Homework 4 Solutions
Homework 4 Solutions
Problem 13.4.
Proposition. If {Tα } is a family of topologies on X, then
T
Tα is a topology on X.
Proof. Since ∅ and X are in each Tα , they must also be elements of
then C ⊂ Tα for all α; it follows that
is a finite subcollection of
T
S
Tα , then
C ∈ Tα for all α, and hence
T
T
Tα . Next, if C ⊂
S
C∈
C ∈ Tα for all α, and therefore
T
T
T
Tα ,
Tα . Finally, if C
C∈
T
Tα .
It is not true in general that the union of two topologies is a topology. For example,
the union T1 ∪ T2 = {∅, X, {a}, {a, b}, {b, c}} of the two topologies from part (c) is not a
topology, since {a, b}, {b, c} ∈ T1 ∪ T2 but {a, b} ∩ {b, c} = {b} ∈
/ T1 ∪ T2 .
Proposition. Let {Tα } be a family of topologies on X. Then there exists a unique smallest
topology on X containing all the collections Tα , and a unique largest topology contained in
all Tα .
Proof. The unique largest topology contained in all the Tα is simply the intersection
For the other statement, observe that the family of all topologies on X that contain
T
S
Tα .
Tα is
nonempty, since it includes the discrete topology on X. Then the intersection of this family
is the unique smallest topology that contains every Tα .
For part (c), the smallest topology containing both T1 and T2 is
{∅, X, {a}, {b}, {a, b}, {b, c}},
and the largest topology contained in T1 and T2 is the intersection
T1 ∩ T2 = {∅, X, {a}}.
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Problem 13.8.
Proposition. The countable collection B = {(a, b) | a < b, a and b rational} is a basis for
the standard topology on R.
Proof. Let U ⊂ R be open, and let x ∈ U . Since the open intervals are a basis for the
topology on R, there exists an open interval (a, b) such that x ∈ (a, b) ⊂ U . Then a < x < b,
so there exist rational numbers q and r such that a < q < x < r < b. Then (q, r) ∈ B and
x ∈ (q, r) ⊂ U , so B is a basis for the standard topology on R by Lemma 13.2.
Proposition. The basis C = {[a, b) | a < b, a and b rational} generates a topology on R
different from the lower limit topology.
√
√
Proof. The set [ 2, 2) is open in the lower-limit topology. However, since 2 is irrational,
there does not exist an [a, b) ∈ C for which
√
√
2 ∈ [a, b) ⊂ [ 2, 2).
Problem 16.4.
A map f : X → Y is said to be an open map if for every open set U of X, the set f (U ) is
open in Y .
Proposition. If X and Y are topological spaces, then the projections π1 : X × Y → X and
π2 : X × Y → Y are open maps.
Proof. Let U be an open set in X × Y . Then U is a union of basic open sets in X × Y ,
say U =
π1 (U ) =
S
S
α
Vα × Wα , where each Vα is open in X and each Wα is open on Y . Then
α π1 (Vα × Wα ) =
S
α
Vα , which is open in X, and similarly π2 (U ) is open in Y .
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Problem 16.9.
Proposition. The dictionary order topology on the set R × R is the same as the product
topology Rd × R, where Rd denotes R in the discrete topology.
Proof. Because singleton sets are a basis for the topology on Rd , the collection
B = {{a} × (b, c) | a, b, c ∈ R and b < c}
is a basis for the product topology on Rd × R. Each of these sets is open in the dictionary
order topology, since {a}×(b, c) = (a×b, a×c), and thus the topology on Rd ×R is contained
in the dictionary order topology.
For the other direction, observe that every ray in the dictionary order topology is open
in Rd × R. In particular,
(a × b, ∞) =
{a} × (b, ∞) ∪ (a, ∞) × R
and
(−∞, a × b) =
(∞, a) × R ∪ {a} × (−∞, b) .
Since the rays are a subbasis for the dictionary order topology, it follows that the dictionary
order topology is contained in the product topology on Rd × R.
The dictionary order topology on R × R contains the standard topology. In particular,
every basic open rectangle (a, b) × (c, d) in the standard topology is also open in Rd × R,
since (a, b) is open in Rd and (c, d) is open in R. This containment of topologies is strict,
e.g. {0} × (0, 1) is open in the dictionary order topology but not in the standard topology.
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