1. science
2. mathematics
3. geometry

[102]
Mayra Quiroga
Let p : X ! Y be a closed continuous surjective map such that p 1 ({y})
is compact, for each y 2 Y . (Such a map is called a perf ect map ). Show
that if Y is compact, then X is compact.
First we prove the hint: If U is an open set containing p
is a neighborhood W of y such that p 1 (W ) ⇢ U.
1
({y}), there
Assume p 1 ({y}) ⇢ U , where U is open in X. Then U c is closed in X, so
c
that p (U c ) is closed in Y , since p is a closed map. And so (p (U c )) is open in Y,
c
c c
and since y 2
/ p (U ) , y 2 (p (U )) , so its a neighborhood of y. By the continuc
ity of p, 8z 2 p 1 ({y}) there is an open set Vz in X such that pS(Vz ) ⇢ (p (U c )) .
c
c c
Since p (Vz )S\ p (U ) = Ø we S
have that Vz ⇢ U and alsoSp ( Vz ) ⇢ (p (U )) .
c
c
c c
Now since ( SVz ) is closed, S
p ( Vz ) is closed inSY, so (p ( Vz ) ) is open.
S And,
c
c c
c c
since y 2 p ( Vz ), y 2
/Sp (( Vz ) ), so y 2S(p ( Vz ) ) . SLet W =(pS( Vz ) ) .
c c
c c
c c
1
1
1
So, p (W ) = p (p ( Vz ) ) = p (p ( Vz ) ) = (( Vz ) ) = Vz ⇢ U .
If Y is compact, then X is compact:
S
({y}). Let {O↵ } be an open cover of X. So, Py ⇢ ( O↵ ).
ny
[
Since Py is compact there exists a finite subcover of Py , that is, Py ⇢
O ↵i
Let Py = p
1
i=1
for some ny 2 N. Then, by the
! hint, 8y 2 Y 9 a neighborhood Wy of y
ny
[
[
such that p 1 (Wy ) ⇢
O↵i . Then
Wy forms an open cover of Y ,
i=1
y2Y
[
which implies that there is a finite subcover
Ws , where S ⇢ Y is finite.
s2S
!
[
[
1
And so p
Ws =
p 1 (Ws ) covers X, since p is surjective, and
s2S !
s2S
ns
[
[[
X⇢
p 1 (Ws ) ⇢
O↵k , which is a finite subcover.
s2S
s2S k=1
1
[105]
1-
Topology
Exercise 114: Munkres pg 186 #3
Let X be a locally compact space. If f : X ! Y is continuous, does it follow that f ( X ) is
locally compact? What if f is both continuous and open? Justify your answer.
To answer the first question, no. Note that since f is continuous then it maps a compact
subspace to a compact subspace. So if X was compact, then f ( X ) would be compact,
which would make it automatically locally compact. So we need to construct X in a way
such that it is locally compact, but not compact, and we need f to map X onto a space Y
that is not locally compact. Let X = Q under the discrete topology. Since it is not finite,
then it is not compact. Further, for any x 2 X, { x } serves as both an open neighorhood of
x and a compact subspace in X (since all subsets of X are both open and closed under the
discrete topology), so X is locally compact. Let Y = Q under the standard topology. Finally, let f : X ! Y by f ( x ) = x, the identity function. Then f is onto, but f is not an open
map. Also, f is continuous, since the preimage of any open subset of Y is open in X. However, by the first exercise (Q is not locally compact) we know that Y is not locally compact.
On the otherhand, if X is locally compact and f is a continuous open map then f ( X )
is locally compact. Let y 2 f ( X ) be arbitrary. Then f 1 (y) = x 2 X, and since X is
locally compact there exists an open neighborhood U of X that contains x and a compact
subspace C of X such that x 2 U ⇢ C. Since f is continuous open map, then f (C ) is a
compact subspace in Y, and f (U ) is open in Y. It follows that y 2 f (U ) ⇢ f (C ). Since y
was arbitrary, we have that f ( X ) is locally compact.
⌅
1
[120]
TOPOLOGY SPRING 2010, PROBLEM 2
[i] Suppose A, B are nonempty, disjoint, closed subsets of a metric space X. Show
that the function defined by f : X ! [0, 1] defined by
dist(x, A)
f (x) =
dist(x, A) + dist(x, B)
is continuous with f (x) = 0 for all x 2 A, f (x) = 1 for all x 2 B, and 0 < f (x) < 1
for all x 2 X\(A [ B). [ii] Show that if X is a connected metric space with at
least two distinct points, then X is uncountable.
PROOF [i].
Recall dist(x, M ) = inf{d(x, m) | m 2 M }.
Given A nonempty, let a 2 A. Then dist(a, A) = 0. By assumption, A \ B = ,
so a 2 B c . Since B c is an open set, there exists an ✏ > 0 such that the open ball
dist(a,A)
0
B(a, ✏) ✓ B c . Then dist(a, B) ✏ > 0. Then f (a) = dist(a,A)+dist(a,B)
= 0+✏
= 0.
This is true for each a 2 A. Similarly, let b 2 B. Then dist(b, B) = 0. It follows
dist(b,A)
that f (b) = dist(b,A)+dist(b,B)
= dist(b,A)
= 1. Finally, for all x 2 X\(A [ B), x 2 Ac
dist(b,A)
and x 2 B c , so dist(x, A) = ↵ > 0 and dist(x, B) = > 0. Then 0 < f (x) <
1.
Now we wish to show f is a continuous function. It suffices to show that dist
is a continuous function. Let M be a nonempty subset of a metric space X.
Let ✏ > 0. Let x, y 2 X. Given ✏ > 0, there exists an m 2 M such that
d(x, m)  dist(x, M ) + ✏. By the Triangle Inequality, d(y, m)  d(y, x) + d(x, m).
Combining the inequalities gives dist(y, M )  d(y, m)  d(y, x) + d(x, m) 
d(y, x) + dist(x, M ) + ✏. We conclude that dist(y, M ) dist(x, M )  d(y, x) + ✏
for any ✏ > 0. A similar argument shows dist(x, M ) dist(y, M )  dist(x, y) + ✏
for any ✏ > 0. Then we have shown that | dist(x, M ) dist(y, M ) |< d(x, y). This
proves Lipschitz continuity, with Lipschitz constant 1. Therefore dist is a Lipschitz continuous function, which is continuous. f (x) is the composition of such
functions, so f is continuous.
[ii]. Let x, y 2 X with x 6= y. Let A = {a} and B = {b}. Then A, B are closed,
dist(x,A)
disjoint sets. By part [i], the function f (x) = dist(x,A)+dist(x,B)
is a continuous
function with f (x) = 0, f (y) = 1. Then 0, 1 2 f (X), X is connected, and f (X) ✓
[0, 1]. Since f (X) is the continuous image of a connected set, f (X) is connected.
Then f (X) = [0, 1], which is uncountable. f maps X onto uncountable set, so X
is uncountable.
1
[121]
Exercise 122
Definition. A subset, A of a metric space, (X, d) is precompact if its closure, cl(A) is compact.
Show that if A is precompact, then for every ✏ > 0, there exists a finite covering of A by open balls of radius
✏ with centers in A.
Proof. Suppose A is as defined above and define B = cl(A) \ (A). For x 2 cl(A) and r > 0 define
B(x, r) = {y 2 cl(A) : d(x, y) < r}.
Note that cl(A) = A [ B and for any ✏ > 0 we have
cl(A) ✓
[
x2cl(A)
✏
B(x, )
2
is an open cover of cl(A) . Since A is precompact, cl(A) is compact so there exists {x1 , . . . , xn } such that
n
[
✏
cl(A) ✓
B(xj , ).
2
j=1
WLOG we can assume that {x1 , . . . , xk } ⇢ B and {x
T k+1 , . . . , xn } ⇢ A. Now, B ✓ @(A) so for each 1  j  k
there exists a corresponding yj such that yj 2 A B(xj , 2✏ ). Moreover, if x 2 B(xj , 2✏ ) we have from the
triangle inequality
d(x, yn )  d(x, xn ) + d(xn , yn ) < ✏,
✏
which shows that B(yj , ✏) B(xj , 2 ). By taking yj = xj for k < j  n we obtain
A✓
n
[
✏
B(xj , ) ✓
B(yj , ✏)
2
j=1
j=1
n
[
where the union on the right is a finite covering of A by open balls of radius ✏ and each yj 2 A as required.
⇤
1
Aaron Hutchinson
General Topology
Problem 123
123. Generalize Exercise 4 on the Spring 2013 Analysis Qualifying Exam: prove
Proposition: Suppose { Xk }nk=1 are compact spaces and f k : Xk → R m are continuous functions. Then
{∑ f k ( xk )} ⊂ R m is compact.
Does the same hold for connectedness? Prove your answer.
We first prove a lemma: if A, B ⊆ R m are compact sets, then A + B ⊆ R m is a compact set. By Theorem 26.7
in Munkres, A × B is a compact set. Define f : A × B → R m by f ( a, b) = a + b. The function f is actually
the sum of the two projection functions π1 , π2 : A × B → R m defined by π1 ( a, b) = a and π2 ( a, b) = b; that
is, f = π1 + π2 . Since projection functions are continuous, and the sum of real valued continuous functions is
continuous, we have that f is continuous. Furthermore,
A + B = { a + b| a ∈ A, b ∈ B} = { f ( a, b)| a ∈ A, b ∈ B} = f ( A × B).
Invoking Theorem 26.5 in Munkres, we now have that A + B is compact and so the lemma is proved.
We now prove the proposition by induction on n. Suppose n = 1; then there is only one set, and so
+1
f 1 ( X1 ) ⊂ R m is compact (by Theorem 26.5 again). Suppose now that the proposition is true for n. Let { Xk }nk=
1
m
be compact spaces, and f k : Xk → R be continuous functions for k = 1, . . . , n + 1. By the induction hypothesis, we have that f 1 ( X1 ) + · · · + f n ( Xn ) is a compact set. In the hypothesis of the lemma proved above, let
n +1
A = f 1 ( X1 ) + · · · + f n ( Xn ) and B = f n+1 ( Xn+1 ). Then A and B are compact sets, and so A + B = ∑ f k ( Xk )
is a compact set. This ends the induction and proves the proposition.
k =1
The same proposition can be proved for connectedness in place of compactness, and the proof is essentially
the same. First prove the lemma: if A, B ⊆ R m are connected sets, then A + B ⊆ R m is a connected set. By
Theorem 23.6 of Munkres, A × B is a connected set. Define the same function f as above. In the same way as
before, f is continuous and f ( A × B) = A + B. Since the image of a connected space under a continuous map
is connected (Theorem 23.5), we have that A + B is connected.
Now we prove the proposition by induction on n. If n = 1, then there is only one set and so f 1 ( X1 ) is
+1
m
connected. Suppose that the proposition holds for n, and let { Xk }nk=
1 be connected spaces, with f k : Xk → R
continuous functions for k = 1, . . . , n + 1. The induction hypothesis gives that f 1 ( X1 ) + · · · + f n ( Xn ) is a
connected set. Now using the second lemma, let A = f 1 ( X1 ) + · · · + f n ( Xn ) and B = Xn+1 . Then A and B are
n +1
connected sets, and so A + B = ∑ f k ( Xk ) is a connected set. This ends the induction, and so the proposition
is proved once again.
k =1
!
1
[125]
Give an example of a plane set that does not have the fixed point property.
R2 doesn’t have the fixed point property since the continuous function f : R2 → R2 defined f (x, y) = (x+1, y)
has no fixed points.
1
[126]
3. If a tetrahedron T , whose vertices have been labeled A, B, C, and D has been divided into subtetrahedra
with verticies labeled so that only three labels appear on each face of the original tetrahedron then T contains
at least one subtetrahedron with all four labels.
Proof:
Let b be the number of complete tetrahedra in a given Sperner labeling. To prove that b is odd, we count the
number of triangles labeled ABC inside and on the faces of the tetrahedron in the following way: let a be
the number of tetrahedra whose labels read ABCA, ABCB or ABCC. Now the tetrahedra of these types
have two faces labeled ABC, while the complete tetrahedra have one face labeled ABC. Other types of
tetrahedra have no faces labeled ABC, so the total number of these faces, counted tetrahedra by tetrahedra,
is 2a + b. But wait! In this total, the faces in the interiorT are counted twice since they belong to two
tetrahedra. Letting c be the number of faces labeled ABC inside the tetrahedra, we have counted 2c + d
faces, where d is the number of faces labeled ABC on the outside of T . Thus 2a + b = 2c + d. According to
the 2-dimensional result d is odd, and therefore b is odd as well. !
1