Topology
Homework Assignment 2
Solutions
1. Show that the composition of continuous functions is continuous.
Suppose that X, Y, and Z are topological spaces, and f : X → Y and g : Y → Z
are continuous functions. We need to show that if W is an open subset of Z, then
(g ◦ f )−1 (W ) is an open subset of X.
The key observation here is that (g ◦ f )−1 (W ) = f −1 [g −1 (W )]. This is essentially the
definition, but it can be shown directly using the definition of an inverse. First suppose
that x ∈ (g ◦ f )−1 (W ). Then (g ◦ f )(x) = g[f (x)] ∈ W. If f (x) = y and g(y) = z, then
y ∈ g −1 (W ) and x ∈ f −1 (y) ⊆ f −1 [g −1 (W )]. Conversely, if x ∈ f −1 [g −1 (W )], then,
g[f (x)] = (g ◦ f )(x) ∈ W. So x ∈ (g ◦ f )−1 (W ).
Thus if W is an open subset of Z, then g −1 (W ) is an open subset of Y, and hence
f −1 [g −1 (W )] is an open subset of X. Since f −1 [g −1 (W )] = (g ◦ f )−1 (W ), (g ◦ f )−1 (W )
is an open subset of X, and g ◦ f is continuous.
2. Suppose X and Y are topological spaces and f : X → Y is a continuous function.
Show that if X is pathwise-connected, then so is f (X).
Suppose that p, q ∈ f (X). We want to find a continuous function g : [0, 1] → f (X)
such that g(0) = p and g(1) = q.
Say f (x) = p and f (y) = q. Then since X is path-connected, there is a continuous
function h : [0, 1] → X such that h(0) = x and h(1) = y. Thus (f ◦ h)(0) = p and
(f ◦ h)(1) = q. But by problem 1, f ◦ h is continuous. So let g : [0, 1] → f (X), by
g(t) = (f ◦ h)(t).
3. Suppose X and Y are topological spaces. A function f : X → Y (possibly not
continuous) is closed if for each closed set C ⊆ X, f (C) is a closed subset of Y.
Suppose that X and Y are subsets of Euclidean spaces.
(a) Find an example of a continuous function f : X → Y that is open but not closed.
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Let X = R2 and Y = R. Also, let f : R2 → R be the projection map f (x, y) = x.
Then by problem 6, f is continuous. It’s not hard to show that f is also open.
However, f is not closed. For consider
S = {(x, y) ∈ R2 : y = 1/x and x > 0}.
Then S is closed because its complement is open. For if (x0 , y0 ) ∈ R2 −S, then the
distance from (x0 , y0 ) is positive. Hence, there is an open ball Br (x0 , y0 ) contained
in R2 − S. However, f (S) = (0, ∞), which is not closed in R.
(b) Find an example of a continuous function f : X → Y that is closed but not open.
Let X = Y = R, and let f : X → Y be a constant map. Say f (x) = 0, for
all x ∈ R. Then if A is any nonempty subset of R, f (A) = {0}, which is a
closed subset of R. In particular, if A is any closed, nonempty subset of R, then
f (A) = {0} is closed. On the other hand, if A is any open nonempty subset of R,
then f (A) = {0}, which is not open. So f is closed but not open.
4. Show that an open ball in Rn is, in fact, an open set in the usual topology. That
is, if r > 0 and x0 ∈ Rn , then for each x ∈ Br (x0 ), there exists s > 0 such that
Bs (x) ⊆ Br (x0 ).
Let x ∈ Br (x0 ). Consider
s = r − ||x − x0 ||.
Since ||x − x0 || < r, s > 0. Furthermore, if y ∈ Bs (x), then
||y − x0 || ≤
<
=
=
||y − x|| + ||x − x0 ||
s + ||x − x0 ||
r − ||x − x0 || + ||x − x0 ||
r
So y ∈ Br (x0 ), Bs (x) ⊆ Br (x0 ), and Br (x0 ) is open.
5. Prove that all finite sets are compact.
Let X be a topological space and F = {x1 , x2 , . . . , xk } a finite subset of X. If A is an
index set, and
U = {Uα : α ∈ A}
is an open cover of F, we need to find a finite subcover
{Uα1 , Uα2 , . . . , Uαn } ⊆ U.
But since U is a cover of F, for i = 1, . . . , k, there exists an αi ∈ A, such that xi ∈ Uαi .
Hence,
F ⊆ ∪ki=1 Uαi ,
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and
{Uα1 , . . . , Uαk }
is a finite subcover.
6. If X and Y are topological spaces, we can make the product
X × Y = {(x, y) : x ∈ X and y ∈ Y },
into a topological space by defining a subset U ⊆ X ×Y to be open if for each (x, y) ∈ U,
there exist open sets V ⊆ X and W ⊆ Y such that x ∈ V, y ∈ W, and V × W ⊆ U.
(a) Show that with this definition X × Y is a topological space.
1a. We need to see that X × Y is open. So let (x, y) ∈ X × Y. Then x ∈ X
and y ∈ Y, and since X is an open subset of X and Y is an open subset of Y,
X × Y is an open subset of X × Y.
1b. We need to see that ∅ is open. But if (x, y) ∈ ∅, then x ∈ ∅ ⊆ X, and
y ∈ ∅ ⊆ Y. So ∅ × ∅ ⊆ ∅ ⊆ X × Y, and ∅ is open.
2. Suppose A is an index set and Uα is an open subset of X × Y, for each
α ∈ A. We need to see that ∪α∈A Uα is open. If
(x, y) ∈ ∪α∈A Uα ,
then for some α0 ∈ A, (x, y) ∈ Uα0 . Since Uα0 is open, there exist open subsets
V ⊆ X and W ⊆ Y such that x ∈ V, y ∈ W, and V × W ⊆ Uα0 . Hence
V × W ⊆ ∪α∈A Uα ,
and ∪α∈A Uα is open.
3. Suppose that {U1 , U2 , . . . , Uk } is a finite collection of open subsets of X ×Y.
We need to show that
∩ki=1 Ui
is also open. So suppose (x, y) ∈ ∩Ui . Since each Ui is open, for i = 1, . . . k,
there exist open subsets Vi ⊆ X, and Wi ⊆ Y such that x ∈ Vi , y ∈ Wi , and
Vi × Wi ⊆ Ui . Consider
V = ∩ki=1 Vi and W = ∩ki=1 Wi .
Then V is an open subset of X, W is an open subset of Y, x ∈ V, and y ∈ W.
Furthermore, for each i = 1, . . . k,
V × W ⊆ Vi × Wi ⊆ Ui .
Hence
V × W ⊆ ∩ki=1 Ui ,
and ∩Ui is open in X × Y.
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(b) Show that with this definition the product topology on R × R is the same as the
usual topology on R2 . (More generally, the product topology on Rm × Rn is the
same as the usual topology on Rm+n .)
Let U denote the usual topology on R2 = R × R, and let P denote the product
topology. We want to see that if U ∈ U, then U ∈ P, and if V ∈ P, then V ∈ U.
So let U ∈ U. Then if (x, y) ∈ U, there exists a real number r > 0, such that
Br (x, y) ⊆ U. Let s = r/2, and define V = (x − s, x + s) ⊆ R, and W =
(y − s, y + s) ⊆ R. Since r > 0, s > 0, and hence both V and W are open subsets
of R. Furthermore
V × W ⊆ Br (x, y).
For if (u, v) ∈ V × W, then |x − u| < s and |y − v| < s. So
||(u, v) − (x, y)|| ≤
=
<
=
=
||(u, v) − (u, y)|| + ||(u, y) − (x, y)||
|v − y| + |u − x|
s+s
r r
+
2 2
r.
Thus (u, v) ∈ Br (x, y), V × W ⊆ U, and hence U ∈ (P ).
Conversely, suppose that V ∈ P. If (x, y) ∈ V, then there exist open subsets
G, H ⊆ R, such that x ∈ G, y ∈ H, and G × H ⊆ V. Since G and H are open
subsets of R (with the usual topology), there exist real numbers t, u > 0, such
that (x − t, x + t) ⊆ G and (y − u, y + u) ⊆ H. Let
s = min{t, u}.
Then s > 0, and
Bs (x, y) ⊆ (x − t, x + t) × (y − u, y + u).
For suppose that (a, b) ∈ Bs (x, y). Then
|a − x| ≤ ||(a, b) − (x, y)|| < s ≤ t
and
|b − y| ≤ ||(a, b) − (x, y)|| < s ≤ u.
So a ∈ (x − t, x + t) and b ∈ (y − u, y + u), and hence
Bs (x, y) ⊆ (x − t, x + t) × (y − u, y + u) ⊆ G × H ⊆ V.
Thus V is open in the usual topology on R2 .
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(c) In a product X×Y, we can define two functions pX : X×Y → X and pY : X×Y →
Y by pX (x, y) = x and py (x, y) = y. Show that pX and pY are continuous if X × Y
has the product topology.
We want to show that pX is continuous. Completely analogous reasoning will
show that pY is continuous. So let U be an open subset of X. We need to show
−1
that p−1
X (U ) is an open subset of X × Y. But observe that pX (U ) = U × Y. For
if (x, y) ∈ p−1
X (U ), then pX (x, y) = x ∈ U, and hence (x, y) ∈ U × Y. Conversely,
if (x, y) ∈ U × Y, then x ∈ U and hence pX (x, y) ∈ U.
Finally observe that U × Y is an open subset of X × Y. For if (x, y) ∈ U × Y, then
x ∈ U, y ∈ Y, U is open in X, Y is open in Y, and U × Y ⊆ U × Y.
(d) Show that if f : W → X × Y is a function, then f is continuous iff both pX ◦ f
and pY ◦ f are continuous.
We know from problem 1 that the composition of continuous functions is continuous, and we know from the preceding part of this problem that pX and pY are
continuous. So if f is continuous, then both pX ◦ f and pY ◦ f are continuous.
So to complete the problem we need to show that if both pX ◦ f and pY ◦ f are
continuous, then so is f. So let U be an open subset of X × Y.
Observe first that U is a union of “open rectangles.” That is, U is a union of open
sets of the form G × H, where G is an open subset of X and H is an open subset
of Y. This is true because for each (x, y) ∈ U, there exist open sets Gx ⊆ X, and
Hy ⊆ Y, such that Gx × Hy ⊆ U. Thus
U = ∪(x,y)∈U Gx × Hy .
(Why is each Gx × Hy open?) Since
f −1 (U ) = f −1 [∪(Gx × Hy )] = ∪f −1 [Gx × Hy ],
it suffices to show that the inverse image of open sets of the form G × H are
open. But we’re assuming that pX ◦ f and pY ◦ f are continuous. So if G is an
open subset of X, (pX ◦ f )−1 (G) and (pY ◦ f )−1 (H) are open subsets of W. Finally
observe that
f −1 (G × H) =
=
=
=
f −1 [(G × Y ) ∩ (X × H)]
f −1 (G × Y ) ∩ f −1 (X × H)
−1 −1
[pY (H)]
f −1 [p−1
X (G)] ∩ f
−1
(pX ◦ f ) (G) ∩ (py ◦ f )−1 (H).
Since the last two sets are both open, their intersection is also open.
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