Download Homework7 - UCSB Math Department

SELECTED SOLUTIONS TO HOMEWORK 7
2. Let O be an open subset of a topological space X . Prove that a subset A of O is relatively open in
O if and only if it is an open subset of X .
Solution. By denition of subspace topology, we have
A open in O ⇐⇒ A = O ∩ B
for some open subset B in X.
Now, if A is relatively open in O, then A = O ∩ B for some B open in X . Since O and B are both open
in X , their interesection A is also open in X . Conversely, if A is open in X , then O ∩ A is open in O by
denition. But O ∩ A = A because A ⊂ O, so A is open in O.
4. Prove that a subspace of a Hausdor space is a Hausdor space.
Solution. Let X be a topological space which is Hausdor and Y ⊂ X a subspace. Let a, b ∈ Y with
a 6= b. Since a, b ∈ X and X is Hausdor, there exist neighborhoods A of a and B of b in X such that
A ∩ B = ∅. Now, by Theorem 6.3
A0 := A ∩ Y
and
B 0 := B ∩ Y
are neighborhoods of a and b in Y , respectively. Since A0 ∩ B 0 ⊂ A ∩ B = ∅, we see that A0 ∩ B 0 = ∅ and
so Y is Hausdor, as claimed.
2. Let X = α∈I Xα be given the product topology. Prove that a function f : Y → X is continuous if
and only if fα = pα f is continuous for each α ∈ I .
Solution. First assume that f is continuous. Since pα is continuous (from p. 99 on the textbook), by
Theorem 5.6 we know that fα = pα f is continuous for each α ∈ I also.
Conversely, assume that fα is continuous for each α ∈ I . By Theorem 5.3, f will be continuous if f −1 (O)
is open in Y whenever O is open in X . By denition 7.6, every open O in X is a union of sets of the form
Q
−1
B = p−1
α1 (Oα1 ) ∩ · · · ∩ pαk (Oαk )
(∗),
where Oαi is open in Xαi . Since f −1 preserves union, and an arbitrary union of open sets is again open, it
suces to show that f −1 (B) is open for any set B of the form (∗). Now, since f −1 preserves intersection
−1 −1
f −1 (B) = (f −1 p−1
pαk )(Oαk )
α1 )(Oα1 ) ∩ · · · ∩ (f
= fα−1
(Oα1 ) ∩ · · · ∩ fα−1
(Oαk ),
1
k
where each fα−1i (Oαi ) is open because fα is continuous for each α ∈ I by hypothesis. Since nite intersection
of open sets is again open, we see that f −1 (B) is indeed open, whence f is continuous.
6. Prove that the family of open intervals with rational and points is a basis for the topology of R.
1
SELECTED SOLUTIONS TO HOMEWORK 7
2
Let O be an open set in R. By Denition 7.3, we need to show that O is a union of open
intervals (a, b) with a, b ∈ Q. Now, since O is open, for each x ∈ O there exists δx > 0 such that
Solution.
(x − δx , x + δx ) ⊂ O.
Since Q is dense in R, there exist ax , bx ∈ Q such that
and
x − δx < ax < x
x < bx < x + δx .
Then x ∈ (ax , bx ) ⊂ (x − δx , x + δx ) ⊂ O and so
[
(ax , bx ) ⊂ O.
x∈O
It is clear that y ∈
S
x∈O (ax , bx )
for each y ∈ O so in fact
O=
[
x∈O
Thus, indeed {(a, b) | a, b ∈ Q} is a basis for R.
(ax , bx ).