A is open in (A, dA ) . Then for every y ∈ Y there
S exists εy > 0 such that
BdA (y, εy ) ⊆ Y . As in the proof of Proposition 2.15 it follows that Y = y∈Y BdA (y, εy ) .
Take
[
U=
B(y, εy ) .
Then U ∩ A =
S
y∈Y
y∈Y
[B(y, εy ) ∩ A] = Y .
Conversely, let U be an open set in X and Y = U ∩ A . For every y ∈ Y ⊆ U there exists εy > 0
so that B(y, εy ) ⊆ U . Then B(y, εy ) ∩ A = BdA (y, εy ) ⊆ U ∩ A = Y .
The statement on closed sets follows immediately from the statement on open sets, and from the
fact that A \ (A ∩ U ) = A ∩ (X \ U ) .
2.3. Convergent sequences and complete metric spaces. We now define convergent sequences
and complete metric spaces.
In the setting of metric spaces, convergent sequences can be used to define most of the important
topological notions (closed sets, continuous functions etc).
Complete metric spaces are spaces in which:
• convergent sequences can be defined without a priori knowing their limit (using the Cauchy
property).
• the Contraction Map Theorem is true. This theorem is an important tool in Analysis, in
particular in solving various types of equations.
Definition 2.26. A sequence (xn ) in a metric space (X, d) converges to a point a ∈ X if for every
ε > 0, there exists N ∈ N such that whenever n > N
d(xn , a) < ε :
equivalently, whenever n > N , xn ∈ B(a, ε) .
The sequence (xn ) is called a convergent sequence; the element a is called the limit of the sequence
(xn ) .
Remark 2.27. In the above one can replace the inequality d(xn , a) < ε by the inequality d(xn , a) ≤ ε .
Indeed, consider the properties:
(strict) for every ε > 0, there exists N ∈ N such that whenever n > N
d(xn , a) < ε .
(non-strict) for every ε > 0, there exists N ∈ N such that whenever n > N
d(xn , a) ≤ ε .
We prove that these properties are equivalent.
Clearly (strict) implies (non-strict). Conversely, assume that (non-strict) is satisfied, and
let us prove (strict). Consider an arbitrary ε > 0. Property (non-strict) applied to 2ε implies
that there exists N such that for every n ≥ N , d(xn , a) ≤ 2ε < ε .
Proposition 2.28. The limit of a convergent sequence in a metric space is unique.
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PART A TOPOLOGY COURSE: HT 2011
Proof. Indeed, assume that (xn ) converges to a , and that (xn ) converges to b . Take ε > 0 arbitrary.
There exists N1 such that for every n ≥ N1 , d(xn , a) < ε , and N2 such that for every n ≥ N2 ,
d(xn , b) < ε . Then for any n ≥ max(N1 , N2 ) ,
d(a, b) ≤ d(xn , a) + d(xn , b) < 2ε .
We have thus found that for every ε > 0, d(a, b) < 2ε . It follows that d(a, b) = 0, whence
a = b.
Examples 2.29.
(a) X = R with the Euclidean metric: convergence is the usual convergence of
(xn ) to a .
(b) X with a discrete metric: a sequence (xn ) is convergent if and only if there exists N such
that xn = xN for every n ≥ N .
(c) X = B(A, R), where A ⊆ R , endowed with the metric
d∞ (f, g) = kfg k∞ = sup |f (a) − g(a)| .
a∈A
The sequence (fn ) of functions converges to f in (X, d∞ ) if and only if (fn ) converges to
f uniformly on A .
The notion of convergent sequence allows to define closed subsets in metric spaces without having
to refer to their complementary subsets.
Proposition 2.30. A subset F in a metric space (X, d) is closed if and only if for every convergent
sequence contained in F its limit is also in F .
Proof. Assume that F is closed and let (xn ) be a convergent sequence in F with limit a .
If a ∈ X \ F , as X \ F is open, there exists ε > 0 such that B(a, ε) ⊆ X \ F .
Since (xn ) converges to a it follows that there exists N such that for every n ≥ N , xn ∈ B(a, ε) ⊆
X \ F . This contradicts the fact that the entire sequence (xn ) is contained in F .
Therefore a cannot be in X \ F , it must be in F .
Conversely, assume that F has the property that for every convergent sequence contained in F its
limit is also in F . We want to prove that X \ F is open.
Assume, on the contrary, that X \ F is not open. Then there exists a ∈ X \ F such that no ball
centred in a is contained in X \ F . Thus, for every ε > 0, B(a, ε) ∩ F 6= ∅.
In particular for every integer n ≥ 1 we can pick a point xn in B a, n1 ∩ F . It is easy to check
that the sequence (xn ) converges to a . Since (xn ) is contained in F , by hypothesis a should be
likewise in F . This contradicts the fact that a ∈ X \ F .
We conclude that X \ F must be open.
We would like to find an intrinsic way of defining a convergent sequence, without having to refer
to an extra object, its limit.
We notice a property satisfied by convergent sequences.
Definition 2.31. A sequence (xn ) in a metric space (X, d) is Cauchy if for every ε > 0 there exists
N ∈ N such that whenever m > n > N
d(xm , xn ) < ε .
Remark 2.32. Inequality d(xm , xn ) < ε may be replaced by d(xm , xn ) ≤ ε without changing the
property (see Remark 2.27).
Proposition 2.33. Any convergent sequence is Cauchy.
PART A TOPOLOGY COURSE: HT 2011
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Proof. Let (xn ) be a sequence in a metric space (X, d) that converges to a point a ∈ X . Let ε be
an arbitrary positive number. There exists N such that for every n ≥ N , d(xn , a) < 2ε . It follows
that for every m ≥ n ≥ N ,
ε ε
d(xn , xm ) ≤ d(xn , a) + d(xm , a) < + = ε .
2 2
Definition 2.34. A metric space X is complete if every Cauchy sequence in X converges (to a point
in X ).
(a) R is complete with the usual metric; so is Z , with the same metric;
(b) R \ {0} with the restricted Euclidean metric is not complete: n1 is a Cauchy sequence in
R \ {0}, but it is not convergent in R \ {0}.
Examples 2.35.
Proposition 2.36. A subset A of a complete metric space is complete (when endowed with the
restriction of the metric) if and only if A is closed.
Proof. Let (X, d) be a complete metric space and let A be a subset in X with the restricted metric
dA .
Assume that (A, dA ) is complete. We shall prove that A is closed using Proposition 2.30.
Let (xn ) be a sequence in A convergent to a point x ∈ X . Then (xn ) is Cauchy in (X, d) , hence
it is Cauchy in (A, dA ) . As (A, dA ) is complete, it follows that (xn ) converges to some a ∈ A .
The uniqueness of the limit of a convergent sequence (Proposition 2.28) implies that x = a ∈ A .
Conversely, assume that A is closed, and let us prove that (A, dA ) is complete.
Let (xn ) be a Cauchy sequence in (A, dA ) . Then (xn ) is a Cauchy sequence in (X, d) , which is a
complete metric space, therefore (xn ) converges to a point x ∈ X .
The fact that A is closed and Proposition 2.30 imply that x ∈ A . Thus, (xn ) converges in (A, dA )
to a point x ∈ A .
Examples 2.37.
(a) the sets Q and R\Q with the restricted Euclidean metric are not complete;
(b) the subspace (0, 1) of R with the usual metric is not complete.
Proposition 2.38. A Cartesian product X × Y of two complete metric spaces is complete (whatever
the product metric chosen on X × Y ).
Proof. Let (X, dX ) and (Y, dY ) be two metric spaces. On X × Y the following three metrics can be
defined (see Exercise 6, Sheet 1):
p
d2 ((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 )2 + dY (y1 , y2 )2 ,
d1 ((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 ) + dY (y1 , y2 ) ,
d∞ ((x1 , y1 ), (x2 , y2 )) = max (dX (x1 , x2 ) , dY (y1 , y2 )) .
It is easy to prove that
(1)
d∞ ≤ d2 ≤ d1 ≤ 2d∞ .
It suffices to show that X × Y is complete when endowed with one of the three metrics, for the
other two metrics completeness will be deduced without difficulty via the sequence of inequalities (1).
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PART A TOPOLOGY COURSE: HT 2011
If (xn , yn ) is a Cauchy sequence in (X × Y, d1 ) , then (xn ) is a Cauchy sequence in (X, dX ) and
(yn ) is a Cauchy sequence in (Y, dY ) . Since X and Y are both complete, (xn ) converges to a point
x ∈ X and (yn ) converges to a point y ∈ Y . It follows that for every ε > 0 there exists N1 such that
dX (xn , x) ≤ 2ε whenever n ≥ N1 , and there exists N2 such that dY (yn , y) ≤ 2ε whenever n ≥ N2 .
Then for every n ≥ max (N1 , N2 ) ,
d1 ((xn , yn ), (x, y)) = dX (xn , x) + dY (yn , y) < ε .
Theorem 2.39. For any non-empty set S , the function space (B(S, R), d∞ ) of bounded real-valued
functions on S with the sup metric is complete.
Remark 2.40. If S is finite, say S = {1, 2, .., n}, the map f 7→ (f (1), ..., f (n)) is a one-to-one and
onto map (i.e. a bijection) from B(S, R) to Rn , and via this map (B(S, R), d∞ ) is identified with
(Rn , d∞ ) . The latter metric space is known to be complete. Therefore what remains to be proved is
the case when S is infinite. This can be seen as an infinite dimensional version of the statement
“(Rn , d∞ ) is complete.”
Proof. Let (fn ) be a Cauchy sequence in (B(S, R), d∞ ) . Then for an arbitrary ε > 0 there exists N
such that for any m ≥ n ≥ N ,
(2)
kfn − fm k∞ = sup |fn (s) − fm (s)| < ε .
s∈S
For an arbitrary s ∈ S , |fn (s) − fm (s)| ≤ kfn − fm k∞ . This and the above imply that for an
arbitrary s ∈ S , (fn (s)) is a Cauchy sequence in (R, | |) . As (R, | |) is a complete metric space, it
follows that (fn (s)) converges to a real number ℓs .
We define the function f : S → R , f (s) = ℓs .
Consider an arbitrary ε > 0 and N such that whenever m ≥ n ≥ N , (2) holds.
Fix ε > 0 and n ≥ N . We have that for any m ≥ n , |fn (s) − fm (s)| < ε for every s ∈ S .
In the inequality corresponding to each s ∈ S we let m → ∞ , and obtain that |fn (s) − f (s)| ≤ ε .
Then for each s ∈ S , |f (s)| ≤ |fn (s) − f (s)| + |fn (s)| ≤ ε + kfn k∞ . This implies that f is a
bounded function.
The fact that |fn (s) − f (s)| ≤ ε for every s ∈ S implies that kfn − f k∞ ≤ ε .
We have thus obtained that for any ε > 0 there exists N (coming from the Cauchy property of
(fn ) ) such that for every n ≥ N , kfn − f k∞ ≤ ε .
This implies that (fn ) converges to f in (B(S, R), d∞ ) .
2.4. Continuous maps of metric spaces.
Definition 2.41. Let (X, dX ) and (Y, dY ) be metric spaces and let f : X → Y be a map.
(a) We say f is continuous at x0 ∈ X if given any ε > 0, there exists δ > 0 such that whenever
dX (x, x0 ) < δ
dY (f (x), f (x0 )) < ε ;
equivalently, such that f (BdX (x0 , δ)) ⊆ BdY (f (x0 ), ε).
(b) We say f is continuous if f is continuous at every x0 ∈ X.
Proposition 2.42. A map f : X → Y between two metric spaces is continuous if and only if for
every x ∈ X and (xn ) sequence in X converging to x , the sequence (f (xn )) converges to f (x) .
PART A TOPOLOGY COURSE: HT 2011
11
Proof. Assume that f is continuous. Let (xn ) be a sequence that converges to x ∈ X . We want to
prove that (f (xn )) converges to f (x) . More precisely, we want to show that for every ε > 0 there
exists N such that f (xn ) ∈ B(f (x), ε) whenever n ≥ N .
The map f is continuous at x , therefore there exists δ > 0 such that f (B(x, δ)) ⊆ B(f (x), ε).
As (xn ) converges to x , there exists N such that for any n ≥ N , xn ∈ B(x, δ) , therefore f (xn ) ∈
f (B(x, δ)) ⊆ B(f (x), ε).
Conversely, assume that for every x ∈ X and (xn ) sequence in X converging to x , the sequence
(f (xn )) converges to f (x) . We want to prove that f is continuous.
Assume that f is not continuous at some point x . Then there exists ε > 0 such that for every δ > 0
the set f (B(x, δ)) has some point outside B(f (x), ε). In particular for δ = n1 there exists xn ∈ B(x, n1 )
such that d(f (xn ), f (x)) ≥ ε . The sequence (xn ) then converges to x while d(f (xn ), f (x)) ≥ ε .
On the other hand, since (xn ) converges to x by hypothesis f (xn ) converges to f (x) . This
contradicts the inequality d(f (xn ), f (x)) ≥ ε .
We may then conclude that f is continuous.
Continuity of maps can also be defined using either open sets or closed sets.
Proposition 2.43. Let X, Y be metric spaces and let f : X → Y be a map. Then f is continuous
if and only if the inverse image of every open set is open (i.e. whenever U is open in Y , the
inverse image f −1 (U ) is open in X ).
Proof. Assume that f is continuous in the sense of Definition 2.41. Let U be an open set in Y . We
prove that f −1 (U ) is open in X . Let x0 be an arbitrary point in f −1 (U ) . Then f (x0 ) ∈ U , and
since U is open there exists ε > 0 such that B(f (x0 , ε)) ⊂ U .
The map f is continuous at x0 therefore for ε > 0 given above there exists δ > 0 such that
f (B(x0 , δ)) ⊆ B(f (x0 , ε)) ⊂ U . This implies that B(x0 , δ) ⊆ f −1 (U ) .
Conversely, assume that f −1 (U ) is open in X whenever U is open in Y . Let us prove that f is
continuous in the sense of Definition 2.41.
Let x0 be an arbitrary point in X , and let ε > 0 be an arbitrary positive number. The set
U = B(f (x0 ), ε) is open, therefore by hypothesis f −1 (U ) is an open set, moreover containing x0 . It
follows that there exists δ > 0 such that B(x0 , δ) ⊆ f −1 (U ) . This implies that f (B(x0 , δ)) ⊆ U =
B(f (x0 ), ε) .
We have thus proved that f is continuous at x0 , and x0 was arbitrary.
Proposition 2.44. Let X, Y be metric spaces and let f : X → Y be a map.
Then f is continuous if and only if the inverse image of every closed set is closed (i.e.
whenever V is closed in Y the inverse image f −1 (V ) is closed in X ).
Proof. This follows from Proposition 2.43 and the formula for complementaries of inverse images:
f −1 (Y \ V ) = X \ f −1 (V )
Remark 2.45. Proposition 2.44 may also be proved using the characterisations of closed sets and of
continuous maps via sequences (Propositions 2.30 and 2.42).
Example 2.46. If A is a subset of the metric space (X, d) then the inclusion map i : A → X is
continuous, when A is endowed with the restricted metric dA .
We investigate the behaviour of continuity with respect to composition of maps, and taking the
inverse map.
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PART A TOPOLOGY COURSE: HT 2011
Proposition 2.47. If f : X → Y and g : Y → Z are continuous maps, where X, Y, Z are metric
spaces, then so is g ◦ f .
Proof. Let U be an open set in Z .
Since g is continuous, g −1 (U ) is an open subset of Y .
Since f is continuous, f −1 g −1 (U ) is an open subset of X .
But the latter is the same as (g ◦ f )−1 (U ) .
Example 2.48. In general it is not true that if a map f : X → Y is a bijection, and f is
continuous then f −1 is continuous.
Consider X = [0, 1) ∪ {2} and Y = [0, 1] with the restricted metric from R .
Let f : X → Y be defined by f (x) = x for any x ∈ [0, 1) , and f (2) = 1. The map f is continuous.
But the map g = f −1 is not continuous at 1.
We have that g(1) = 2.
BX 2, 21 = X ∩ 32 , 25 = {2}.
For any 1 ≥ δ > 0, BY (1, δ) = (1 − δ, 1 + δ) ∩ Y = (1 − δ, 1] .
The image g(1 − δ, 1] is (1 − δ, 1) ∪ {2}. It can never be contained in BX 2, 21 = {2}.
Definition 2.49. A homeomorphism between metric spaces X and Y is a bijection f : X → Y such
that f and the inverse function f −1 are both continuous.
If a homeomorphism f : X → Y exists then X, Y are called homeomorphic.
Example 2.50. The surface of a cylinder (without the ends)
X = {(x, y, z) ∈ R3 | x2 + y 2 = 1 , 1 < z < 2}
and an annulus lying in the xy -plane, without the boundary,
Y = {(x, y, 0) | 1 < x2 + y 2 < 4}
are homeomorphic.
Indeed, the maps
f : X → Y , f (x, y, z) = (zx, zy, 0)
and
f
−1
are both continuous.
:Y →X, f
−1
(x, y, 0) =
p
x
x2 + y 2
, p
y
x2 + y 2
,
p
x2
+
y2
!
Exercise 1. Check that f (X) ⊆ Y , f −1 (Y ) ⊆ X and that f ◦ f −1 = idY , f −1 ◦ f = idX .
Some sets may be defined by inequalities or equalities relating continuous functions. In that case
we can detect whether they are open or closed (or neither).
Proposition 2.51 (defining open/closed sets via continuous functions). Let X be a metric space and
let f : X → R be a continuous function. Then for every λ ∈ R
(1) the sets {x ∈ X | f (x) > λ} and {x ∈ X | f (x) < λ} are open;
(2) the sets {x ∈ X | f (x) ≥ λ}, {x ∈ X | f (x) ≤ λ} and {x ∈ X | f (x) = λ} are closed.
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13
Proof. (1) follows from Proposition 2.43 applied to U = (λ, +∞) , and to U = (−∞, λ) , open sets in
R.
(2) follows from Proposition 2.44 applied to F = [λ, +∞) , to F = (−∞, λ] , and to F = {λ},
closed sets in R .
Examples 2.52.
(1) The set {x ∈ R ; ex > cos x} is open in R with the Euclidean metric.
(2) The planar set {(x, y) ∈ R2 ; x3 ≤ y 3 + y} is closed in R2 with the Euclidean metric (or the
metric d1 or the metric d∞ ).
(3) Let C[0, 1] denote the metric space of continuous real-valued functions on the closed interval
[0, 1] equipped with the sup metric
d∞ (f, g) = sup |f (x) − g(x)|.
x∈[0,1]
The function I : C[0, 1] → R defined by I(f ) =
It follows that for any a ∈ R the sets
Z 1
f (x)dx > a
f ∈ C[0, 1] ;
and
0
are open, while
Z
f ∈ C[0, 1] ;
are closed.
1
0
f (x)dx ≤ a
and
Z
1
f (x)dx is continuous (see Ex. 8, Sheet 1).
0
Z
f ∈ C[0, 1] ;
Z
f ∈ C[0, 1] ;
f (x)dx < a
f (x)dx = a
1
0
1
0
2.5. The space of continuous functions. Continuous functions compose a very interesting metric
space.
Let (X, d) be a metric space. Consider the set Cb (X, R) of bounded continuous real-valued functions on X . It is a subset of B(X, R) , the function-space of bounded real-valued functions on X .
Recall that the latter is endowed with the sup-norm, inducing the sup-metric
d∞ (f, g) = sup |f (x) − g(x)|.
x∈X
Theorem 2.53. The subspace Cb (X, R) is a closed subset in B(X, R) .
Proof. We prove that Cb (X, R) is closed in B(X, R) by proving that whenever it contains a convergent
sequence it contains its limit as well.
Let (fn ) be a sequence in Cb (X, R) convergent to f in B(X, R) . Then for every ε > 0 there exists
N such that for every n ≥ N ,
kfn − f k∞ = sup |fn (x) − f (x)| < ε .
x∈X
We want to prove that f is continuous at an arbitrary point x0 ∈ X .
Let ε > 0. There exists N such that
ε
kfN − f k∞ < .
3
By hypothesis fN is continuous at x0 , therefore there exists δ > 0 such that d(x, x0 ) < δ implies
|fN (x) − fN (x0 )| < 3ε .
Assume that d(x, x0 ) < δ . Then
2ε ε
ε
+ = ε.
|f (x)− f (x0 )| ≤ |f (x)− fN (x)|+ |fN (x)− fN (x0 )|+ |fN (x0 )− f (x0 )| < 2kfN − f k∞ + <
3
3 3
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PART A TOPOLOGY COURSE: HT 2011
Corollary 2.54. The space (Cb (X, R), d∞ ) with the sup metric is complete.
Proof. This follows from:
• the fact that (B(X, R), d∞ ) is complete (Theorem 2.39);
• the fact that a closed subset in a complete metric space is complete (Proposition 2.36);
• Theorem 2.53.
Corollary 2.55. The space C[a, b] of continuous real-valued functions on [a, b] (where a < b are
real numbers) with the sup metric is complete.
Proof. We only need to recall that any continuous real-valued function on [a, b] is bounded, and apply
Corollary 2.54.
Remark 2.56. We shall see that, in some sense, the metric spaces (Cb (X, R), d∞ ) and their closed
subsets are the only examples of complete metric spaces.
2.6. Various strong versions of continuity. We now define some strong versions of continuity.
Definition 2.57. Let (X, dX ), (Y, dY ) be metric spaces.
A map f : X → Y is uniformly continuous if for every ε > 0 there exists δ > 0 such that
dX (x, y) < δ ⇒ dY (f (x), f (y)) < ε .
Remark 2.58. Uniform continuity is stronger than continuity.
Indeed ‘ f continuous’ means that for every x0 ∈ X and for every ε > 0 there exists δx0 > 0
(depending on x0 ) such that dX (x0 , y) < δx0 ⇒ dY (f (x0 ), f (y)) < ε .
Uniform continuity requires the existence of a uniform δ , the same for all points x0 .
A particular case of uniformly continuous map is the following.
Definition 2.59. Let (X, dX ), (Y, dY ) be metric spaces. A map f : X → Y is K –Lipschitz (or
simply Lipschitz ), for some constant K ∈ (0, ∞) , if
dY (f (x), f (y)) ≤ KdX (x, y) , ∀x, y .
Proposition 2.60. Any K –Lipschitz map is uniformly continuous.
Proof. For every ε > 0 one can take δ =
ε
K
.
Examples 2.61.
(a) Let f : I → R be a differentiable function, where I is an open interval in R ,
such that f ′ is bounded, that is |f ′ (x)| ≤ M for every x ∈ I . Then f is M –Lipschitz.
Indeed, according to the Mean Value theorem, for every x < y in I , f (x) − f (y) =
f ′ (c)(x − y) , where c ∈ (x, y) .
(b) The function x 7→ xα with α ∈ [0, 1] is Lipschitz on [1, +∞) .
(c) The function x 7→ 2x is Lipschitz on any interval [a, b] but not on R.
Indeed if the map were K –Lipschitz then for any x ∈ R , 2x+1 − 2x = 2x would have to
be at most K , which is impossible.
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15
(d) Given a metric space (X, d) and a point a ∈ X , the function f : X → R , f (x) = d(x, a) is
1–Lipschitz. This is an immediate consequence of the Triangle Inequality.
Example (d) and Proposition 2.51 imply that open balls and complementaries of closed balls are
open sets. Indeed B(a, r) = f −1 (−1, r) and X \ B ′ (a, r) = f −1 (r, +∞) .
Definition 2.62. A map f : X → X of a metric space (X, d) to itself is called a K –contraction (or
simply a contraction) if it is K –Lipschitz for some constant K < 1.
Example 2.63. The cosine function on [0, 1].
According to the Mean Value Theorem, for every x, y ∈ [0, 1],
| cos x − cos y| ≤ sin 1 |x − y| .
Definition 2.64. Let (X, dX ), (Y, dY ) be metric spaces.
A map f : X → Y is an isometric embedding if and only if for every x, x′ ∈ X
dY (f (x), f (x′ )) = dX (x, x′ ) .
If moreover f is onto then f is an isometry, and the metric spaces X and Y are isometric.
Example 2.65. The map i : Rn → Rn+m , i(¯
x) = (¯
x, 0, ..., 0) is an isometric embedding of (Rn , k ki )
n+m
into (R
, k ki ) for any i ∈ {1, 2, ∞}.
Remark 2.66. Every isometric embedding is an injective continuous map.
Indeed if f (x) = f (x′ ) then 0 = dY (f (x), f (x′ )) = dX (x, x′ ) , whence x = x′ .
Remark 2.67. If f : X → Y is an isometry then f −1 : Y → X is an isometry.
Indeed for two arbitrary points y, y ′ ∈ Y , there exist x, x′ such that y = f (x), y ′ = f (x′ ) . This
implies that f −1 (y) = x, f −1 (y ′ ) = x′ . It follows that dY (y, y ′ ) = dY (f (x), f (x′ )) = dX (x, x′ ) =
dX (f −1 (y), f −1 (y ′ )) .
In particular, every isometry is a homeomorphism.
Not every homeomorphism is an isometry. Example: id : (Rn , k k2 ) → (Rn , k k1 ) (or id :
n
(R , k k2 ) → (Rn , k k∞ ) ).
Proposition 2.68. If (X, dX ), (Y, dY ) are two isometric metric spaces and X is complete then Y
is complete as well.
Proof. Let f : Y → X be an isometry. If (yn ) is a Cauchy sequence then so is (f (yn )) . Since
X is complete (f (yn )) converges to some point x ∈ X . Since f −1 is continuous, yn converges to
f −1 (x) .
Remarks 2.69.
(1) In Proposition 2.68 the hypothesis can be weakened to: “there exists a bijection
f : Y → X , f Lipschitz, f −1 continuous.”
(2) In Proposition 2.68 the hypothesis cannot be weakened to “there exists f : Y → X homeomorphism.”
See for instance the function tan : − π2 , π2 → R .
Theorem 2.70 (Contraction Map Theorem). Suppose that X is a complete metric space and
that f : X → X is a K –contraction, for some constant K < 1. Then
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PART A TOPOLOGY COURSE: HT 2011
(a) f has a unique fixed point, i.e. there exists a unique p ∈ X such that f (p) = p ;
(b) for any x0 ∈ X , the sequence (xn ) defined inductively by xn = f (xn−1 ) for all n > 1
converges to p ;
Kn
d(x0 , x1 ) for any n > 1.
(c) (error estimate) d(xn , p) 6
1−K
Remarks 2.71.
(1) If X is not complete then Theorem 2.70 is not true. For instance the
function f : (0, 1] → (0, 1] , f (x) = x2 is a 12 –contraction but it has no fixed point in (0, 1] .
(2) The hypothesis ‘ f is a contraction’ cannot be weakened to ‘ f satisfies the inequality:
(3)
d(f (x1 ), f (x2 )) < d(x1 , x2 ) , for all x1 6= x2 in X .’
√
For instance the function f : R → R , f (x) = x2 + 1 satisfies (3), due to the Mean Value
Theorem, R is complete, but f has no fixed point.
Remark 2.72. Theorem 2.70 is useful when trying to solve an equation x = f (x) . When f is a
contraction:
• Theorem 2.70, (a), implies that the equation admits a unique solution;
• Theorem 2.70, (b) and (c) give a way of approximating the solution of the equation x = f (x) ,
and of estimating the error in the approximation.
See Exercise 8 on Problem Sheet 1, where a functional equation y = F (y) , with y ∈ C[0, 1] , is thus
discussed.
Proof of Theorem 2.70.
Step 1: Existence of a fixed point.
(xn ) inductively by xn+1 = f (xn ) .
Let x0 be an arbitrary point in X , and define the sequence
We prove by induction that d(xn+1 , xn ) ≤ K n d(x1 , x0 ) . For n = 0 it is an equality.
Assume that the inequality is satisfied for n . Then
d(xn+2 , xn+1 ) = d(f (xn+1 ), f (xn )) ≤ Kd(xn+1 , xn ) ≤ K n+1 d(x1 , x0 ) .
For m ≥ n ≥ N we have that
d(xn , xm ) ≤ d(xn , xn+1 ) + d(xn+1 , xn+2 ) + · · · + d(xm−1 , xm ) ≤ (K n + K n+1 + · · · + K m−1 )d(x1 , x0 ) .
The last term is equal to K n
1−K m−n
1−K
≤
Kn
1−K
≤
KN
1−K
.
N
K
For any ε > 0 there exists N large enough so that 1−K
< ε . It follows that (xn ) is a Cauchy
sequence. Since X is a complete metric space, (xn ) converges to some point p ∈ X .
The function f is continuous. In the equality f (xn ) = xn+1 by letting n → ∞ we obtain that
f (p) = p .
Step 2: Uniqueness of the fixed point. Let q be another point such that f (q) = q . Then d(p, q) =
d(f (p), f (q)) ≤ Kd(p, q) . If p 6= q this yields a contradiction. It follows that p = q .
This in particular implies that any sequence constructed as in Step 1 converges to the same point
p , the unique fixed point of f .
Moreover, in Step 1 it was proved that for m ≥ n , d(xn , xm ) ≤
Kn
.
fix n and let m → ∞ , we obtain d(xn , p) ≤ 1−K
Kn
1−K
. If in the last inequality we
PART A TOPOLOGY COURSE: HT 2011
17
3. Topological spaces
3.1. Definitions and examples.
Definition 3.1. A topological space (X, T ) consists of a non-empty set X together with a family T
of subsets of X satisfying:
(T1) X, ∅ ∈ T ;
(T2) U, V ∈ T ⇒ U ∩ V ∈ T ;
[
(T3) Ui ∈ T for all i ∈ I ⇒
Ui ∈ T .
i∈I
The family T is called a topology for X . The sets in T are called the open sets of X . When T is
understood we talk about the topological space X .
There are several good reasons to generalise the setting from metric spaces to topological spaces:
• it is clear that two isometric metric spaces are identical from the metrical point of view; a
natural question to ask is what are the common properties of two homeomorphic metric spaces
(like the cylinder and the annulus in the example 2.50) ?
• some types of convergence cannot be defined in the setting of metric spaces, but can be defined
in the setting of topological spaces; this is the case for point-wise convergence of a sequence
of functions;
• some constructions like that of quotient with respect to some equivalence relation or of infinite
Cartesian product may be endowed much easier with a topology than with a metric;
• an important part of the theory of convergence and continuity, as well as other important
notions, may be recovered using the three axioms (T1), (T2), (T3) only. This setting may
simplify some arguments and emphasize what is really essential to them.
Examples 3.2.
(1) Any metric space (X, d) gives rise to a topological space (X, Td ) , where Td
consists of those subsets of X which are open in X with respect to the metric d .
(2) (Discrete spaces) Let X be any non-empty set. The discrete topology on X is the set of all
subsets of X .
(3) (Indiscrete spaces) Let X be any non-empty set. The indiscrete topology on X is the family
of subsets {X, ∅}.
(4) Let X be any non-empty set. The co-finite topology on X consists of the empty set together
with every subset U of X such that X \ U is finite.
Definition 3.3.
(1) We call a topological space (X, T ) metrizable if it arises as in Example 3.2,
(1), from (at least one) metric space (X, d) i.e. there is at least one metric d on X such that
T = Td .
(2) Two metrics on a set are topologically equivalent if they give rise to the same topology.
Example 3.4. The metrics d1 , d2 , d∞ on Rn are all topologically equivalent - see Sheet 1, Ex. 5, (3).
We shall call the topology defined by the above three metrics the standard (or canonical ) topology
on Rn .
Definition 3.5. Given two topologies T1 , T2 on the same set, we say T1 is coarser than T2 if T1 ⊆ T2 .
Remark 3.6. For any space (X, T ) , the indiscrete topology on X is coarser than T which in turn is
coarser than the discrete topology on X .
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PART A TOPOLOGY COURSE: HT 2011
Definition 3.7. Let (X, T ) be a topological space.
A subset V of X is closed in X if X \ V is open in X (i.e. X \ V ∈ T ).
Examples 3.8.
(a) You have to be careful which space you’re saying a set is closed in.
For example in the space [0, 1) with the usual topology coming from the Euclidean metric,
[1/2, 1) is closed.
(b) In a discrete space, all subsets are closed since their complements are open.
(c) In the co-finite topology a set is closed if and only if it is finite.
Proposition 3.9. Let X be a topological space. Then
(C1) X, ∅ are closed in X ;
(C2) if V1 , V2 are closed in X then V1 ∪ V2 is closed in X ;
\
(C3) if Vi is closed in X for all i ∈ I then
Vi is closed in X .
i∈I
Proof. Properties (C1),(C2), (C3) follow from (T1), (T2), (T3) and from the De Morgan laws (see
Appendix A).
Definition 3.10. A sequence (xn ) in a topological space X converges to a point x ∈ X if given any
open set U containing x there exists an integer N such that xn ∈ U for all n > N.
Examples 3.11.
(a) In a metric space this is equivalent to the metric definition of convergence.
Indeed, in the metric setting (xn ) converges to x if for any ε > 0 there exists an integer N
such that xn ∈ B(x, ε) for all n > N.
This is clearly implied by the definition of convergence in (X, Td ) , as open balls are in Td .
On the other hand every open set U in Td containing x contains an open ball B(x, ε) .
Thus, the metric definition of convergence in its turn implies the topological one.
(b) In a (non-empty) indiscrete topological space X any sequence converges to any point x ∈ X .
(c) In an infinite space X with the co-finite topology any sequence (xn ) of pairwise distinct
elements (i.e. such that xn 6= xm when n =
6 m ) converges to any point x ∈ X .
Remark 3.12. Note that the uniqueness of the limit for a convergent sequence is not granted in a
topological space.
In particular (b) and (c) show that both the indiscrete topology and the co-finite topology on an
infinite metric space are not metrizable, as in a metric space any convergent sequence has only one
limit.
In order to have the uniqueness of the limit for a convergent sequence, one has to add an extra
axiom (Hausdorff ). We shall discuss this axiom later on.
Proposition 3.13. If a subset F in a topological space X is closed then for any convergent sequence
contained in F any limit of it is also in F .
Proof. Let (xn ) be a convergent sequence, contained in F , and let x ∈ X be a limit of (xn ) . If
x ∈ X \ F then, as X \ F is open, it follows that there exists N such that xn ∈ X \ F for every
n ≥ N . This contradicts the hypothesis that (xn ) is contained in F .
PART A TOPOLOGY COURSE: HT 2011
19
Remark 3.14. Unlike in the case of metric spaces, the converse of the statement in Proposition 3.13
might not be true.
Loose Remark. We shall see many implications of the following form:
topological property ⇒ property in terms of sequences.
In metric spaces the converse implication may sometime be true (i.e. the topological property may
be characterized in terms of sequences).
In the general topological setting the converse implication is never true.
Definition 3.15. Suppose that (X, TX ) and (Y, TY ) are topological spaces and that f : X → Y is
a map. We say that f is continuous if U ∈ TY ⇒ f −1 (U ) ∈ TX i.e. ‘inverse images of open sets are
open’.
If there are other topologies around we may say ‘ f is continuous with respect to TX , TY ’ or ‘ f is
( TX , TY )-continuous’.
Proposition 3.16. A map f : X → Y of topological spaces is continuous if and only if f −1 (V ) is
closed in X whenever V is closed in Y .
Proof. This follows from the definition of a continuous map and from the formula
f −1 (Y \ V ) = X \ f −1 (V ) .
Proposition 3.17. If f : X → Y and g : Y → Z are continuous maps, X, Y, Z topological spaces,
then so is g ◦ f .
The proof is identical to the one of Proposition 2.47.
Definition 3.18. A homeomorphism between topological spaces X and Y is a bijection f : X → Y
such that f and f −1 are continuous. The spaces X, Y are said to be homeomorphic.
Proposition 3.19. If a map f : X → Y is a continuous map between two topological spaces then for
any sequence (xn ) in X converging to a point x , (f (xn )) converges to f (x) .
Proof. Let U be an open set containing f (x) . Then f −1 (U ) is an open set ( f is continuous) containing x .
Since (xn ) converges to x there exists N such that for n ≥ N , xn ∈ f −1 (U ) . It follows that
f (xn ) ∈ U .
Remark 3.20. The converse of Proposition 3.19 might not be true.
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PART A TOPOLOGY COURSE: HT 2011
3.2. Closure of a set. Property (C3) of closed sets, that is:
\
Vi is closed in X .’
‘if Vi is closed in X for all i ∈ I then
i∈I
suggests the idea of approximating from above an arbitrary set by a closed set.
Definition 3.21. Consider a topological space X and a subset A ⊆ X .
The closure of A in X is the set
A=
\
F.
F closed,A⊆F
A point in A is sometimes called an adherent point of A .
Proposition 3.22. The closure A is the smallest (with respect to inclusion) closed subset
of X containing A , that is:
(a) A is closed in X and it contains A ;
(b) if A ⊆ V where V is closed in X then A ⊆ V .
Proof. (a) is immediate from the definition of the closure and property (C3) of closed sets.
A closed subset V as in (b) appears in the intersection defining A , hence A ⊆ V .
Proposition 3.23. Let A, B be subsets of a topological space X . Then
(a) A ⊆ B implies that A ⊆ B ;
(b) A is closed in X if and only if A = A ;
(c) A = A .
Proof. (a) We have A ⊆ B ⊆ B , we apply Proposition 3.22, (b).
(b) A ⊆ A . If A is closed then Proposition 3.22, (b), with V = A implies A ⊆ A , whence equality.
If A = A then A is closed.
(c) We apply (b) to A .
The closure can also be described by the following property of its points:
Proposition 3.24.
A = {x ∈ X : for every open U ⊆ X with x ∈ U , U ∩ A 6= ∅}.
Proof. Consider A1 = {x ∈ X : for every open U ⊆ X with x ∈ U , U ∩ A 6= ∅}.
We first prove that A ⊆ A1 using Proposition 3.22, (b). Clearly A1 contains A . It remains to
prove that A1 is closed. We prove that X \ A1 = B1 is open. Note that
B1 = {x ∈ X : there exists an open set Ux such that x ∈ Ux ⊆ X \ A} .
For every element y in some Ux one can S
take Uy = Ux and have y ∈ Uy ⊆SX \ A . It follows that
each Ux is S
entirely contained in B1 . Then x∈B1 Ux ⊆ B1 , and since B1 ⊆ x∈B1 Ux , we conclude
that B1 = x∈B1 Ux , hence B1 is open.
We have thus proved that A1 is closed, which implies that A ⊆ A1 .
Assume that there exists x ∈ A1 \ A . Since x 6∈ A it follows that there exists a closed set F
containing A such that x 6∈ F . Then x ∈ X \ F , and X \ F is an open set which does not intersect
A . This contradicts the fact that x ∈ A1 .
PART A TOPOLOGY COURSE: HT 2011
Examples 3.25.
21
(1) The closure of (a, b) in (R, | |) is [a, b] .
(2) The closure of Q in (R, | |) is R . Same for R \ Q .
(3) The closure of any set A in a space X endowed with the co-finite topology is either X if A
is infinite, or A if A is finite.
Exercise 2. Let A be a bounded subset of R .
Prove that both its least upper bound sup A and its greatest lower bound inf A are in A .
Definition 3.26. A subset A in X is called dense if A = X .
We shall see in what follows that the closure of a set A contains two types of points:
• points in A that are isolated;
• points in A or outside A but near which points in A accumulate (i.e. accumulation
points).
Definition 3.27. A point x ∈ X such that for any open U ⊆ X with x ∈ U , (U \ {x}) ∩ A 6= ∅ is
called an accumulation point (or limit point) of A .
The set of accumulation points of A is sometimes denoted by A′ .
Note that A = A ∪ A′ . The points in A \ A′ are points x ∈ A such that for some open set U
containing x , U ∩ A = {x}. Such points are called isolated points of A .
Examples 3.28. In the metric space (R, | |)
(1) if A = (a, b) then A′ = A = [a, b] ;
(2) if A = Q or R \ Q then A′ = A = R ;
(3) if A = Z then A = Z while A′ = ∅.
(4) if A = (0, 1) ∪ {9, 10} then A = [0, 1] ∪ {9, 10}, A′ = [0, 1] and 9, 10 are isolated points.
Proposition 3.29. Let (X, d) be a metric space and let A ⊆ X .
(1) The closure A is the set of limits of convergent sequences (an ) in A .
(2) The set of accumulation points A′ is the set of limits of convergent sequences (an ) of pairwise
distinct elements in A (i.e. such that an 6= am if n 6= m ).
Proof.
(1) Let Aℓ be the set of all limits of convergent sequences (an ) in A .
Since A is closed and it contains A , it contains all the limits of convergent sequences (an ) in A ,
due to Proposition 3.13. Thus Aℓ ⊆ A .
On the other hand, let x be a point in A . According to Proposition 3.24, for every integer n ≥ 1,
the open ball B x, n1 intersects A in a point an . The sequence (an ) converges to x . Hence x ∈ Aℓ .
We have thus proved that A ⊆ Aℓ .
(2) Let A′ℓ be the set of all limits of convergent sequences (an ) in A such that an 6= am whenever
n 6= m .
Let x be a point in A′ℓ , limit of a sequence (an ) . At most one element in the sequence can be
equal to x , thus without loss of generality we may assume that an 6= x for all n .
For every open set U containing x there exists N such that an ∈ U for all n ≥ N . Moreover
an ∈ U \ {x}. In particular (U \ {x}) ∩ A 6= ∅. Thus x ∈ A′ .
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PART A TOPOLOGY COURSE: HT 2011
We conclude that A′ℓ ⊆ A′ .
Conversely let x be a point in A′ . We construct inductively a sequence (an ) in A such that
d(x, an+1 ) < d(x, an ) and 0 < d(x, an ) < n1 .
Since x ∈ A′ , B(x, 1) \ {x} intersects A in a point a1 .
Assume that we found the elements
in the sequence a1 , ..., an satisfying the required hypotheses.
1
Let rn+1 = min n+1 , d(x, an ) .
Since x ∈ A′ , B(x, rn+1 ) \ {x} intersects A in a point an+1 .
The sequence (an ) thus constructed converges to x , and for n < m , d(x, am ) < d(x, an ) , whence
am 6= an . Therefore x ∈ A′ℓ .
We have proved that A′ ⊆ A′ℓ , which together with the converse inclusion implies that A′ = A′ℓ .
Proposition 3.30. A map f : X → Y of topological spaces is continuous if and only if f (A) ⊆ f (A)
for every A ⊆ X .
Proof. Assume that f is continuous. Since f (A) is closed, it follows that f −1 f (A) is closed. The
latter subset also contains A , therefore it contains A . We have thus proved that A ⊆ f −1 f (A) ,
which implies that f (A) ⊆ f (A) .
Conversely, assume that f (A) ⊆ f (A) for every A ⊆ X .
Let F be a closed subset in Y and let A = f −1 (F ) . Then f (A) ⊆ F , whence f (A) ⊆ f (A) ⊆ F .
It follows that A ⊆ f −1 (F ) = A , therefore A = A . This implies that A is closed.
3.3. Interior of a set. Property (T3) of open sets, that is:
S
‘if Ui are open for all i ∈ I then i∈I Ui is open’
suggests the idea of approximating from below an arbitrary set by an open set.
Definition 3.31. Let X be a topological space X and A a subset in X . The interior of A in X is
the set
[
˚=
A
U.
U open,U ⊆A
˚ may be empty even if A is not. For instance consider the subset A = [0, 1] × {0} in
Note that A
R2 . Clearly A contains no open ball B((x, y), ε) , therefore A contains no non-empty open subset.
˚ is the largest (with respect to inclusion) open subset of
Proposition 3.32. The interior A
X contained in A , that is:
˚ is open contained in A ;
(a) A
˚.
(b) every open subset U of A is contained in A
Both (a) and (b) follow easily from the definition of the interior.
Proposition 3.33. Let A, B be subsets of a topological space X . Then
˚⊆ B
˚;
(a) A ⊆ B implies that A
˚ = A;
(b) A is open in X if and only if A
˚
˚= A
˚;
(c) A
PART A TOPOLOGY COURSE: HT 2011
23
˚ ⊆ A ⊆ B , and A
˚ is open. Therefore A
˚⊆ B
˚.
Proof. (a) A
˚, whence equality. The converse implication is
(b) If A is open, since A ⊆ A it follows that A ⊆ A
immediate.
˚.
(c) follows from (b) applied to A
Examples 3.34.
• The interior of [a, b] (or (a, b] ) in (R, | |) is (a, b) ;
Indeed (a, b) is open and contained in [a, b] , therefore in its interior.
If a was in the interior then there would exist ε > 0 such that (a − ε, a + ε) is in the
interior, therefore in [a, b] . This is impossible, hence a is not in the interior. Likewise for b .
It follows that the interior is in (a, b) , therefore it is equal to it.
• The interiors of Z , Q , R \ Q in (R, | |) are ∅.
None of the sets above contains an open interval, therefore none contains a non-empty open
subset.
˚ is
• Let X be a space endowed with the co-finite topology, and let A ⊆ X . The interior A
either equal to A if X \ A is finite (i.e. A is open), or it is empty.
˚ would be non-empty, then X \ A
˚ would be finite.
Indeed if A
˚ is contained in A , X \ A is contained in X \ A,
˚ therefore it is also finite.
As A
Next we note that the operation of taking the interior and of taking the closure are in some sense
complementary to each other.
Proposition 3.35. Let A be a subset in a topological space X .
The closure of the complement of A is the complement of the interior, i.e.
˚.
X \A=X \A
Proof. Denote X \ A by B .
˚ is open, and A
˚ ⊆ A , its complement X \ A
˚ is closed, and it contains B . Therefore B ⊆ X \ A
˚.
As A
On the other hand B is closed and it contains B = X \ A . It follows that X \ B is open and
˚ This is equivalent to X \ A
˚⊆ B.
contained in A . Therefore X \ B ⊆ A.
˚.
From the two opposite inclusions above we deduce that B = X \ A
If in Proposition 3.35 we denote X \ A by B we obtain:
Proposition 3.36. Let B be a subset in a topological space X .
The interior of the complement of A is the complement of the closure, i.e.
˚
\
X
\B =X \B.
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PART A TOPOLOGY COURSE: HT 2011
3.4. Boundary of a set.
˚ is called the boundary of A .
Definition 3.37. The set ∂A = A \ A
Proposition 3.38. The boundary of A equals the intersection A ∩ X \ A, and it equals the boundary
of X \ A .
˚ = A ∩ (X \ A)
˚ = A ∩ (X \ A) . The last inequality is due to Proposition
Proof. Indeed ∂A = A \ A
3.35.
Then by the above ∂(X \ A) = (X \ A) ∩ A .
We conclude that ∂A = ∂(X \ A) .
Example 3.39. Let A = [a, b) , where a < b . Then A = [a, b] .
On the other hand R \ A = (−∞, a) ∪ [b, +∞) , and its closure is R \ A = (−∞, a] ∪ [b, +∞) .
It follows that ∂A = {a, b}.
3.5. Completion of a metric space. Let (X, d) be a metric space.
Recall that the space Cb (X, R) of bounded continuous real-valued functions on X endowed with
the metric defined by the sup-norm
kf k = sup |f (x)|
x∈X
is a complete metric space.
Theorem 3.40.
(1) The metric space X can be isometrically embedded into Cb (X, R) , that is
there exists a map F : X → Cb (X, R) such that
kF (x) − F (y)k = d(x, y) .
b.
(2) The space X is isometric to a dense subspace in a complete metric space X
b1 , X
b2 as in (2), i.e. complete and containing a dense subspace isometric
(3) Two metric spaces X
to X , are isometric.
b as in (2) is called a completion of X .
Definition 3.41. A metric space X
Proof. (1) Let a be a fixed point in X .
For every p ∈ X define the function fp : X → R , fp (x) = d(p, x) − d(a, x) .
The map p 7→ fp is an isometric embedding of (X, d) into (Cb (X, R), d∞ ) .
Indeed, consider p, q ∈ X ; for every x ∈ X |fp (x) − fq (x)| = |d(p, x) − d(q, x)| ≤ d(p, q) . It follows
that
kfp − fq k = sup |fp (x) − fq (x)| ≤ d(p, q) .
x∈X
On the other hand for x = q we obtain |fp (x) − fq (x)| = d(p, q) , whence kfp − fq k ≥ d(p, q) .
We have thus obtained that kfp − fq k = d(p, q) .
b = F (X) . It is a closed subset in the complete metric
(2) For the map F defined in (1) consider X
space Cb (X, R) , therefore it is complete with the restricted metric, according to Proposition 2.36. By
b.
construction F (X) is isometric to X and dense in X
bi and an isometry Fi : X → Yi .
(3) For i ∈ {1, 2} there exists Yi dense subspace of X
Then G = F2 ◦ F1−1 : Y1 → Y2 is an isometry.
PART A TOPOLOGY COURSE: HT 2011
25
b1 every point x1 ∈ X
b1 is limit of some sequence (yn ) in Y1 . In particular this
Since Y 1 = X
implies that (yn ) is a Cauchy sequence. As G is an isometry, (G(yn )) is also a Cauchy sequence.
b2 being complete, it follows that (G(yn )) converges to some point x2 ∈ X
b2 .
The ambient space X
The point x2 does not depend on the choice of the sequence (yn ) in Y1 . Indeed, for another
sequence (yn′ ) in Y1 with the same argument as above we obtain that (G(yn′ )) converges to some
b2 .
point x′ ∈ X
2
On the other hand as both (yn ) and (yn′ ) converge to x1 , d(yn , yn′ ) converges to 0 (because it is
at most d(yn , x1 ) + d(x1 , yn′ ) ).
Since G is an isometry, d(G(yn ), G(yn′ )) also converges to 0, whence d(x2 , x′2 ) = 0 ⇒ x2 = x′2 .
b1 → X
b2 by H(x1 ) = x2 .
This allows us to define the map H : X
b1 and (yn ) and (yn′ ) sequences in
The map H is an isometry. Indeed given x1 and x′1 in X
′
′
Y1 converging to x1 respectively x1 , H(x1 ) and H(x1 ) are the limits of the sequences (G(yn )) ,
respectively (G(yn′ )) .
Then d (H(x1 ) , H(x′1 )) = limn→∞ d(G(yn ) , G(yn′ )) = limn→∞ d(yn , yn′ ) = d(x1 , x′1 ) .
b2 . As X
b2 is the closure
Finally we prove that H is surjective. Let x2 be an arbitrary point in X
−1
of Y2 there exists a sequence (zn ) in Y2 converging to x2 . Let yn = G (zn ) . Since (zn ) is Cauchy
b1 .
and G−1 is an isometry (yn ) is also Cauchy, therefore it converges to some point x1 ∈ X
According to the definition of the map H , H(x1 ) = x2 .
Remark 3.42. An intrinsic construction of the completion of a metric space involves equivalence classes
of Cauchy sequences in the metric space. See for instance W. Sierpi´
nski, General Topology, Chapter
VII, §74, Theorem 96.
3.6. Separation axioms.
Definition 3.43. A topological space satisfies the first separation axiom if for any two distinct points
a 6= b there exists an open set U containing a and not b .
Proposition 3.44. A topological space satisfies the first separation axiom if and only if singletons
are closed subsets in it.
Proof. If every singleton {x} is closed in X then for any two distinct points a 6= b consider U =
X \ {b}.
Conversely, assume that X satisfies the first separation axiom. We prove that V = X \ {x} is
open, for any x ∈ X .
Let y ∈ V arbitrary. Then y 6= x , hence by the first separation axiom there exists Uy open,
containing
contained in V . S
S y , not containing x , hence
S
Thus y∈V Uy ⊆ V and V ⊆ y∈V Uy , whence V = y∈V Uy is open.
Definition 3.45. A topological space X is said to be Hausdorff (or to satisfy the second separation
axiom) if given any two distinct points x, y in X , there exist disjoint open sets U, V with x ∈ U, y ∈
V.
Examples 3.46.
(a) Any metric space is Hausdorff.
(b) An indiscrete space X with more than one point does not satisfy the first separation axiom.
(c) An infinite set with the co-finite topology is not Hausdorff (consequently it is not a metrizable
topological space), but it satisfies the first separation axiom because singletons are closed in
it.
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PART A TOPOLOGY COURSE: HT 2011
Proposition 3.47. If f : X → Y is an injective continuous map of topological spaces X, Y and Y
satisfies the first (second) separation axiom then so is X .
Proof. Assume that Y satisfies the first separation axiom. Let a, b be two distinct points in X .
Since f is injective, f (a) 6= f (b) . By the first separation axiom in Y there exists U open such that
f (a) ∈ U and f (b) 6∈ U . This implies that a ∈ f −1 (U ) , and b 6∈ f −1 (U ) . As f is continuous, f −1 (U )
is open. We have thus proved that X satisfies the first separation axiom.
A similar argument shows that if Y is Hausdorff then X is Hausdorff. Note that we must use the
fact that U ∩ V = ∅ ⇒ f −1 (U ) ∩ f −1 (V ) = ∅. See Appendix.
Corollary 3.48. If spaces X, Y are homeomorphic then X satisfies the first (second) separation
axiom if and only if Y is satisfies the first (respectively the second) separation axiom.
Proposition 3.49. In a Hausdorff space, a sequence converges to at most one point.
Proof. Indeed assume that a convergent sequence (xn ) has two distinct limits a 6= b . Then by the
Hausdorff property there exist U, V open and disjoint (i.e. U ∩V = ∅) such that a ∈ U and b ∈ U . As
(xn ) converges to a there exists N such that xn ∈ U for every n ≥ N . Similarly, (xn ) convergent to
b implies that there exists N ′ such that xn ∈ V for every n ≥ N ′ . Then for every n ≥ max(N, N ′ ) ,
xn ∈ U ∩ V , contradicting U ∩ V = ∅.
Remark 3.50. Note that the first separation axiom does not suffice to ensure the uniqueness of the
limit of a convergent sequence. Indeed, an infinite space with the co-finite topology satisfies the first
separation axiom (see above), but contains sequences with several limits (see Example 3.11, (c)).
3.7. Subspace of a topological space. We saw that a subspace of a metric space is naturally
endowed with a metric structure.
Now we shall see that the same thing happens for topological structures.
Definition 3.51. Let (X, T ) be a topological space and let A be a non-empty subset of X .
The subspace or induced topology on A is TA = {A ∩ U : U ∈ T }.
The three properties (T1), (T2), (T3) that a topology must satisfy are easily checked:
(T1) ∅ = ∅ ∩ A ∈ TA ; A = X ∩ A ∈ TA .
(T2) Given V1 , V2 in TA , V1 = A ∩ U1 and V2 = A ∩ U2 , where U1 and U2 are both in T .
Then V1 ∩ V2 = U1 ∩ U2 ∩ A ∈ TA .
(T3) Let (Vi )i∈I be such that Vi ∈ TA for every i ∈ I . Then for every i ∈ I there exists Ui open
subset in X such that Vi = A ∩ Ui .
S
S
S
It follows that i∈I Vi = A ∩ i∈I Ui which is in TA because i∈I Ui is in T .
Proposition 3.52. A subset W in A is closed in TA if and only if there exists F closed in (X, T )
such that W = A ∩ F .
PART A TOPOLOGY COURSE: HT 2011
27
Proof. Assume that W ⊆ A is closed in TA . Then A \ W is open, i.e. there exists U open in X such
that A \ W = A ∩ U .
The last equality may be re-written as W = A \ A ∩ U = A \ U = A ∩ (X \ U ) . The set F = X \ U
is closed in X .
Conversely, let F be closed in (X, T ) , and let W = A ∩ F .
Then A \ W = A \ F = A ∩ (X \ F ) . Since F is closed in X , X \ F is open in X , hence A \ W
is open in A . It follows that W is closed in A .
Proposition 3.53. Let (A, TA ) be a subspace of (X, T ) , and let B ⊆ A .
A
(a) The closure of B in (A, TA ) , B , coincides with the intersection between A and the closure
X
of B in X , B .
˚A , contains the interior of B in X , B
˚X .
(b) The interior of B in (A, TA ) , B
Proof. (a)
For simplicity let us denote B
X
by B , and B
A
′
by B .
′
The set B is closed in X , therefore B ∩A is closed in A , and it contains B . Therefore B ⊆ B ∩A .
′
On the other hand B is a closed set in A .
′
′
It follows by Proposition 3.53 that B = F ∩ A , where F is closed in X . Since B ⊆ B ⊆ F it
′
follows that B ⊆ F , hence B ∩ A ⊆ F ∩ A = B .
′
We conclude that B ∩ A = B .
˚X is open in X and contained in B ⊆ A , it is an open set in A . Therefore B
˚X ⊆
(b) Since B
A
˚ .
B
Remarks 3.54.
˚X ⊆ B
˚A may be strict.
(1) The inclusion B
Consider for instance B = {0} ⊂ A = {0, 1} ⊂ X = R .
˚R is empty.
As B contains no open interval, B
˚A = B .
On the other hand B = A ∩ (−1, 1) is an open set in A , therefore B
˚X = B
˚A for every B ⊆ A .
(2) If A is open then B
˚A is open in A , hence B
˚A = U ∩ A for some U open in X .
Indeed B
˚A is open in X . It is also contained in B , therefore
Since A is also open, it follows that B
A
X
˚
˚
B ⊆B .
This and Proposition 3.53, (b), imply equality.
We finish our discussion of subspaces of topological spaces with some easy to see but important
remarks.
Remarks 3.55.
(1) If A ⊆ B ⊆ X and (X, T ) is a topological space then the topology induced on
A by T coincides with the topology induced on A by TB .
(2) If A ⊆ X and (X, d) is a metric space then the topology on A induced by Td coincides with
the topology on A induced by the restricted metric dA .
This follows immediately from the fact that for any a ∈ A , the ball centred in a and of
radius r > 0 in (A, dA ) coincides with B(a, r) ∩ A , where B(a, r) is the ball centred in a and
of radius r > 0 in (X, d) .
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PART A TOPOLOGY COURSE: HT 2011
(3) If A is open in X then TA is contained in T .
(4) If A is closed in X then any set closed in A is closed in X .
3.8. Basis for a topology. As in the case of vector spaces, we can consider a minimal amount of
data (a basis) via which we may recover the whole space endowed with the considered structure.
This notion is particularly useful in two settings:
• that of topologies induced by metrics;
• that of products of topological spaces (to be studied further on).
Definition 3.56. Given a topological space (X, T ) , a subfamily B of T is called a basis for T if
every set in T can be expressed as a union of sets in B .
Example 3.57. In a metric space (X, d) the family of open balls
B = {B(x, r) : x ∈ X, r > 0}
is a basis for Td .
In particular in R the family of open intervals B = {(a, b) | a, b ∈ R , a < b} is a basis for the
standard topology.
Moreover, the family of open intervals with rational endpoints BQ = {(a, b) | a, b ∈ Q , a < b} is a
basis for the standard topology in R (Ex. 9, (2), Sheet 2).
Criterion 3.58. Let (X, T ) be a topological space, and B a basis for T .
A subset U is open in (X, T ) if and only if for every x ∈ U there exists B ∈ B such that
x ∈ B ⊆ U.
Proof.
S Indeed, if U is open then there exists (Bi )i∈I collection of subsets in the basis B such that
U = i∈I Bi . This accounts for the direct implication.
Conversely
assume that for every
S
S x ∈ U there exists Bx ∈ B such that x ∈ Bx ⊆ U . Then
U ⊆ x∈U Bx ⊆ U , whence U = x∈U Bx .
Proposition 3.59. Let X, Y be topological spaces, and let B be a basis for the topology on Y .
A map f : X → Y is continuous if and only if for every B in B its inverse image f −1 (B) is open
in X .
Proof. The necessary part follows from the fact that B ⊆ T , i.e. a basis is composed of open sets.
The sufficient part follows from:
• the fact that every open set in Y can be written as
S
S
−1
• the fact that f −1
(Bi ) .
i∈I Bi =
i∈I f
S
i∈I
Bi , with Bi ∈ B for every i ;
Proposition 3.60. Let (X, T ) be a topological space, B a basis for T and A ⊆ X . Then
(4)
A = {x ∈ X :
for every B ∈ B with x ∈ B , B ∩ A 6= ∅} ,
(5)
˚ = {x ∈ X :
A
there exists B ∈ B such that x ∈ B ⊆ A}.
PART A TOPOLOGY COURSE: HT 2011
29
Proof. (4) Denote by AB the set
{x ∈ X : B ∩ A 6= ∅ for any B ∈ B with x ∈ B} .
Recall that according to Proposition 3.24,
A = {x ∈ X : U ∩ A 6= ∅ for any U ∈ T
with x ∈ U } .
This and the fact that B ⊆ T imply that A ⊆ AB .
Let x be an arbitrary point in AB . Let U be an open set containing x . According to the Criterion
3.58, there exists B ∈ B such that x ∈ B ⊆ U .
Since x ∈ AB , it follows that B ∩ A 6= ∅, whence U ∩ A 6= ∅.
We conclude that x ∈ A .
(5) We denote by IB the set
{x ∈ X :
there exists B ∈ B such that x ∈ B ⊆ A}.
Let x be an arbitrary point in IB . Then there exists B ∈ B such that x ∈ B ⊆ A . Note that the
whole B is contained in IB . Thus x ∈ B ⊆ IB . This, according to the Criterion 3.58, implies that
IB is open.
˚.
Also IB is a subset of A by definition, therefore IB ⊆ A
˚, which is an open set. The Criterion 3.58 implies that there exists
Let x be an arbitrary point in A
˚ ⊆ A . It follows that x ∈ IB .
B ∈ B such that x ∈ B ⊆ A
˚ ⊆ IB , whence the two sets are equal.
We may therefore conclude that A
Remark 3.61. In a metric space, (5) can be made even more precise:
˚ = {x ∈ X :
A
(6)
there exists ε > 0 such that B(x, ε) ⊆ A}.
Proof. Let Ad be the set
{x ∈ X :
there exists ε > 0 such that B(x, ε) ⊆ A}.
˚.
According to (5), Ad is contained in A
The converse inclusion follows from the fact that if a point x is contained in some ball B(y, r) then
B(x, ε) ⊂ B(y, r) for ε = r − d(x, y) .
Now we shall see:
• how to recognise when a family of subsets is a basis of some topology;
• how to re-construct the topology when a basis is given.
Proposition 3.62. Let X be a set and B a family of subsets of X such that
(B1) X is a union of sets in B ;
(B2) the intersection B ∩ B ′ of any B, B ′ ∈ B can be expressed as a union of sets in B .
Then the family TB of all unions of sets in B is a topology for X . Note that B is a basis for TB .
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PART A TOPOLOGY COURSE: HT 2011
Proof. (T1) The set X is contained in TB due to (B1).
The empty set ∅ is contained in TB as the union of no sets from B .
S
S
(T2) Let U, V be two sets in B . Then U = i∈I Bi and V = j∈J Bj′ , where Bj , Bj′ ∈ B .
S
S
The intersection U ∩ V is equal to i∈I, j∈J Bi ∩ Bj′ . According to (B2), Bi ∩ Bj′ = k∈Kij Bk′′ ,
for some collection (Bk′′ )k∈Kij in B . It follows that
[
[
U ∩V =
Bk′′ ,
i∈I, j∈J k∈Kij
therefore U ∩ V is in TB .
Property (T3) is immediate from the definition of TB .
3.9. Product of topological spaces. One of the main ways of constructing new topological spaces
out of given ones is by taking Cartesian products. This is what we shall discuss now.
Proposition 3.63. Let (X, TX ), (Y, TY ) be topological spaces. The family of subsets
BX×Y = {U × V : U ∈ TX , V ∈ TY } ,
satisfies the conditions (B1) and (B2) in Proposition 3.62.
Proof. Indeed X × Y is itself in BX×Y . This accounts for (B1).
Also, consider U × V and U ′ × V ′ , with U, U ′ ∈ TX and V, V ′ ∈ TY . Then
(U × V ) ∩ (U ′ × V ′ ) = (U ∩ U ′ ) × (V ∩ V ′ )
is in BX×Y . This accounts for (B2).
Definition 3.64. The family of all unions of sets from BX×Y is a topology TX×Y for X×Y (according
to Proposition 3.62).
We call this topology the product topology.
The space (X × Y, TX×Y ) is called the topological product of (X, TX ) and (Y, TY ) .
NB One of the commonest errors about products is to assume that any open set in the product is a
‘rectangular’ open set such as U × V .
Criterion 3.65. A subset W ⊆ X ×Y is open in the product topology if and only if for any (x, y) ∈ W
there exist U ∈ TX , V ∈ TY such that (x, y) ∈ U × V ⊆ W.
Proposition 3.66. Let (X × Y, TX×Y ) be the topological product of (X, TX ) and (Y, TY ) .
If A is closed in X and B is closed in Y then A × B is closed in X × Y .
Proof. Since A is closed, its complement U = X \ A is open.
Likewise B closed implies V = Y \ B open.
The complement (X × Y ) \ (A × B) is not U × V , but
{(x, y) ∈ X × Y ; x 6∈ A or y 6∈ B} = (U × Y ) ∪ (X × V ) .
Therefore it is open.
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Remark 3.67. For finitely many topological spaces X1 , ..., Xn one can define a topology on X1 × · · · ×
Xn either by considering a basis composed of products of open sets U1 × · · · × Un , or by induction
on n and by identifying X1 × · · · × Xn with (X1 × · · · × Xn−1 ) × Xn .
Proposition 3.68. Let (X, dX ) and (Y, dY ) be two metric spaces.
The product topology of the two metric topologies TdX and TdY coincides with the topology induced
by any of the product metrics (recall that they are all topologically equivalent, by Ex. 6, Sheet 1).
Proof. Consider the product metric
d∞ ((x, y), (x′ , y ′ )) = max [dX (x, x′ ) , dY (y, y ′ )] .
For every U ∈ TdX and V ∈ TdY , U × V is in Td∞ . It follows that BX×Y ⊆ Td∞ , whence
TX×Y ⊆ Td∞ , due to property (T3) of Td∞ .
Conversely, let W be a set in Td∞ . We prove that W is in TX×Y using the Criterion 3.65.
Let (x, y) ∈ W . Then there exists ε such that Bd∞ ((x, y), ε) ⊆ W . But
Bd∞ ((x, y), ε) = BdX (x, ε) × BdY (y, ε) ,
and the latter set is in BX×Y .
Proposition 3.69. Let (X × Y, TX×Y ) be the topological product of (X, TX ) and (Y, TY ) , and let
A × B be a subset of X × Y . Then
(i) A × B = A × B ;
˚
˚× B
˚.
\
(ii) A
×B =A
Proof. (i) According to Proposition 3.66, A × B is a closed set. Since it also contains A × B , we have
that A × B ⊆ A × B .
Conversely, let (x, y) be a point in A × B . Then for every U open in X containing x , U ∩ A 6= ∅.
Likewise for every V open in Y containing y , V ∩ B 6= ∅. Then (U × V ) ∩ (A × B) 6= ∅.
It follows that (x, y) satisfies the property described in Proposition 3.60, (4), for the set A×B with
respect to the basis BX×Y . Therefore (x, y) ∈ A × B . We have thus proved that A × B ⊆ A × B ,
whence the equality.
˚
˚× B
˚ is open and contained in A × B , therefore it is contained in A
\
× B.
(ii) The set A
˚
\
Conversely, let (x, y) be a point in A
× B . Proposition 3.60, (5), applied to the set A × B with
respect to the basis BX×Y implies that there exist U open in X and V open in Y such that
˚ and y ∈ B
˚,
(x, y) ∈ U × V ⊆ A × B . This implies that x ∈ U ⊆ A and y ∈ V ⊆ B , whence x ∈ A
˚
˚
that is (x, y) ∈ A × B .
˚
˚× B
˚ , which together with the opposite inclusion yields the
\
×B ⊆ A
We have proved that A
equality.
Proposition 3.70.
(1) If X and Y satisfy the first separation axiom then so does their topological product X × Y .
(2) If X and Y are Hausdorff spaces, so is their topological product X × Y .
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PART A TOPOLOGY COURSE: HT 2011
Proof. (1) Let (x, y) 6= (x′ , y ′ ) . Then either x 6= x′ or y 6= y ′ .
Assume that x 6= x′ . Then there exists U open in X containing x and not x′ . It follows that
U × Y contains (x, y) and not (x′ , y ′ ) . The case y 6= y ′ is similar.
The proof of (2) is done along the same lines, using the fact that
U ∩ V = ∅ ⇒ (U × Y ) ∩ (V × Y ) = ∅ .
3.10. Product of topological spaces and continuous maps.
Proposition 3.71.
(1) When X × Y is endowed with the product topology TX×Y , the projection
maps pX : X × Y → X , pX (x, y) = x, and pY : X × Y → Y , pY (x, y) = y , are continuous.
(2) ( a second way of defining the product topology) The product topology TX×Y is the coarsest
topology making pX and pY continuous: any topology T on X × Y with respect to which
pX and pY are continuous contains TX×Y .
Proof. (1) Indeed for any open set U in X , p−1
X (U ) = U × Y ; likewise for any open set V in Y ,
p−1
(V
)
=
X
×
V
.
Y
(2) According to the above any topology T with respect to which pX and pY are continuous
must contain all sets U × Y with U ∈ TX , and all sets X × V with V ∈ TY .
Therefore, by property (T2), T must contain all sets U ×V = (U ×Y )∩(X ×V ) . Thus BX×Y ⊆ T
whence, by property (T3) of T , TX×Y ⊆ T .
Proposition 3.72. Let X, Y be two topological spaces.
(1) Let Z be a topological space, and let f : Z → X and g : Z → Y be two maps.
The map (f, g) : Z → X × Y is continuous, with X × Y endowed with the product topology,
if and only if both f and g are continuous.
(2) ( a third way of defining the product topology) The product topology is the only topology for
which (1) holds for all possible Z and f, g .
Remark 3.73. Proposition 3.72, (1) can be reformulated as follows:
A map F : Z → X ×Y from a topological space Z into the topological product X ×Y is continuous
if and only if both pX ◦ F : Z → X and pY ◦ F : Z → Y are continuous.
Proof of Proposition 3.72. (1) Assume that the map (f, g) is continuous. Then f = pX ◦ (f, g)
and g = pY ◦ (f, g) are also continuous, by Proposition 1.
Conversely, assume that both f and g are continuous.
For every U open in X and V open in Y ,
is open.
(f, g)−1 (U × V ) = f −1 (U ) ∩ g −1 (V )
This and Proposition 3.59 applied to the map (f, g) and the basis BX×Y allow to conclude that
(f, g) is continuous.
(2) Let T be another topology on X × Y for which (1) holds for all possible Z and f, g .
We take Z = X × Y with the product topology TX×Y and the maps f = pX and g = pY
continuous. Then (pX , pY ) = idX×Y is a continuous map between (X × Y , TX×Y ) and (X × Y , T ) .
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33
It follows that for every open set U ∈ T , its inverse image by idX×Y , which is U , is contained in
TX×Y .
We have thus proved that T ⊆ TX×Y .
Clearly idX×Y = (pX , pY ) is a continuous map between (X × Y , T ) and (X × Y , T ) . This and
the property (1) of T implies that pX and pY are continuous also when X × Y is endowed with the
topology T .
Proposition 3.71, (2), implies that TX×Y ⊆ T , whence equality.
4. Connected spaces
4.1. Definition and properties of connected spaces. We now define the important notion of
connected space.
• This notion allows to define intervals in R in terms of the topology on R (instead of order on
R ). This was seen last term and shall be recalled here.
• In various problems in Analysis and Geometry one needs to work in a set A such that one
can move continuously inside A from an arbitrary point of A to another (i.e. a set A that is
path-connected ).
• In real life too such questions naturally occur. If one can only travel by car for instance, one
is bound to consider the topological space X composed of all the points above the sea level
and make one’s travels plans within a path-connected component A of X .
• A weaker condition which is also much used in Analysis and Geometry is that A cannot be
written as the disjoint union of two open non-empty sets (i.e. A is connected, as opposite to
disconnected ).
We shall see that the two conditions of connectedness and path-connectedness are not equivalent
in general.
Definition 4.1. Suppose that A is a subset
[ of a topological space X . A family {Ui : i ∈ I} of
subsets of X is called a cover for A if A ⊆
Ui .
i∈I
We also sometimes say that the sets Ui , i ∈ I , cover the set A .
If each Ui is open (closed) in X then U is called an open cover (respectively a closed cover ) for A .
Proposition 4.2. Let X be a topological space. The following properties are equivalent:
(1) the only subsets of X which are both open and closed are X and ∅;
(2) if {U, V } is an open cover of X such that U ∩ V = ∅ then either U = ∅ or V = ∅;
(3) any continuous map from X to {0, 1} (with the discrete topology) is constant.
Proof. (1) ⇒ (2) If U, V are open sets in X such that U ∩ V = ∅ and U ∪ V = X , then U is both
open and closed (as X \ U = V ). It follows by (1) that either U = ∅ or U = X , hence V = ∅.
(2) ⇒ (3) Let f : X → {0, 1} be a continuous map. Then U = f −1 ({0}) and V = f −1 ({1})
compose an open cover of X , and U ∩ V = ∅.
It follows that either U = ∅, in which case f is constant 1, or V = ∅, in which case f is constant 0.
(3) ⇒ (1) Let A be a subset of X both open and closed. Then the map f : X → {0, 1}
such that f (x) = 1 if x ∈ A and f (x) = 0 if x 6∈ A is continuous: f −1 ({1}) = A which is open,
f −1 ({0}) = X \ A , which is open.
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PART A TOPOLOGY COURSE: HT 2011
According to (3) either f is constant 1, which means that A = X , or f is constant 0, which
means that A = ∅.
Definition 4.3. A topological space X is connected if it satisfies one of the three properties in
Proposition 4.2.
Remark 4.4. Statement (2) in Proposition 4.2 can be reformulated as follows:
• there exists no open cover {U, V } of X composed of non-empty disjoint sets;
• for any open cover {U, V } of X composed of non-empty sets we have U ∩ V 6= ∅;
• for any closed cover {U, V } of X composed of non-empty sets we have U ∩ V 6= ∅.
Definition 4.5. A (non-empty) subset A of a topological space X is connected if A with the subspace
topology is connected.
Conventionally an empty set is assumed to be connected.
Remark 4.6. The connectedness of a subset A can be formulated in terms of the topology T on X
(rather than the subspace topology) as follows:
A is connected if and only if for every open cover {U, V } ⊂ T of A such that U ∩ V ∩ A = ∅ either
U ∩ A = ∅ or V ∩ A = ∅ .
Proposition 4.7. Let R be endowed with the standard topology and let A be a subset in R . The
following statements are equivalent:
(1) A is a connected subset of R ;
(2) A is an interval, i.e. a set of the form
(−∞, b), (−∞, b], (a, b), [a, b), (a, b], [a, b], (a, ∞), [a, ∞), ∅, and R,
where b > a for intervals with endpoints a, b except for [a, b] , where b > a.
Proof. (2) ⇒ (1) Consider and interval I . Let V, W be two non-empty closed subsets such that
V ∪ W = I . Since both are non-empty, there exists v ∈ V and w ∈ W . By eventually swapping V
and W we may assume that v ≤ w .
Since I is an interval, [v, w] ⊆ I .
The set V ∩ [v, w] is a non-empty closed set with upper bound w . Therefore it has a least upper
bound z . The set V ∩ [v, w] is closed therefore z ∈ V ∩ [v, w] , by Exercise 2.
Assume that z 6= w . According to the choice of z , (z, w] ∩ V = ∅. On the other hand (z, w] ⊂
I
I = V ∪ W , therefore (z, w] ⊆ W . It follows that (z, w] = [z, w] ∩ I = [z, w] is in W , in particular
z ∈W.
Since z ∈ V , we have thus proved that V ∩ W 6= ∅.
(1) ⇒ (2) Let A be a connected subset of R . We prove that A is an interval. Let x ≤ y ≤ z
be such that x, z ∈ A . We prove that y is in A .
The sets V = A ∩ (−∞, y] and W = A ∩ [y, +∞) are closed in A , non-empty ( V contains x and
W contains z ), and V ∪ W = A .
Since A is connected, it follows that V ∩ W is non-empty. The latter set is equal to A ∩ {y},
therefore y ∈ A .
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We saw that several topological properties (openness, closedness etc) are preserved under finite
intersections and unions. This is no longer true for connectedness: one can easily draw examples
disproving the statement both for finite intersections and unions.
Still, under some extra conditions, connectedness can be granted for unions of connected sets.
Proposition 4.8. Suppose that {Ai : i ∈ I} is a family of connected subsets of a topological space X
with Ai ∩ Aj 6= ∅ for each pair i, j ∈ I .
S
Then i∈I Ai is connected.
S
Proof. Let f : i∈I Ai → {0, 1} be a continuous map. Its restriction to Ai , f |Ai , is also continuous.
Since Ai is connected it follows that f |Ai is constant equal to ci ∈ {0, 1}.
For each pair i, j ∈ I there exists a point x in Ai ∩ Aj , and f (x) = ci = cj . Therefore all the
constants ci are equal, and f is constant.
Corollary 4.9. If {Ci : i ∈ I} andS B are connected subsets of a topological space X such that
Ci ∩ B 6= ∅ for each i ∈ I , then B ∪ i∈I Ci is connected.
Proof. Proposition 4.8 implies that Ai = Ci ∪ B is connected for every i ∈ I .
Since
S Ai ∩ Aj contains B for every i, j , again by Proposition 4.8 it follows that
B ∪ i∈I Ci is connected.
S
i∈I
Ai =
Proposition 4.10. If f : X → Y is a continuous map of topological spaces X, Y and A ⊂ X is
connected then so is f (A) . (“The continuous image of a connected set is connected.”)
Proof. Note that fA : A → f (A), fA (x) = f (x) for every x ∈ A is also continuous, moreover onto.
Indeed for any open set U in Y , fA−1 (U ∩ f (A)) = f −1 (U ) ∩ A .
Assume that {U, V } is an open cover of f (A) composed of non-empty, disjoint subsets.
Then fA−1 (U ), fA−1 (V ) are non-empty (because fA is onto), open, disjoint, and fA−1 (U ) ∪ fA−1 (V ) =
A . This contradicts the fact that A is connected.
Corollary 4.11 (the intermediate value theorem). Let f : X → R be a continuous function,
where X is a connected topological space. If f (x), f (y) are two values of f , and r is a real number
between f (x), f (y) then there exists u ∈ X such that f (u) = r .
In particular this is satisfied when X is an interval in R .
Proof. According to Proposition 4.10, f (X) is a connected subset in R . This and Proposition 4.7
imply that f (X) is an interval.
Theorem 4.12. The topological product X × Y is connected if and only if both X and Y are
connected.
Remark 4.13. An inductive argument allows to extend Theorem 4.12 to any finite product of topological spaces: a topological product X1 × · · · × Xn is connected if and only if X1 , ..., Xn are connected.
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PART A TOPOLOGY COURSE: HT 2011
Proof of Theorem 4.12. If X × Y is connected then X = pX (X × Y ) and Y = pY (X × Y ) are
connected by Proposition 4.10.
Assume that X and Y are connected. Then X × {y} is connected for every y ∈ Y , and {x} × Y
is connected for every x ∈ X .
Fix a point y0 ∈ Y . Corollary 4.9 applied to SB = X × {y0 } and to Cx = {x} × Y , connected and
such that B ∩ Cx = {(x, y0 )} implies that B ∪ x∈X Cx = X × Y is connected.
Theorem 4.14. Suppose that A is a connected subset of a topological space X and A ⊆ B ⊆ A .
Then B is connected.
Proof. Assume that B has an open cover {U, V } composed of non-empty disjoint sets.
Then {U ∩ A , V ∩ A} is an open cover of A , composed of disjoint sets.
The set U is open in B therefore U = U ′ ∩ B where U ′ is open in X . We have that U ′ ∩ B 6= ∅,
whence U ′ ∩ A 6= ∅.
We know that given an open set U ′ in X , U ′ ∩ A 6= ∅ if and only if U ′ ∩ A 6= ∅ (Exercise 7, (1),
Sheet 2).
Therefore U ′ ∩ A = U ∩ A is non-empty. Similarly we prove that V ∩ A is non-empty.
We have thus obtained an open cover {U ∩ A , V ∩ A} of A , composed of non-empty disjoint sets.
This contradicts the fact that A is connected.
4.2. Path-connected spaces.
Definition 4.15. A path connecting two points x, y in a topological space X is a continuous map
p : [0, 1] → X with p(0) = x, p(1) = y.
Remark 4.16. If p : [0, 1] → X is a path then p ([0, 1]) is a connected set.
Remark 4.17. If p : [0, 1] → X is a path connecting x, y and q : [0, 1] → X is a path connecting y, z
then we may construct a path connecting x, z as follows: r : [0, 1] → X ,

if t ∈ 0, 21 ,
 p(2t)
r(t) =

q(2t − 1) if t ∈ 12 , 1 .
The restriction of r to 0, 12 is continuous because p is continuous and the map t 7→ 2t is
continuous. We denote this restriction by r1 .
Likewise, the continuity of q and of the map t 7→ 2t − 1 implies that the restriction of r to 12 , 1
is continuous. We denote the latter restriction by r2 .
−1
For any closed set F in X , r−1 (F ) = r−1
1 (F ) ∪ r2 (F ) .
−1
1
We have that r1 (F ) is closed in 0, 2 , which is closed in [0, 1] , therefore r−1
1 (F ) is closed in
(F
)
is
closed
in
[0,
1]
.
[0, 1] . A similar argument yields that r−1
2
We conclude that r−1 (F ) is closed in [0, 1] .
Definition 4.18. A topological space X is path-connected if any two points in X are connected by
a path in X .
A subset A ⊆ X is path-connected if with the subspace topology it satisfies the previous condition.
Conventionally the empty set is path-connected.
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37
Examples 4.19.
(1) Given a continuous function f : I → R , I ⊆ R interval, the graph Gf =
{(x, f (x)) | x ∈ I} is path-connected.
For instance consider the function f : R → R ,
x sin x1 if x > 0 ,
f (x) =
0
if x = 0 .
(2) Any normed vector space (V , k k) is path-connected.
Indeed, consider u, v two vectors in V . Let p : [0, 1] → V be defined by
p(t) = (1 − t)u + tv .
Clearly p(0) = u and p(1) = v . Moreover
kp(s) − p(t)k = k(t − s)u − (t − s)vk = |t − s|ku − vk .
It follows that p is K –Lipschitz, with K = ku − vk.
(3) Any ball in a normed vector space is path-connected.
According to Remark 4.17 it suffices to prove that any vector u in a ball B(a, R) is
connected by a path to a .
We consider the path p : [0, 1] → V defined by p(t) = (1 − t)u + ta .
For every t ∈ [0, 1] , kp(t) − ak = |1 − t| ku − ak ≤ ku − ak < R . Thus the image of the
path p is entirely contained in B(a, R) .
Proposition 4.20. Any path-connected space is connected.
Proof. Let X be a path-connected space and let a be a fixed point in X . For every x ∈ X there
exists a path px : [0, 1] → X such that px (0) = a and px (1) = x . By Remark 4.16, Px = px [0, 1] is a
connected set.
We apply Corollary 4.9
S to the collection of connected sets {Px ; x ∈ X} and to the connected set
{a}, and conclude that x∈X Px ∪ {a} = X is connected.
Remark 4.21. There exist connected spaces that are not path-connected.
Example 4.22. Let Gf be the graph of the function f : (0, ∞) → R , f (x) = sin x1 , that is the set
1
|x>0 .
Gf =
x, sin
x
Clearly Gf is path-connected, hence connected.
The closure of Gf in R2 with the standard topology, Gf , is equal to Gf ∪ ({0} × [−1, 1]) , it is
connected but not path-connected.
Remark 4.23. The example 4.22 also shows that ‘ A path-connected’ does not in general imply ‘ A
path-connected’.
In the above statement two things must be proved:
(i) that Gf = Gf ∪ ({0} × [−1, 1]) ;
(ii) that Gf = Gf ∪ ({0} × [−1, 1]) is not path-connected.
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PART A TOPOLOGY COURSE: HT 2011
(i) First we prove that Gf ⊆ Gf ∪ ({0} × [−1, 1]) .
Let (a, b) be an arbitrary point in Gf .
Proposition 3.29, (1), implies that (a, b) is the limit of a sequence
2
xn , sin x1n
with xn > 0 (we
are in the metric space R ). This implies that (xn ) converges to a , therefore a must be in [0, +∞) .
If a ∈ (0, +∞) then by the continuity of the map x 7→ sin x1 we have that the limit b of sin x1n
must be sin a1 . Thus in this case (a, b) ∈ Gf .
Assume that a = 0. The sequence sin x1n converges to b and it is contained in [−1, 1] closed in
R , therefore b ∈ [−1, 1] .
We have thus found that Gf ⊆ Gf ∪ ({0} × [−1, 1]) .
For the converse inclusion, we take an arbitrary point (0, b) ∈ {0} × [−1, 1] .
The sequence xn , sin x1n
in Gf with xn = arcsin1b+2nπ converges to (0, b) . Therefore (0, b) ∈
n∈N
Gf , hence {0} × [−1, 1] ⊆ Gf .
(ii) Assume that Gf = Gf ∪ ({0} × [−1, 1]) is path-connected. In particular there is a path
p : [0, 1] → G connecting (0, 0) and (1, sin 1) .
Since p is continuous and F = {0} × [−1, 1] is closed, we have that p−1 (F ) is closed in [0, 1] . Let
T be the supremum of p−1 (F ) . This in particular implies that p(T, 1] ∩ F = ∅.
As p−1 (F ) is closed, its supremum T must be contained in it, therefore p(T ) = (0, b) for some
b ∈ [−1, 1] .
The path p is continuous at T hence for ε =
1
4
there exists δ such that
∀s ∈ [T, T + δ] , kp(s) − p(T )k <
1
.
4
We may assume that δ is such that T + δ < 1 and we denote T + δ by S .
We have that p(S) ∈ Gf , thus p(S) = x0 , sin x10 for some x0 > 0.
Let π : R2 → R be the projection π(x, y) = x . It is a continuous map, and so is p , therefore π ◦ p
is continuous. The interval [T, S] is a connected set, therefore π ◦ p[T, S] is connected.
It follows that π ◦ p[T, S] is an interval containing 0 and x0 > 0, thus containing [0, x0 ] .
Then for every x ∈ (0, x0 ] there exists a point in p[T, S] contained in π −1 (x) .
−1
On the
other hand p[T, S] is contained in Gf and the latter intersects π (x) in a unique point,
1
x, sin x .
We then conclude that p[T, S] contains
1
; x ∈ (0, x0 ] .
x, sin
x
According to the choice of S , p[T, S] should be contained in the ball B p(T ) ,
any two points in p[T, S] should be at distance ≤ 12 , by the triangle inequality.
1
4
. In particular
1
1
are both in (0, x0 ] .
and yn = 3π +2nπ
On the other hand, for n large enough xn = π +2nπ
2
2
It follows that p[T, S] contains the points xn , sin x1n = (xn , 1) and yn , sin y1n = (yn , −1) ,
which are at distance at least 2. This gives a contradiction.
Nevertheless the implication ‘connected ⇒ path-connected’ holds for a particular type of topological
spaces, and for open sets.
PART A TOPOLOGY COURSE: HT 2011
39
Proposition 4.24. A connected open subset U of a normed vector space is path-connected.
Remark 4.25. The hypothesis ‘ U open’ in Proposition 4.24 cannot be removed, as shown by the
Example 4.22 ( U = Gf is a connected subset of the normed vector space (R2 , k k) ).
Proof. Let a be a fixed point in U and let C be the set of elements x ∈ U connected to a by a path
in U .
The set C contains a , therefore it is non-empty. We prove that C is both open and closed in U .
First, we show that C is open. Indeed, consider a point x in C . Since x ∈ U open, there exists
ε such that B(x, ε) ⊆ U . All points in B(x, ε) can be connected by a path to x , by Example 4.19,
(3). This and Remark 4.17 imply that B(x, ε) ⊆ C .
U
We prove that C is closed in U by proving that C = C , that is C
granted).
U
⊆ C (the other inclusion is
U
For simplicity we write C for C .
Let x be a point in C .
The point x is in U open, therefore there exists ε > 0 such that B(x, ε) ⊆ U .
As x ∈ C it follows that B(x, ε) ∩ C contains a point y .
• y can be connected by a path to a because y ∈ C ;
• y can be connected by a path to x because y ∈ B(x, ε) ⊆ U .
Remark 4.17 implies that x can be connected by a path to a , therefore x ∈ C .
We have thus proved that C ⊆ C , whence C = C is closed.
Thus C is a non-empty, closed and open subset in U connected, therefore C = U . This and
Remark 4.17 imply that U is path-connected.
Proposition 4.26. If f : X → Y is a continuous map of topological spaces X, Y and A ⊂ X is
path-connected then so is f (A) . (“The continuous image of a path-connected set is pathconnected.”)
Proof. For every two elements x, y ∈ f (A) let a, b be two elements in A such that f (a) = x and
f (b) = y .
The set A is path-connected, therefore there exists a path p : [0, 1] → A connecting a and b . Then
f ◦ p is a path connecting x and y .
5. Compact spaces
Compact sets are a particular type of bounded closed set, very much used in all branches of
mathematics. Here are two reasons why this notion is interesting:
• in Analysis it is important to know whether there exists a certain object minimizing a given
functional or not (for instance in calculus of variations problems);
• from the topological point of view, as it will become clear from the theory, compact sets are
‘the next best thing, after singletons and finite sets’.
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PART A TOPOLOGY COURSE: HT 2011
5.1. Definition and properties of compact sets.
Definition 5.1. A subcover of a cover {Ui : i ∈ I} for a space X is a subfamily {Uj : j ∈ J} for
some subset J ⊆ I such that {Uj : j ∈ J} is still a cover for X .
We call it a finite (or countable) subcover if J is finite (or countable).
Proposition 5.2. Let X be a topological space. The following are equivalent:
(1) (the Borel-Lebesgue property) every open cover for X has a finite subcover;
\
(2) if {Vi : i ∈ I} is an indexed family of closed subsets of X such that
Vj 6= ∅ for any finite
subset J ⊆ I then
\
i∈I
j∈J
Vi 6= ∅.
Proof. (1) ⇒ (2) Assume that {Vi : i ∈ I} is a family of closed subsets of X such that
for any finite subset J ⊆ I , while
\
i∈I
Vi = ∅. Then
S
i∈I (X
\ Vi ) = X .
According to (1) there exists J finite subset in I such that
T
to j∈J Vj = ∅, contradicting the hypothesis.
S
j∈J (X
(2) ⇒ (1) Let {Ui ; i ∈ I} be an open cover of X . Then X =
S
i∈I
\
j∈J
Vj 6= ∅
\ Vj ) = X . This is equivalent
Ui , whence
T
i∈I (X
\ Ui ) = ∅.
According to property (2) applied
to the family of closed setsS{X \ Ui ; i ∈ I} there exists some
T
finite subset J of I such that j∈J (X \ Uj ) = ∅. Equivalently j∈J Uj = X .
Definition 5.3. A topological space X is compact if it satisfies any (each) of the two properties in
Proposition 5.2.
Definition 5.4. A subset A of a topological space X is compact if it is compact when endowed with
the induced topology.
Remark 5.5. In view of the way in which the induced topology is defined, a subset A of a topological
S
space X is compact if and only if for every family {U
S i ; i ∈ I} of open sets in X such that A ⊆ i∈I Ui
there exists a finite subset J ⊆ I such that A ⊆ j∈J Uj .
Examples 5.6.
(1) Any finite space X is compact.
(2) If (xn )n∈N is a sequence converging to x in a topological space then the set
L = {xn | n ∈ N} ∪ {x}
is compact.
Indeed consider {Ui ; i ∈ I} cover of L . There exists k ∈ I such that x ∈ Uk .
The sequence (xn ) converges to x therefore there exists N such that xn ∈ Uk for every
n ≥ N . For every n ∈ {1, ..., N }, xn is contained in some Uin , in ∈ I . Therefore L is
contained in Ui1 ∪ · · · ∪ UiN ∪ Uk .
PART A TOPOLOGY COURSE: HT 2011
41
(3) Any (non-empty) set with the cofinite topology is compact. Likewise for the indiscrete topology.
Thus any set X can be endowed with a topology T such that (X, T ) is compact.
The statement for the indiscrete topology follows from the fact that there are only two
open sets in that topology.
The statement for the co-finite topology is easier to check with Proposition 5.2, (2). Indeed
consider a family {Fi ; i ∈ I} of closed sets in X such that any intersection of finitely many
of those sets is non-empty.
Pick k ∈ I and consider Fk = {x1 , ..., xm }. Assume that for every xj there exists Fij not
containing it. Then Fk ∩ Fi1 ∩ · · · ∩ Fim is empty. This contradicts the hypothesis.
T
Therefore some xj is contained in all Fi with i ∈ I . It follows that i∈I Fi contains xj .
Proposition 5.7. (the Heine-Borel Theorem) Any interval [a, b] in R is compact.
Proof. Indeed let U = {Ui ; i ∈ I} be an open cover of [a, b] . Let A be the set of points x in [a, b]
such that [a, x] has a finite subcover in U .
The set A is non-empty (it contains a ) and has the upper bound b . Therefore it has a least upper
bound M . We shall simultaneously prove that M ∈ A and that M = b .
Clearly M ∈ [a, b] , therefore there exists k ∈ I such that M ∈ Uk . As Uk is open, there exists
ε > 0 such that (M − ε, M + ε) ⊆ Uk . The number M − ε is no longer an upper bound of A , therefore
there exists c ∈ A with M − ε < c ≤ M .
S
The fact that c ∈ A implies that there exists J finite subset of I such that [a, c] ⊆ j∈J Uj . Then
S
[a, M + ε) ⊆ [a, c] ∪ (M − ε, M + ε) ⊆ j∈J Uj ∪ Uk .
It follows that [a, M + ε) ∩ [a, b] ⊆ A , whence [a, M + ε) ∩ [a, b] ⊆ [a, M ] . This implies that M = b .
Then the above inclusion becomes [a, b] ⊆ A .
In particular b ∈ A , whence [a, b] has a finite subcover in U .
Remark 5.8. An open interval (a, b) , where a < b , is not compact in R (with the Euclidean topology).
Nor are the intervals [a, ∞) or (−∞, b] .
S
Indeed (a, b) = n∈N a + n1 , b − n1 . Any finite subfamily a + n11 , b − n11 , ......, a + n1k , b − n1k
(for which we may assume that n1 < ... < nk , otherwise we change the order) covers only a + n1k , b − n1k .
S
S
A similar argument can be done for [a, ∞) = n∈N [a, n) and for (−∞, b] = n∈N (−n, b] .
Proposition 5.9. Any closed subset A of a compact space X is compact.
\
Vj 6= ∅ for
Vi 6= ∅.
Proof. Indeed let {Vi : i ∈ I} be an indexed family of closed subsets of A such that
any finite subset J ⊆ I .
As A is closed in X , Vi are also closed in X , which is compact. Therefore
\
i∈I
j∈J
Remark 5.10. The converse of Proposition 5.9 may not be true: a singleton in X is always compact
but may not be closed.
As the example suggest, a separation axiom must be added in order to have the converse of
Proposition 5.9.
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PART A TOPOLOGY COURSE: HT 2011
Proposition 5.11. Let X be a Hausdorff space.
If K is a compact subset of X and x ∈ X \ K then there exists U, V disjoint open sets such that
K ⊆ U and x ∈ V .
Proof. For every y ∈ K , according to the Hausdorff property, there exists a pair of disjoint open sets
Uy and Vy , y ∈ Uy , x ∈ Vy .
The collection {Uy ; y ∈ K} is an open cover for K . Therefore there exist y1 , ..., ym such that
K ⊆ Uy1 ∪ · · · ∪ Uym . Let U = Uy1 ∪ · · · ∪ Uym .
Sm
The set V = Vy1 ∩ · · · ∩ Vym is open, it contains x , and V ∩ U = V ∩ j=1 Uyj = ∅.
Corollary 5.12. Any compact subset K of a Hausdorff space X is closed in X .
Proof. According to Proposition 5.11, for every x ∈ X \ K , there exists V open such that x ∈ V ⊆
X \K.
This implies that X \ K is open (for instance by Criterion 3.58 with B = T ).
Proposition 5.13. Let X be a topological space.
(1) If K1 , ..., Kn are compact subsets in X then K1 ∪ · · · ∪ Kn is compact in X .
(2) If {Ki ; i ∈ I} is a non-empty family of compact subsets and X is Hausdorff then
is compact.
T
i∈I
Ki
Proof.S(1) Let {Ui ; i ∈ I} be an open cover of K1 ∪ · · · ∪ Kn . In particular for everySt ∈ {1, ..., n},
Kt ⊆ i∈I Ui . As Kt is compact, there exists a finite subset Jt of I such that Kt ⊆ j∈Jt Uj .
Sn S
It follows that K1 ∪ · · · ∪ Kn ⊆ t=1 j∈Jt Uj .
(2)T Since X is Hausdorff, every Ki is a closed set according to Corollary 5.12. Therefore
L = i∈I Ki is also closed. It is also contained in some (each) compact space Ki , hence by Proposition
5.9, L is compact.
Remark 5.14.
(1) Property (1) is no longer true for infinite unions.
S
Example: n∈N [0, n] = [0, ∞) .
(2) Property (2) is no longer true if X is not Hausdorff (see Ex. 6, (2), Sheet 4).
Proposition 5.15. A compact subset A of a metric space X is bounded.
Proof. Let x be a point in X . We have that A ⊆ X =
S
n∈N
B(x, n) .
It follows that there exist n1 < ... < nk such that A ⊆ B(x, n1 ) ∪ · · · ∪ B(x, nk ) = B(x, nk ) .
Corollary 5.16. A compact subset of a metric space Y is bounded and closed in Y .
PART A TOPOLOGY COURSE: HT 2011
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5.2. Compact spaces and continuous maps.
Proposition 5.17. If f : X → Y is a continuous map of topological spaces and A ⊂ X is compact
then so is f (A) . (‘The continuous image of a compact set is compact.’)
Proof. Let {Ui ; i ∈ I} be an open cover of f (A) .
Then {f −1 (Ui ) ; i ∈ I} is an open cover of A .
The Borel-Lebesgue property of A implies that there exists a finite subcover {f −1 (Uj ) ; j ∈ J},
J ⊆ I.
It follows that {Uj ; j ∈ J} is a subcover for f (A) .
Corollary 5.18. Suppose that X is a compact space, Y is a metric space and f : X → Y is
continuous.
Then f (X) is bounded and closed in Y . (When f (X) is bounded we say ‘ f is bounded’.)
Corollary 5.19 (the extreme value theorem). If f : X → R is a continuous real-valued function on
a compact space X then f is bounded and attains its bounds, i.e. there exist m, M ∈ X such that
f (m) ≤ f (x) ≤ f (M ) for every x ∈ X .
Proof. Since X is compact, f (X) is compact, in particular it is bounded and closed.
Let a, A be the highest lower bound, respectively the least upper bound of f (X) .
According to Exercise 2 it follows that both a and A are in f (X) , since f (X) is closed.
Remark 5.20. Another consequence of Proposition 5.17 is the following: if A ⊂ X ⊂ Y , where Y is a
topological space and X is endowed with the induced topology, then A compact in X ⇒ A compact
in Y . This follows by using Proposition 5.17 for the inclusion map i : X → Y .
The same implication does not hold when ‘compact’ is replaced by ‘closed’: (0, 1] is closed in (0, 2)
but not in R .
The converse implication holds both for ‘closed’ and for ‘compact’, i.e. A compact (closed) in
Y ⇒ A compact (closed) in X .
Theorem 5.21. The product X × Y is compact if and only if both X and Y are compact.
Proof. If the product X × Y is compact then X = pX (X × Y ) and Y = pY (X × Y ) are compact,
due to Proposition 5.17.
Conversely, assume that X and Y are compact.
Let {Wi ; i ∈ I} be an open cover of X × Y .
Since the rectangular open sets U × V compose a basis for the product topology, we may write
[
[
X ×Y =
Wi =
U j × Vj ,
i∈I
j∈J
where each Uj × Vj is contained in some Wi .
Therefore it is enough if we prove that the open cover {Uj × Vj ; j ∈ J} has a finite subcover.
For every y ∈ Y , the compact set X × {y} is covered by {Uj × Vj ; j ∈ J}. Therefore there exists
a finite subset Fy ⊆ J such that
[
X × {y} ⊆
U j × Vj .
j∈Fy
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PART A TOPOLOGY COURSE: HT 2011
T
The set Vy = j∈Fy Vj is an open set containing y . The family {Vy ; y ∈ Y } is an open cover of
Y , which is compact.
It follows that there exist y1 , ..., ym such that Y = Vy1 ∪ · · · ∪ Vym .
Sm S
We state that X × Y = k=1 j∈Fy Uj × Vj . Indeed consider an arbitrary element (x, y) in the
k
product. Then there exists k ∈ S
{1, ..., m} such that y ∈ Vyk . In particular y ∈ Vj for all j ∈ Fyk .
On the other hand X × {yk } ⊆ j∈Fy Uj × Vj , whence there exists j0 ∈ Fyk such that x ∈ Uj0 . It
k
follows that (x, y) ∈ Uj0 × Vj0 .
Theorem 5.22. (general Heine-Borel) Any closed bounded subset of (Rn , k ki ) , where i ∈
{1, 2, ∞}, is compact.
Proof. Let A be a closed bounded subset of Rn . Since A is bounded, there exists x = (x1 , ..., xn )
and r > 0 such that A ⊆ Bd∞ (x, r) .
The ball Bd∞ (x, r) in (Rn , k k∞ ) can be rewritten as
Bd∞ (x, r) = (x1 − r, x1 + r) × · · · × (xn − r, xn + r) ⊆ [x1 − r, x1 + r] × · · · × [xn − r, xn + r] .
Each interval [xi −r, xi +r] is compact by Proposition 5.7, and their product is compact by Theorem
5.21.
It follows that A is a closed subset contained in a compact space, therefore it is compact by
Proposition 5.9.
N.B. Theorem 5.22 fails in general when (Rn , k ki ) is replaced by an arbitrary metric
space (X, d) :
• an easy example of this is (0, 1] - it’s bounded, and closed in (0, 2) , but it’s not compact;
• another such example, in which the ambient space is a normed vector space, moreover complete,
will be given in Exercise 3, Sheet 4.
The inverse of a continuous map is not necessarily continuous, as Example 2.48 shows. Still, the
statement is true under some extra conditions.
Theorem 5.23. Suppose that X is a compact space, that Y is a Hausdorff space and that f : X → Y
is a continuous bijection. Then f is a homeomorphism (i.e. f −1 is continuous too).
Proof. Let V be a closed subset of X . Then V is compact, by Proposition 5.9. It follows that f (V )
is compact, by Proposition 5.17. Since Y is Hausdorff, this implies that f (V ) is closed, according to
Corollary 5.12.
Corollary 5.24.
(a) Let X be a space, and let TH ⊆ TK be two topologies such that (X, TH ) is
Hausdorff and (X, TK ) compact. Then TH = TK .
(b) Let (X, T ) be a Hausdorff compact space. Then any strictly coarser topology T ′ ( T is no
longer Hausdorff, and any strictly finer topology T ′′ ) T is no longer compact.
Proof. (a)
We apply Theorem 5.23 to the map id : (X, TK ) → (X, TH ) .
(b) is a reformulation of (a).
Quirky application 5.25. No topology on [0, 1] which is strictly coarser than the Euclidean topology
can be Hausdorff. No topology which is strictly finer than the Euclidean topology can be compact.
PART A TOPOLOGY COURSE: HT 2011
45
Corollary 5.26. Suppose that X is a compact space, that Y is a Hausdorff space and that f : X → Y
is injective and continuous.
Then f determines a homeomorphism of X onto f (X) .
Remark 5.27. This generalises the Mods result: if f : [0, 1] → R is continuous and monotonic then
its inverse function f −1 : f ([0, 1]) → [0, 1] is continuous.
An important property of continuous maps defined on compact metric spaces is the following.
Theorem 5.28. If f : X → Y is a continuous map of metric spaces and X is compact then f is
uniformly continuous.
Proof. Let ε be an arbitrary positive number (which we consider as fixed in the sequel).
Let x be an arbitrary point in X . The map f is continuous at x . Therefore, for the given ε > 0
there exists δx such that
ε
(7)
d(y, x) < δx ⇒ d(f (y), f (x)) < .
2
S
δx
We have that
X=
x∈X B x, 2 . The space X being compact, there exist x1 , ..., xn such that
Sn
δxi
X = i=1 B xi , 2 .
n
o
δ
δ
Let δ = min x21 , ...., x2n . We shall prove that d(y, z) < δ ⇒ d(f (y), f (z)) < ε .
Sn
δ
Indeed, let y, z be two points such that d(y, z) < δ . As y ∈ i=1 B xi , x2i , there exists
δ
δ
k ∈ {1, ..., n} such that y ∈ B xk , x2k , that is d(y, xk ) < x2k .
Then d(z, xk ) ≤ d(z, y) + d(y, xk ) < δ +
δxk
2
< δxk .
It follows according to (7) that d(f (y), f (xk )) <
ε
2
and d(f (z), f (xk )) <
ε
2
, whence
d(f (y), f (z)) ≤ d(f (y), f (xk )) + d(f (z), f (xk )) < ε .
5.3. Sequential compactness. Loose Remark. We already saw that in a metric space many
topological properties can be characterised in terms of sequences.
In what follows we show that one of these properties is compactness.
Definition 5.29. A topological space X is sequentially compact if every sequence in X has a subsequence converging to a point in X .
A (non-empty) subset A of a topological space X is sequentially compact if, with the induced
topology, A is sequentially compact. (Conventionally, the empty set is sequentially compact.)
Theorem 5.30.
(1) (Bolzano-Weierstrass theorem) In a compact topological space X every
infinite subset has accumulation points.
(2) Every compact metric space is sequentially compact.
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PART A TOPOLOGY COURSE: HT 2011
Proof. (1) We prove that if a subset A of X has the set of accumulation points A′ empty then A is
finite.
Recall that the closure A is equal to A ∪ A′ . Therefore if A′ is empty then A = A = A \ A′ .
The set A \ A′ is the set of isolated points of A , hence all points of A are isolated: for every a ∈ A
there exists Ua open set such that Ua ∩ A = {a}.
S
We have that A = A ⊆ a∈A Ua . The set A is a closed set in a compact metric space, therefore it
is compact by Proposition 5.9. TheSBorel-Lebesgue property of compact sets implies that there exist
n
a1 , ..., an in A such that A = A ⊆ i=1 Uai .
Sn
Sn
We may then write A = i=1 (Uai ∩ A) = i=1 {ai } = {a1 , ..., an }.
(2) Consider (xn ) a sequence in a compact metric space X . Let A = {xn ; n ∈ N}.
Assume that A is finite, A = {a1 , ..., ak }. For every ai we consider the set Ni = {n ∈ N ; xn = ai }.
Since N = N1 ⊔ N2 ⊔ · · · ⊔ Nk , at least one of the sets Ni is infinite. It follows that (xn ) has a subsequence which is constant, therefore convergent.
Assume that A is infinite. Then according to (1), A′ 6= ∅. Let a be a point in A′ . For every open
set U containing a , (U \ {a}) ∩ A 6= ∅.
In particular there exists
• a term xn1 of the sequence contained in [B(a, 1) \ {a}] ∩ A ,
• a term xn2 in B a, 21 \ {a, x1 , ..., xn1 } ∩ A , etc.
• a term xnk in B a, k1 \ {a, x1 , ..., xnk−1 } ∩ A .
In other words, we construct inductively a subsequence of (xn ) converging to a .
Theorem 5.31. Any compact metric space is complete.
Proof. Let (xn ) be a Cauchy sequence in a compact metric space (X, d) .
According to Theorem 5.30, (2), (xn ) has a subsequence (xϕ(n) ) convergent to a point a , where
ϕ : N → N is a one-to-one increasing map. We prove that (xn ) converges to a .
Consider an arbitrary ε > 0. The sequence (xn ) is Cauchy, therefore there exists N such that for
every n, m ≥ N , d(xn , xm ) < 2ε .
The subsequence
(xϕ(n) ) converges to a , therefore there exists M such that for every n ≥ M ,
ε
d xϕ(n) , a < 2 .
Let k ≥ M large enough so that ϕ(k) ≥ N . Then for every n ≥ N ,
ε ε
d(xn , a) ≤ d xn , xϕ(k) + d xϕ(k) , a < + = ε .
2 2
Theorem 5.32. Any sequentially compact metric space is compact.
N.B. This and Theorem 5.30, (2), imply that a metric space is compact if and only if it is
sequentially compact.
The proof is done in two steps, which we formulate as two separate Propositions.
Proposition 5.33. Let X be a sequentially compact metric space. For any open cover U of X there
exists ε > 0 such that for every x ∈ X, B(x, ε) is entirely contained in some single set from U .
PART A TOPOLOGY COURSE: HT 2011
47
Definition 5.34. The number ε in Proposition 5.33 is called the Lebesgue number for the cover U .
Proof. Assume that for every n ∈ N there exists a point xn ∈ X such that the ball B xn ,
contained in no subset U ∈ U .
1
n
is
The sequence (xn ) has a subsequence (xϕ(n) ) converging to some point a . There exists a subset
U in the cover U containing a . As U is open, it also contains a ball B(a, 2ε) , for ε > 0.
The subsequence (xϕ(n) ) converges to a , hence there exists N such that if n ≥ N then xϕ(n) ∈
B(a, ε) . This implies that B xϕ(n) , ε ⊆ B(a, 2ε) ⊆ U .
1
1
For n large enough ϕ(n)
< ε , whence B xϕ(n) , ϕ(n)
⊆ B xϕ(n) , ε ⊆ U .
The last inclusion contradicts the choice of the sequence (xn ) .
Definition 5.35. Given a real number ε > 0, an ε -net for a metric space X is a subset N ⊆ X
such that {B(x, ε) ; x ∈ N } is a cover of X .
Example 5.36. For every ε >
√
n
2
, Zn is an ε -net for Rn with the Euclidean norm k k2 .
Proposition 5.37. Let X be a sequentially compact metric space and let ε be an arbitrary positive
number.
Then there exists a finite ε -net for X .
Remark 5.38. Thus compact metric spaces can be approximated by finite sets.
This illustrates once more our remark that from the topological point of view compact sets are ‘the
next best thing, after singletons and finite sets’.
Proof. Assume that there exists no finite ε -net for X .
Let x1 be a point in X . According to the assumption there exists x2 ∈ X \ B(x1 , ε) .
We argue by induction and assume that there exist x1 , ..., xn points in X such that d(xi , xj ) ≥ ε
for every i 6= j in {1, 2, ..., n}.
SnSince {x1 , ..., xn } is finite, it cannot be a ε -net. Therefore there exists a point xn+1 in X \
i=1 B(xi , ε) .
In this manner we construct a sequence (xn ) such that
(8)
d(xi , xj ) ≥ ε ∀i 6= j, i, j ∈ N .
The space X is sequentially compact, therefore (xn ) has a convergent subsequence (xϕ(n) ) , where
ϕ : N → N is a strictly increasing function.
In particular (xϕ(n) ) is a Cauchy sequence. It follows that there exists N such that if m ≥ n ≥ N ,
d(xϕ(n) , xϕ(m) ) < ε . This contradicts (8).
We now finish the proof of the fact that a sequentially compact metric space is compact.
Indeed, let U be an open cover of X and let ε be the Lebesgue number for the cover U provided
by Proposition 5.33.
Proposition 5.37 implies that there exists a finite ε -net, {x1 , ..., xn }.
By the definition of the Lebesgue number, each ball B(xi , ε) is contained in some set Ui ∈ U .
Sn
Sn
Thus X = i=1 B(xi , ε) ⊆ i=1 Ui .
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PART A TOPOLOGY COURSE: HT 2011
6. Quotient spaces, quotient topology
In many cases, a space X on which an equivalent relation R is defined is a topological space (e.g.
the group R and the equivalent relation defined by a subgroup, say Z or Q ).
Natural questions to ask are:
• whether the quotient space X/R is a topological space too;
• which of the topological properties of X are inherited by X/R .
We shall see that a quotient space always has a natural topology, but that this topology may not
have good properties - not even the basic separation properties, even though the ambient space X is
a very good space, e.g. R .
6.1. Definition and examples. First we remind some basic notions on equivalence relations and
partitions.
Let X be a space. Recall that an equivalence relation R on X is a relation that is
reflexive xRx for every x ∈ X ;
symmetric xRy ⇒ yRx ;
transitive if xRy and yRz then xRz .
The graph of the equivalence relation R is the set
GR = {(x, y) ∈ X × X ; xRy} .
(9)
The equivalence class of an element x ∈ X is the set
[x] = {y ∈ X ; yRx} .
The set of equivalence classes is denoted by X/R , and it is called the quotient space of X with
respect to R .
Example 6.1. (from Mods)
(a) Let (G, ∗) be a group and H a subgroup of it.
The relation xRy ⇔ x−1 ∗ y ∈ H is an equivalence relation.
An equivalence class is a left coset gH , g ∈ G .
The quotient space is G/H .
(b) For instance we may consider the group (R, +) and the subgroup Z . Or the subgroup Q .
Definition 6.2. A cover {Ui ; i ∈ I} of a space X is called a partition if
• Ui 6= ∅ for every i ∈ I ;
• Ui ∩ Uj = ∅ for every i 6= j , i, j ∈ I .
Remark 6.3. (from Mods) Defining an equivalence relation on a space X is equivalent to defining
a partition of X .
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49
Indeed, given an equivalence relation on a space X , the equivalence classes compose a partition
of X .
Conversely, given a partition {Ui ; i ∈ I} of X , one can define an equivalence relation by
xRy ⇔ ∃ i ∈ I such that {x, y} ⊆ Ui .
Let p : X → X/R be the map that assigns to each point x in X the equivalence class [x] (the set
in the partition containing x ).
Proposition 6.4. Let (X, T ) be a topological space, and R an equivalence relation on X .
˜ in X/R such that p−1 (U
˜ ) ∈ T is a topology for X/R , called the
The family T˜ of subsets U
quotient topology.
Proof. (T1) p−1 (∅) = ∅ and p−1 (X/R) = X are both in T .
(T2) and (T3) follow from the fact that
˜ ∩ V˜ = p−1 U
˜ ∩ p−1 V˜
p−1 U
and that
p
−1
[
i∈I
˜i
U
!
=
[
˜i ) .
p−1 (U
i∈I
Proposition 6.5. A subset V˜ in X/R endowed with the quotient topology is closed if and only if
p−1 (V˜ ) is closed in X .
Proof. This follows from the fact that
p−1 (X/R) \ V˜ = X \ p−1 V˜ .
Example 6.6. Consider the quotient space R/Q defined in Example 6.1, (b).
Let U be an open non-empty set in R/Q . Then p−1 (U ) is open and non-empty in R .
−1
For every
, the set
x ∈ R−1
−x + p (U ) is open and non-empty in R . Since Q is dense in R , there
exists q ∈ −x + p (U ) ∩ Q .
Equivalently, there exists q ∈ Q such that x+q ∈ p−1 (U ) . The set p−1 (U ) contains any equivalence
class that it intersects, therefore x + Q ⊆ p−1 (U ) ⇔ p(x) ∈ U .
We have thus proved that for every x ∈ R , p(x) ∈ U . It follows that U = R/Q .
Thus the quotient topology on R/Q is the indiscrete topology.
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PART A TOPOLOGY COURSE: HT 2011
6.2. Separation axioms.
Proposition 6.7. A quotient topological space satisfies the first separation axiom if and only if every
equivalence class is closed.
Proof. The first separation axiom is equivalent to the statement that every singleton {[x]} in X/R
is closed.
According to Proposition 6.5, the latter is equivalent to the fact that the inverse image of {[x]} by
p , which is [x] seen as a subset of X , is closed.
In order to discuss the Hausdorff property we need to define a particular type of sets.
Lemma 6.8. Let R be an equivalence relation on a space X , let p : X → X/R be the corresponding
quotient map and let A be a subset in X . The following are equivalent:
(1) if A intersects an equivalence class [x] (i.e. A ∩ [x] = ∅) then A contains [x] ;
(2) whenever x ∈ A and yRx the element y is also in A ;
(3) A = p−1 (p(A)) .
The proof of the equivalence is easy and left as an exercise.
Definition 6.9. A subset A of X is saturated with respect to the equivalence relation R if it satisfies
one of the equivalent conditions in Lemma 6.8.
Remark 6.10. If A is a saturated subset of X with respect to R then its complement X \ A is also
saturated with respect to R .
Proof. Let x ∈ X \ A and let yRx . If y ∈ A then x ∈ A , contradiction.
Therefore y ∈ X \ A , hence X \ A is saturated.
Example 6.11. For every subset B in X/R , p−1 (B) is saturated.
Example 6.12.
(a) A map f : X → Y defines the equivalence relation xRf y ⇔ f (x) = f (y) .
The equivalence classes are f −1 (y) , y ∈ f (X) .
For instance, given the map f : R → C , f (t) = e2iπt = cos(2πt)+i sin(2πt) the equivalence
relation Rf is the same as that in Example 6.1, (b), defined by the subgroup Z of R .
(b) We may say that a set A is saturated with respect to f or f -saturated if it is saturated with
respect to Rf .
This is equivalent to the fact that A = f −1 (f (A)) .
Proposition 6.13. A quotient topological space is Hausdorff if and only if any two distinct equivalence
classes are contained in two disjoint open saturated sets.
Proof. The proof is Exercise 5, Sheet 4.
Examples 6.14.
(1) Let S = n1 ; n ∈ N . Consider the partition of R composed of S and all
the singletons {x} with x ∈ R \ S , and the corresponding equivalence relation R .
The quotient space X/R does not satisfy the first separation axiom.
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51
(2) Let X = [0, 1] × [0, 1] , and let ℘ be the partition composed of
{(0, y), (1, 1 − y)} with y ∈ [0, 1], {(x, y)} with x ∈ (0, 1) .
The quotient X/℘ is called the M¨
obius band; it is Hausdorff.
Indeed, consider two distinct equivalent classes {(0, y), (1, 1 − y)} and {(a, b)} with a ∈
(0, 1) .
For ε > 0 and δ > 0 small enough, the open saturated sets
are disjoint.
[B((0, y), ε) ∪ B((1, 1 − y), ε)] ∩ X and B((a, b), δ) ⊆ X
Similar arguments work for any two distinct equivalent classes in X/℘ .
Proposition 6.15.
(1) If X/R is Hausdorff then the graph of R is closed.
(2) If the graph of R is closed and for every U open in X , p(U ) is open then X/R is Hausdorff.
Proof. (1) Let (x, y) ∈ X × X be a point outside the graph GR . It follows that [x] 6= [y] , hence there
e , Ve open disjoint subsets of X/R such that [x] ∈ U
e and [y] ∈ Ve . Then p−1 (U
e ) and p−1 (Ve )
exist U
are open and contain x , respectively y .
e ) × p−1 (Ve ) intersects GR . Then there exist a ∈ p−1 (U
e ) and b ∈ p−1 (Ve ) such
Assume that p−1 (U
e
e
that aRb . It follows that [a] = [b] is in U ∩ V , contradiction.
e ) × p−1 (Ve ) ⊆ (X × X) \ GR .
We have thus found (x, y) ∈ p−1 (U
(2) Consider two distinct points [x] 6= [y] in X/R .
The point (x, y) is in (X × X) \ GR which is open, therefore there exist U, V open in X such that
(10)
(x, y) ∈ U × V ⊆ (X × X) \ GR .
Then [x] ∈ p(U ) and [y] ∈ p(V ) , p(U ), p(V ) open.
If p(U ) ∩ p(V ) contains some element [z] then there exists u ∈ U and v ∈ V such that uRz and
vRz . By transitivity uRv , therefore (u, v) ∈ (U × V ) ∩ GR . This contradicts the inclusion in (10).
Thus p(U ) ∩ p(V ) = ∅. We have proved that X/R is Hausdorff.
Remark 6.16. The converse of Proposition 6.15, (2), is not true.
Indeed, consider the M¨obius band X/℘ , a Hausdorff quotient.
The graph of the equivalence relation corresponding to the partition ℘ is closed, as it is composed
of the diagonal
∆ = {(v, v) ∈ X × X ; v ∈ X}
and of the two graphs
{((0, x), (1, 1 − x)) ; x ∈ [0, 1]} and {((1, x), (0, 1 − x)) ; x ∈ [0, 1]} .
But it is not true that for every U open in X , p(U ) is open in X/℘ (equivalently, p−1 (p(U )) is
open in X ).
Indeed for U = B((0, 1/2), 1/4) ∩ X , p−1 (p(U )) is
p−1 (p(U )) = [B((0, 1/2), 1/4) ∩ X] ∪ [{1} × (1/4, 3/4)] .
The set p−1 (p(U )) is not open, because it does not contain an open ball centred in (1, 1/2) .
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PART A TOPOLOGY COURSE: HT 2011
Examples 6.17. R/Z is Hausdorff.
Indeed, the graph of the equivalence relation is the union of parallel lines
[
{(x, y) ; y = x + n} ,
n∈Z
and it is closed.
For any open set in U , p−1 (p(U )) =
that p(U ) is open in R/Z .
S
n∈Z (n
+ U ) which is open as union of open sets. It follows
Proposition 6.15, (2), implies that R/Z is Hausdorff.
A similar argument shows that Rn /Zn is Hausdorff.
Definition 6.18. The real n -dimensional projective space P Rn is the quotient space of Rn+1 \
{0} with respect to the equivalence relation R where xRy if and only if there exists λ 6= 0 such that
x = λy .
In other words P Rn is the set of lines in Rn+1 through 0.
Example 6.19. For any n ≥ 1, P Rn is Hausdorff (Ex. 7, Sheet 4).
6.3. Quotient spaces and continuous maps.
Proposition 6.20 (a second way of defining the quotient topology). The quotient topology is the
largest topology on X/R making p continuous.
Proof. Let T˜ ′ be a topology on X/R such that p : X → X/R , T˜ ′ is continuous. Then for every
U ∈ T˜ ′ , p−1 (U ) is in T .
It follows that U ∈ T˜ . We have thus proved that T˜ ′ ⊆ T˜ .
Proposition 6.21. A map g from X/R to another space Z is continuous if and only if g ◦ p is
continuous.
Proof. If g is continuous then g ◦ p is continuous.
−1
Conversely,
assume that g ◦ p is continuous. For every open subset V of Z , (g ◦ p) (V ) =
−1
p
g (V ) is open in X .
−1
According to the definition of the quotient topology, this implies that g −1 (V ) ∈ T˜ .
We conclude that g is continuous.
We now define the notion of quotient map, useful when trying to prove that a quotient space
is homeomorphic to another topological space.
Proposition 6.22. Let p be a map of a topological space X onto a topological space Y . The following
conditions are equivalent:
(1) U ⊆ Y is open (closed) in Y if and only if p−1 (U ) is open (respectively closed) in X ;
(2) p is continuous, and every image by p of a p -saturated open (closed) set is an open set
(respectively a closed set) in X/R .
Definition 6.23. A quotient map is a surjective map of topological spaces p : X → Y satisfying
any/each of the equivalent properties in Proposition 6.22.
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53
Proof. Statement (1) for open sets is equivalent to the same statement for closed sets. This follows
from the fact that
p−1 (Y \ V ) = X \ p−1 (V ) .
Likewise, statement (2) for open sets is equivalent to the same for closed sets due to:
(a) the fact that the complement in X of a p -saturated set is p -saturated;
(b) the fact that if p is onto and A is p -saturated then p(X \ A) = Y \ p(A) .
Indeed since p is onto Y \ p(A) ⊆ p(X \ A) . On the other hand as A (hence X \ A ) is p -saturated
we have that p(X \ A) ⊆ Y \ p(A) .
The implication (1) ⇒ (2) follows from the fact that a p -saturated set A satisfies p−1 (p(A)) = A .
Indeed if A is an open saturated set then p−1 (p(A)) is also an open set, therefore by (1), p(A) is
an open set.
We prove the implication (2) ⇒ (1).
Assume that p−1 (U ) is open. It is also p -saturated therefore according to (2) p p−1 (U ) = U is
open. In the last equality we used the fact that p is onto.
Proposition 6.24.
(1) Let R be an equivalence relation on a space X , endow X/R with the
quotient topology, and let p : X → X/R be the ‘natural’ map sending each point of X to its
equivalence class. Then p is a quotient map.
(2) Suppose that p : X → Y is a quotient map and that R is the equivalence relation on X
corresponding to the partition {p−1 (y) : y ∈ Y }. Then X/R and Y are homeomorphic.
Proof. (1) The map p satisfies the property (1) in Proposition 6.22 due to the way in which the
quotient topology is defined.
(2) Consider the map p : X/R → Y , p p−1 (y) = y . It is onto because p is onto, and it is injective
by construction.
If π : X → X/R is the standard quotient map, then by the definition of p we have that p ◦ π = p .
In particular, as p is continuous, p is continuous according to Proposition 6.21.
Let U be an open set in X/R . Then π −1 (U ) is an open p -saturated set in X . According to
Proposition 6.22, (2), p(π −1 (U )) is open.
The relation p = p ◦ π implies that p(π −1 (U )) = p(U ) , therefore p(U ) is open.
We have thus proved that p is a continuous bijection, with continuous inverse, therefore a homeomorphism.
Examples 6.25.
(1) R/Z is homeomorphic to the planar unit circle S1 .
Consider the map f : R → S1 , f (t) = e2iπt = cos(2πt) + i sin(2πt) .
It defines the same equivalence relation as the subgroup Z .
The map f is onto and continuous.
For every x ∈ R and ε < 21 , the image f (x − ε, x + ε) is the intersection of an open
half-plane with the unit circle S1 .
For every U open in R , consider an arbitrary point f (x) in f (U ) .
As x ∈ U and U open, there exists ε < 21 such that (x − ε, x + ε) ⊆ U . It follows that
f (x) ∈ f (x − ε, x + ε) ⊆ f (U ) , and f (x − ε, x + ε) is open in S1 .
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PART A TOPOLOGY COURSE: HT 2011
We have thus proved that f is a quotient map. Proposition 6.24, (2), implies that R/Z is
homeomorphic to S1 .
(2) Rn /Zn is homeomorphic to S1 × · · · × S1 . This space is denoted by Tn and it is called the
{z
}
|
n times
n -dimensional torus. Note that it is a compact space.
To prove the homeomorphism it suffices to consider the quotient map
F (t1 , .., tn ) = e2iπt1 , ..., e2iπtn
and argue as in (1).
(3) Let X = [0, 1] × [0, 1] , and let ℘ be the partition composed of:
• {(x, 0), (x, 1)}, with x arbitrary number in the interval (0, 1) ,
• {(0, y), (1, y)}, with y arbitrary in the interval (0, 1) ,
• {(0, 0), (1, 0), (0, 1), (1, 1)}
• and {(x, y)} with x, y two arbitrary numbers in the interval (0, 1) .
The quotient space [0, 1] × [0, 1]/℘ is homeomorphic to the 2-dimensional torus.
To see this it suffices to take the quotient map F (t1 , t2 ) = e2iπt1 , e2iπt2 restricted to
[0, 1] × [0, 1] , and use Proposition 6.24, (2).
(4) Consider the closed planar unit disk
D2 = {(x, y) ∈ R2 ; x2 + y 2 ≤ 1} .
Let ℘ be the partition composed of:
• all the singletons in the open disk {(x, y)} with x2 + y 2 < 1
• and the boundary circle S1 .
The quotient space D2 /℘ is homeomorphic to the 2-dimensional sphere
S2 = {(x, y, z) ; x2 + y 2 + z 2 = 1} .
This is proved by applying Proposition 6.24, (2), to the map f : D2 → S2 , f (0, 0) = (0, 0, 1) ,
and for (x, y) 6= (0, 0)
!
p
p
p
y
x
.
sin π x2 + y 2 , p
sin π x2 + y 2 , cos π x2 + y 2
f (x, y) = p
x2 + y 2
x2 + y 2
Indeed, the map f is continuous and onto.
Moreover every closed set V in D2 is compact, as D2 is compact, therefore f (V ) is compact
by Proposition 5.17. Since S2 is Hausdorff it follows that f (V ) is closed. Thus for every closed
set V in D2 , f (V ) is closed.
Proposition 6.22, (2), implies that f is a quotient map.
The partition defined by f is the partition ℘ .
Proposition 6.26. There is a (topological) embedding of the 2-dimensional torus T2 in R3 , i.e. T2
is homeomorphic to a subset of R3 with the induced topology.
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55
Proof. Consider the map p : R2 → R3 ,
p(x, y) = ((2 + cos(2πy)) cos(2πx) , (2 + cos(2πy)) sin(2πx) , sin(2πy)) .
One can easily check that p(x, y) = p(x′ , y ′ ) if and only if (x, y) − (x′ , y ′ ) ∈ Z2 .
Indeed (2 + cos α) cos β = (2 + cos α′ ) cos β ′ and (2 + cos α) sin β = (2 + cos α′ ) sin β ′ imply that
the sums of the squares are equal therefore (2 + cos α)2 = (2 + cos α′ )2 .
The latter is equivalent to |2 + cos α| = |2 + cos α′ | , but since 2 + cos α ≥ 0 for every α we obtain
that cos α = cos α′ .
Therefore p(x, y) = p(x′ , y ′ ) implies that cos(2πy) = cos(2πy ′ ) and sin(2πy) = sin(2πy ′ ) , whence
y − y ′ ∈ Z . Also cos(2πx) = cos(2πx′ ) and sin(2πx) = sin(2πx′ ) , whence x − x′ ∈ Z .
The map p therefore induces a bijective map p¯ : R2 /Z2 → p(R2 ) defined by p¯ (x, y) + Z2 =
p(x, y) . If we denote by π : R2 → R2 /Z2 the standard quotient map then p¯ ◦ π = p . Since p is
continuous it follows by Proposition 6.21 that p¯ is continuous.
The quotient R2 /Z2 is compact and p(R2 ) is a subspace of R3 , therefore Hausdorff. Theorem 5.23
implies that p¯ is a homeomorphism.
Definition 6.27. The Klein bottle K is the quotient space of the rectangle [0, 2π] × [0, π] by the
equivalence relation which identifies the points (x, 0), (2π − x, π) for each x ∈ [0, 2π] and the points
(0, y), (2π, y) for each y ∈ [0, π].
Note that, since [0, 2π] × [0, π] is compact, the Klein bottle is compact.
Proposition 6.28. There is an embedding of the Klein bottle K in R4 .
Proof. This embedding is defined by the map g : [0, 2π] × [0, π] → R4 ,
g(x, y) = ((2 + cos x) cos 2y , (2 + cos x) sin 2y , sin x cos y , sin x sin y) .
Indeed g is continuous.
Assume that g(x, y) = g(x′ , y ′ ) .
Then the same calculation as in the proof of Proposition 6.26 implies that cos x = cos x′ .
It follows that cos 2y = cos 2y ′ and sin 2y = sin 2y ′ whence y = y ′ modulo π .
If y = y ′ ∈ (0, π) then sin x = sin x′ hence either x = x′ or x = 0, x′ = 2π or x = 2π, x′ = 0.
If y = 0 and y ′ = π then sin x cos y = sin x′ cos y ′ implies that sin x = − sin x′ . Together with
cos x = cos x′ this implies x′ = 2π − x .
Thus we proved that the equivalence relation on [0, 2π] × [0, π] defined by g coincides to the one
with quotient space the Klein bottle.
Consider the continuous bijection g¯ : K → g¯(K) ⊆ R4 defined by g .
The Klein bottle K is compact and g¯(K) is a subspace of R4 , therefore it is Hausdorff. Theorem
5.23 implies that g¯ is a homeomorphism.
For the next proposition we use the following notation:
(a) The unit sphere in R3 is S2 = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 1};
(b) The upper hemisphere in S2 is D+ = {(x, y, z) ∈ S2 ; z > 0};
(c) The closed unit disc in R2 is D2 = {(x, y) ∈ R2 : x2 + y 2 6 1}.
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PART A TOPOLOGY COURSE: HT 2011
Proposition 6.29. The following are all homeomorphic to the projective plane P R2 .
(a) The quotient space S2 /R where R identifies each pair of antipodal points of S .
(b) The quotient space D+ /R where R identifies each pair of antipodal points on the boundary
of D+ .
(c) The quotient space D2 /R where R identifies each pair of antipodal points on the boundary of
D.
(d) The quotient space of the square [0, 1] × [0, 1] by the equivalence relation which identifies
(s, 0), (1 − s, 1) for each s ∈ [0, 1] and (0, t), (1, 1 − t) for each t ∈ [0, 1].
Remarks 6.30.
(1) Proposition 6.29, (a), implies that the projective plane is compact.
(2) A proposition similar to Proposition 6.29 can be formulated for every projective space P Rn ,
n ≥ 2.
Proof. To prove (a) and (b) it suffices to restrict the quotient map p : R3 → P R2 to S2 , respectively
D+ , and apply Proposition 6.24, (2).
(c) The stereographic projection π : R2 → S2 ,
2y
1 − x2 − y 2
2x
,
,
,
π(x) =
1 + x2 + y 2 1 + x2 + y 2 1 + x2 + y 2
restricted to the unit disk D2 defines a homeomorphism between D2 and the upper hemisphere.
We apply Proposition 6.24, (2) to p|D+ ◦ π|D2 .
(d) follows from the fact that the unit disk D2 and the unit square [0, 1] × [0, 1] are homeomorphic.
Proposition 6.31. There is an embedding of the real projective plane P R2 in R4 , i.e. P R2 is
homeomorphic to a subset of R4 .
Proof. See Exercise 8, Sheet 4.
APPENDIX A: USEFUL IDENTITIES
Let X and Y be sets, {Ui }i∈I a set of subsets of X
and {Vj }j∈J a set of subsets of Y .
(1) De Morgan laws
\
[
X\
Ui =
(X \ Ui )
i∈I
(2) Distributivity of
T
X\
over
S
A
\
[
i∈I
Ui =
i∈I
[
i∈I
Ui
\
i∈I
!
=
(X \ Ui )
[ \ A Ui
i∈I
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57
(3) Images and inverse images
Let f : X → Y be a map. Recall that for any subset A in Y its inverse image is
Then
f −1 (A) = {x ∈ X ; f (x) ∈ A} .
f (U ) ⊆ V if and only if U ⊆ f −1 (V )
f −1 (Y \ V ) = X \ f −1 (V )
[
f
Ui
i∈I
f

f −1 
\
i∈I
[
j∈J

f −1 
\
j∈J
Ui
!
!
=
f (Ui )
i∈I
⊆

Vj  =

[
Vj  =
\
f (Ui )
[
f −1 (Vj )
\
f −1 (Vj )
i∈I
j∈J
j∈J
If A ∩ B = ∅, A, B subsets of Y , then f −1 (A) ∩ f −1 (B) = ∅.
Indeed assume that there exists x ∈ f −1 (A) ∩ f −1 (B) . Then f (x) ∈ A and f (x) ∈ B ,
which contradicts A ∩ B = ∅.