University of Toronto
Faculty of Arts and Sciences
MAT327H1 - Y: Introduction to Topology
Summer 2020
Homework 2 Solutions
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Problems to be submitted
1. (5 points) Consider the function f : R → R given by f (x) = x2 . Is this function open? Is it closed?
Solution: It is not open since f (−1, 1) = [0, 1) which is not open in R.
We now prove it is closed: Let C ⊂ R is a closed and define:
C+ = C ∩ [0, ∞),
C− = C ∩ (−∞, 0].
Both of them are closed and C = C+ ∪ C− . Since f (C) = f (C+ ) ∪ f (C− ), it is enough to prove f (C+ )
and f (C− ) are closed.
In other words, we can assume without loss of generality C consists entirely of nonnegative real numbers
or of nonnpositive real numbers. Furthermore, if C consists entirely of nonpositive real numbers then
−C consists of nonnegative real numbers, and since squaring dissappears the sign, we have f (C) =
f (−C). Since C is closed if and only if −C is closed, we deduce that we can assume without loss of
generality C consists of nonnegative real numbers.
Under this assumption, consider x1 , x2 , ... be memebers of f (C) with
lim xn = x.
n→∞
√
Hence, xn = f (yn ) for some yn ∈ C, that is yn = xn , where we have taken the nonnegative square
root (since C consists of nonnegative real numbers). Then we have, since the square root is continuous
in [0, ∞) that
√
√
lim xn = x.
n→∞
√
√
(Notice that x is√nonnegative, as each xn is). But since xn ∈ C, and C is closed, then x ∈ C. We
conclude x = f ( x) is also in f (C). That is, f (C) is closed.
2. (5 points) Let X be a first countable space. Prove that every point x ∈ X has a countable basis {Ui }
such that U1 ⊇ U2 ⊇ U3 ⊇ ....
Solution: Let x be any point of X. By first countability we can find a countable local basis at x,
{Bn }n . Define:
Ui = B1 ∩ ... ∩ Bi .
We claim {Ui }i is a local basis at x as well. Since each one of its members is obtained intersecting
a finite number of open sets, it consists of open sets. Furthermore, everyone includes x as one of its
points, since each Bi does.
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Given x ∈ V any open set around x, we know there exists Bj ⊆ V by definition of local basis. Then
we have Uj ⊆ Bj ⊆ V . This finishes the proof that it is a local basis at x.
Finally, it is clear that by definition Un+1 = Un ∩ Bn+1 ⊂ Un , proving it is nested, as desired.
3. (10 points) Let X be a Hausdorff space, A ⊆ X, and A0 the set of limit points of A. Prove that A0 is
closed in X.
Solution: Let us recall the definition of limit point of A: p ∈ X is a limit point of A if and only if
every neighborhood of p has a point of A different from p.
Suppose p is a point not in A0 . By negating the definiton of limit point this means there exists an open
set V , p ∈ V , such that the only point of A in V is possible p. We want to prove that p ∈ Ext(A0 ).
If V ∩ A0 = ∅, then we are done, since this neighborhood belongs to Ext(A). If q ∈ V ∩ A0 then, firstable
q 6= p, as we have supposed p is not a limit point of A. Because of this, as p is the only possible point
of A in V , we must have q ∈
/ A.
As p and q are different, and X is Hausdorff, we can find disjoint open sets B, C with p ∈ B and q ∈ C.
Since C ∩ V is an open set around q, and q ∈ A0 , then there exists a point of A different from q in
C. This is impossible since C ⊂ V , and in V the only point of A is p which is not in C. This is a
contradiction and so V ∩ A0 = ∅.
We have proved (A0 )c is open, hence A0 is closed, as desired.
4. (10 points) Let X and Y be topological spaces, f : X → Y a function between them and x ∈ X any
point. We say f is continuous at x if for every open set V around f (x) in Y , there exists an open set
U in X around x, such that f (U ) ⊆ V .
Prove that f : X → Y is continuous if and only if it is continuous at every point of X.
Solution: We do the two implications separately:
Suppose f is continuous: Let f : X → Y be continuous and x ∈ X be any point. Consider V ⊆ Y
a neighborhood of f (x), then by continuity f −1 (V ) is an open set around x.
Of course we have f (f −1 (V )) ⊆ V (notice that since f might not be surjective, this might not
be an equality). That is, f is continuous at x.
Suppose f is continuous at any of its points: Let V ⊆ Y be an open set and consider x ∈
f −1 (V ). Since f (x) ∈ V , we can find by local continuity an open set Ux around x with f (Ux ) ⊆ V .
Notice then that x ∈ Ux ⊆ f −1 (V ). Hence, when we vary x ∈ f −1 (V ) we can write:
[
f −1 (V ) =
Ux ,
x∈f −1 (V )
which is a union of open sets, hence open. This proves f is continuous.
5. (15 points) Let X be a topological space and B be a basis for it. We are given for each B ∈ B a function
fB : B → Y , where Y is some fixed topological space. Suppose that
• For each open set V ⊆ Y and B ∈ B, we have fB−1 (V ) ⊆ B is an open set of X.
• For any two B1 , B2 ∈ B with B1 ∩ B2 6= ∅ we have fB1 (x) = fB2 (x) for all x ∈ B1 ∩ B2 .
Prove that the function f : X → Y given by f (x) = fB (x), where x ∈ B ∈ B, is a well defined function
at all points of X which is continuous.
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Solution: We first explain why it is well defined. Suppose x ∈ B1 ∩ B2 , where B1 and B2 are basic
sets. Then, by the second property given above, we have, fB1 (x) = fB2 (x). Hence, by definition of f
f (x) = fB1 (x) = fB2 (x),
so that it doesn’t matter which basic set we use to define its value.
To prove it is continuous pick any open set V ⊆ Y . We claim
[
f −1 (V ) =
fB−1 (V ).
B∈B
This would indeed imply continuity as the righ hand side is an arbitrary union of open sets (by the
first assumption we have on the functions fB ) and hence open. Let us prove the equality by double
inclusion: if x ∈ f −1 (V ) then, on the one hand f (x) ∈ V and on the other hand x ∈ B for some open
basic set. Hence f (x) = fB (x), implies x ∈ fB−1 (V ). This proves the left hand side is inside the right
hand side.Viceversa, if x ∈ fB−1 (V ) for some basic set B, then of course f (x) = fB (x) ∈ V , and so
x ∈ f −1 (V ), proving the other inclusion.
6. (15 points) Let X be a second countable topological space. Prove that every open cover of X has a
countable subcover.
Solution: Since X is second countable we know there exists a countable basis B = {Bn }n . Consider
an arbitrary cover {Uα }, α ∈ Λ for some indexing set Λ.
Define B 0 as
B 0 = {Bn ∈ B | there exists α ∈ Λ with Bn ⊆ Uα }.
We claim B 0 is an open cover of X. If x ∈ X by definition of cover there is α ∈ Λ with x ∈ Uα .
Furthermore, by definition of basis, there exists n such that x ∈ Bn ⊆ Uα . Finally, by definition,
Bn ∈ B 0 . This proves B 0 is a cover.
Now for each Bi ∈ B 0 pick αi ∈ Λ such that Bi ⊆ Uαi (it must exist by definition of B 0 ). Notice that
the Uαi are at most a countable subcollection of {Uα }α , since B 0 is countable. Furthermore
[
[
X=
Bi ⊆
Uαi ,
Bi ∈B0
we conclude Uαi is actually a countable subcover. This concludes the proof.
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