Download Math 535 - General Topology Fall 2012 Homework 2 Solutions

Math 535 - General Topology
Fall 2012
Homework 2 Solutions
Problem 1. (Brown Exercise 2.4.5) Consider X = [0, 2] \ {1} as a subspace of the real line
R. Show that the subset [0, 1) ⊂ X is both open and closed in X.
Solution. [0, 1) is open in X because we can write
[0, 1) = (−8, 1) ∩ X
and (−8, 1) is open in R.
On the other hand, [0, 1) is closed in X because we can write
[0, 1) = [0, 1] ∩ X
and [0, 1] is closed in R.
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Problem 2. (Bredon Exercise I.3.8) Let X be a topological space that can be written as a
union X = A ∪ B where A and B are closed subsets of X. Let f : X → Y be a function,
where Y is any topological space. Assume that the restrictions of f to A and to B are both
continuous. Show that f is continuous.
Solution.
Lemma. Let A ⊆ X be a closed subset. If C ⊆ A is closed in A, then C is also closed in X.
e ∩ A for some closed subset C
e ⊆ X.
Proof. Since C is closed in A, it can be written as C = C
Therefore C is an intersection of closed subsets of X, and thus is closed in X.
Let C ⊂ Y be a closed subset. Its preimage under f is the union
f −1 (C) = f −1 (C) ∩ A ∪ f −1 (C) ∩ B
= (f |A )−1 (C) ∪ (f |B )−1 (C).
Since the restriction f |A : A → Y is continuous, (f |A )−1 (C) is closed in A, and thus closed in
X by the lemma. Likewise, (f |B )−1 (C) is closed in X. Therefore their union
f −1 (C) = (f |A )−1 (C) ∪ (f |B )−1 (C).
is closed in X, so that f in continuous.
Remark. The same proof shows that the statement still holds if A and B are both open in X.
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Problem 3. A map between topological spaces f : X → Y is called an open map if for every
open subset U ⊆ X, its image f (U ) ⊆ Y is open in Y .
a. (Munkres Exercise 2.16.4) Let X and Y be topological spaces. Show that the projection
maps pX : X × Y → X and pY : X × Y → Y are open maps.
Solution.
Lemma. A map f : X → Y is open if and only if f (B) ⊆ Y is open in Y for every B ∈ B
belonging to some basis B of the topology on X.
Proof. (⇒) Each member B ∈ B is open in X.
(⇐) Let U ⊆ X be open in X. Then U is a union U =
Its image under f is
!
[
f (U ) = f
Bα
S
α
Bα of basic open subsets Bα ∈ B.
α
=
[
f (Bα )
α
where each f (Bα ) is open in Y by assumption. Thus f (U ) is a union of open subsets and hence
open.
Take an “open box” U × V ⊆ X × Y , where U ⊆ X is open and V ⊆ Y is open. Its projection
onto the first factor is
pX (U × V ) = U ⊆ X
which is open in X. Since open boxes form a basis of the topology on X × Y , the lemma
guarantees that pX is an open map, and likewise for pY .
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b. Find an example of metric spaces X and Y , and a closed subset C ⊆ X × Y such that the
projection pX (C) ⊆ X is not closed in X.
In other words, the projection maps are (usually) not closed maps.
Solution. Take X = Y = R and consider the hyperbola in R × R
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C = {(x, ) | x 6= 0} = {(x, y) ∈ R × R | xy = 1}.
x
Its projection onto the first factor is
pX (C) = R \ {0}
which is not closed in R.
To show that C is closed in R×R, note that the function f : R×R → R defined by f (x, y) = xy
is continuous, and C is the preimage C = f −1 ({1}). Since the singleton {1} is closed in R, C
is closed in R × R.
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Problem 4. (Munkres Exercise 2.19.7) Consider the set of sequences of real numbers
Y
RN = {(x1 , x2 , . . .) | xn ∈ R for all n ∈ N} ∼
R
=
n∈N
and consider the subset of sequences that are “eventually zero”
R∞ := {x ∈ RN | xn 6= 0 for at most finitely many n}.
a. In the box topology on RN , is R∞ a closed subset?
Solution. Yes, R∞ is closed in the box topology.
Let x ∈ RN \ R∞ , which means that the sequence x has infinitely many non-zero entries xn 6= 0.
For all those indices n, pick an open neighborhood UQ
n of xn ∈ R which does not contain 0. For
other values of n, take Un = R. Then the open box n Un is an open neighborhood of x which
does not intersect R∞ .
Q
Indeed, for any y ∈ n Un and every index n such that xn 6= 0, we have yn ∈ Un so that yn 6= 0
by construction. Because there are infinitely many such indices, we conclude y ∈
/ R∞ .
b. In the product topology on RN , is R∞ a closed subset?
Solution. No, R∞ is not closed in the product topology.
Q
Let x ∈ RN \ R∞ and consider any open neighborhood U = n Un of x which is a “large box”,
i.e. Un ⊆ R is open for all n and Un = R except for finitely many n. In particular, there is a
number N such that Un = R for all n ≥ N . Consider a sequence y with yn = 0 for all n ≥ N
and yn ∈ Un for 1 ≤ n < N . Then we have y ∈ U ∩ R∞ .
Because “large boxes” form a basis of the product topology, every open neighborhood of x
intersects R∞ . Therefore R∞ is not closed.
Remark. In fact, the argument shows that x is not an interior point of RN \ R∞ , so that the
interior of RN \ R∞ is empty. Equivalently, the closure of R∞ is all of RN , i.e. R∞ is dense in
RN .
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Problem 5. Let X be a topological space, S a set, and f : X → S a function. Consider the
collection of subsets of S
T := {U ⊆ S | f −1 (U ) is open in X}.
a. Show that T is a topology on S.
Solution.
1. The preimage f −1 (S) = X is open in X, so that the entire set S is in T . Likewise,
f −1 (∅) = ∅ is open in X, so that the empty set ∅ is in T .
2. Let Uα be a family of members of T . Then we have
!
[
[
f −1
Uα =
f −1 (Uα )
α
α
S
where each f −1 S
(Uα ) is open in X by assumption. Thus f −1 ( α Uα ) is also open in X, so
that the union α Uα is in T .
3. Let U and U 0 be members of T . Then we have
f −1 (U ∩ U 0 ) = f −1 (U ) ∩ f −1 (U 0 )
where f −1 (U ) and f −1 (U 0 ) are open in X by assumption. Thus f −1 (U ∩ U 0 ) is also open
in X, so that the finite intersection U ∩ U 0 is in T .
b. Show that T is the largest topology on S making f continuous.
Solution. Note that T makes f continuous by construction: for all U ∈ T , the preimage
f −1 (U ) ⊆ X is open in X.
Let T 0 be a topology on S making f continuous. Then for every U ∈ T 0 , the preimage f −1 (U )
is open in X, which means U ∈ T . This proves T 0 ≤ T .
c. Let Y be a topological space. Show that a map g : S → Y is continuous if and only if the
composite g ◦ f : X → Y is continuous.
Solution. (⇒) The maps f and g are continuous, hence so is their composite g ◦ f .
(⇐) Assume g ◦ f is continuous; we want to show that g is continuous. Let U ⊆ Y be open
and take its preimage g −1 (U ) ⊆ S. To check that this subset is open, consider its preimage
f −1 g −1 (U ) = (g ◦ f )−1 (U ) ⊆ X
which is open in X since g ◦ f is continuous. By definition of T , g −1 (U ) is indeed open in S.
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d. Show that T is the smallest topology on S with the property that a map g : S → Y is
continuous whenever g ◦ f is continuous.
Solution. Let T 0 be a topology on S with said property. We know that f : X → (S, T ) is
continuous, but it can be written as the composite
f
id
X→
− (S, T 0 ) −
→ (S, T ).
By the property of T 0 , the composite id ◦ f being continuous guarantees that the identity
id : (S, T 0 ) → (S, T ) is continuous, i.e. T ≤ T 0 .
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Problem 6. Consider the subset X = {0} ∪ { n1 | n ∈ N} ⊂ R viewed as a subspace of the real
line R. As a set, X is the disjoint union of the singletons {0} `
and { n1 } for all n ∈ N. However,
show that X does not have the coproduct topology on {0} q n∈N { n1 }.
Solution. In the coproduct topology on {0} q
topology), the summand {0} is open.
`
1
n∈N { n }
(which happens to be the discrete
However, in the subspace topology on X, the singleton {0} is not open. Indeed, any open ball
Br (0) around 0 will contain other points n1 ∈ Br (0), for all n such that n1 < r.
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