Download Topology Homework Assignment 1 Solutions

Topology
Homework Assignment 1
Solutions
1. Prove that Rn with the “usual topology” satisfies the axioms for a
topological space.
Let U denote the usual topology on Rn .
1(a) Rn ∈ U because if x ∈ Rn , then B1 (x) ⊆ Rn .
1(b) ∅ ∈ U because if x ∈ ∅, then B1 (x) ⊆ Rn .
2. Suppose that A is an index set (possibly infinite), and
V = {Vα : α ∈ A} ⊆ U.
Then if
x ∈ ∪ V = ∪α∈A Vα ,
there is some α0 ∈ A, such that x ∈ Vα0 . Since Vα0 is open, there
is a real number r > 0, such that Br (x) ⊆ Vα0 . But then
Br (x) ⊆ ∪α∈A Vα = ∪ V,
and hence ∪ V is open.
3. Suppose that
{V1 , V2 , . . . Vk } ⊆ U,
and
x ∈ ∩ki=1 Vi .
Since each Vi is open, for each i = 1, . . . k, there is a real number
ri such that
Bri (x) ⊆ Vi .
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Furthermore, since
Bri (x) ⊆ Brj (x)
when ri < rj , if r = min{ri : i = 1, . . . , k},
Br (x) ⊆ Bri (x) ⊆ Vi ,
for i = 1, . . . k. Hence
Br (x) ⊆ ∩ki=1 Vi ,
and ∩ki=1 Vi , is open.
2. Prove that the − δ definition of continuity is equivalent to the open
set definition of continuity (i.e., f : Rn → Rm is continuous iff f −1 (V )
is open for each open subset V ⊆ Rm ).
We first show that f is continuous at x0 ∈ Rn in the − δ definition iff
for each open set V containing f (x0 ), there is an open set U containing
x0 such that f (U ) ⊆ V.
So suppose first that f is continuous at x0 in the −δ definition, and let
V be an open set containing f (x0 ). Since Rm has the usual topology,
there exists a real number > 0 such that B [f (x0 )] ⊆ V. Since we’re
assuming f is continuous at x0 in the − δ definition, there is some real
number δ > 0 such that
||x − x0 || < δ ⇒ ||f (x) − f (x0 )|| < .
But
{x ∈ Rn : ||x − x0 || < δ} = Bδ (x0 ),
and
{y ∈ Rn : ||y − f (x0 )|| < } = B [f (x0 )].
So if x ∈ Bδ (x0 ), then f (x) ∈ B [f (x0 )]. Since B [f (x0 )] ⊆ V, we have
that
x ∈ Bδ (x0 ) ⇒ f (x) ∈ V.
That is,
f [Bδ (x0 )] ⊆ V.
Since Bδ (x0 ) is open, we can let U = Bδ (x0 ).
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Conversely, suppose that for each open set V containing f (x0 ), there
is an open set U containing x0 such that f (U ) ⊆ V. Let > 0. Then
B [f (x0 )] is an open set containing f (x0 ). Hence there is an open set
U containing x0 such that f (U ) ⊆ V. Since Rn has the usual topology,
there is a δ > 0, such that Bδ (x0 ) ⊆ U. But then
f [Bδ (x0 )] ⊆ f (U ) ⊆ B [f (x0 )].
Since
Bδ (x0 ) = {x ∈ Rn : ||x − x0 || < δ},
and
B [f (x0 )] = {y ∈ Rn : ||y − f (x0 )|| < },
we have that
||x − x0 || < δ ⇒ ||f (x) − f (x0 )|| < .
Now suppose that f is continuous at each point x of Rn in the − δ
definition. We want to show that if V is an open subset of Rm , then
f −1 (V ) is an open subset of Rn . Let V be an open subset of Rm . Then
for each f (x) ∈ V, by our preceding result, there is a δx > 0, such that
f [Bδx (x)] ⊆ V.
But then
f −1 (V ) = ∪x∈f −1 (V ) Bδx (x)
because every x ∈ f −1 (V ) belongs to the right-hand side, and every
Bδx is contained in the left-hand side. Since f −1 (V ) is a union of open
sets, f −1 (V ) is open.
Conversely suppose that for each open subset V ⊆ Rm , f −1 (V ) is
open in Rn , and let x0 ∈ Rn and > 0. Then B [f (x0 )] is an open
subset of Rm . So f −1 {B [f (x0 )]} is an open subset of Rn . Thus, since
x ∈ f −1 {B [f (x0 )]}, there must exist an open ball
Bδ (x0 ) ⊆ f −1 {B [f (x0 )]}.
Thus if ||x − x0 || < δ, (i.e., if x ∈ Bδ (x0 ), then ||f (x) − f (x0 )|| < —
since f (x) ∈ B [f (x0 ).
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3. Suppose (X, U) is a topological space and Y ⊆ X. Define a family
of subsets V of Y by V ∈ V iff there exists a set U ∈ U such that
V = U ∩ Y. Prove that V is a topology on Y. V is called the relative
topology on Y, and unless we state otherwise, you should assume that
a subset of a topological space has the relative topology.
1(a) Since Y ⊆ X, Y = Y ∩ X, and since X is an element of U, Y
is open in the relative topology.
1(b) Since ∅ = ∅ ∩ Y, and since ∅ is an element of U, ∅ is open in
the relative topology.
2. Let A be an index set and let
V = {Vα : α ∈ A}
be a family of elements of the relative topology. Then for each
element Vα ∈ V, there is an open set Uα ∈ U, such that
Vα = Uα ∩ Y.
Thus
∪α∈A Vα = ∪α∈A (Uα ∩ Y ) = (∪α∈A Uα ) ∩ Y.
But ∪Uα is open in X. So ∪Vα = (∪Uα ) ∩ Y is open in the relative
topology.
3. Let {V1 , V2 , . . . Vk } be a finite subset of the relative topology.
Then for i = 1, . . . k, there is an element Ui ∈ U, such that Vi =
Ui ∩ Y. Thus
∩ki=1 Vi = ∩ki=1 (Ui ∩ Y ) = (∩ki=1 Ui ) ∩ Y.
Since ∩Ui is open in X, ∩Vi = (∩Ui ) ∩ Y is open in the relative
topology.
4. Using subsets of euclidean spaces, find topological spaces X and Y and
a function f : X → Y, such that
(a) f is continuous but not open.
(b) f is open but not continuous.
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Can you find a function f satisfying 4a or 4b that is both one-to-one
and onto?
Let f : [0, 2π) → S 1 by f (t) = (cos(t), sin(t)). Then f is both one-toone and onto, for if (x, y) ∈ S 1 , we can rewrite (x, y) in exponential
form eit , where t is a unique value in [0, 2π), and f (t) = (x, y). Thus,
f has a unique inverse g : S 1 → [0, 2π).
Since the component functions, cos(t) and sin(t), are continuous, f is
continuous. (Technically, we need to know about product spaces in
order to prove this result.) However, f is not open. For open subsets
of S 1 (in the relative topology on S 1 as a subset of R2 ) are unions of
“curved open intervals.” To see this observe that any open subset of
R2 is a union of open balls. Since an open ball intersects S 1 in a curved
open interval (or all of S 1 ), any open subset of S 1 is a union of open
intervals. But open sets in [0, 2π) are unions of intersections of open
intervals with [0, 2π). Thus, [0, π) is open in [0, 2π), and
f ([0, π)) = {(x, y) ∈ S 1 : y > 0} ∪ {(1, 0)},
which isn’t a curved open interval, and hence isn’t open.
The function g, on the other hand is open, but it’s not continuous. Since
f is continuous, if U is open in S 1 , f −1 (U ) = g(U ) is open in [0, 2π).
However, g −1 ([0, π)) = f ([0, π)) isn’t open. So g is not continuous.
5. Suppose X and Y are topological spaces, and f : X → Y is continuous.
If A is a compact subset of X, show that f (A) is a compact subset of
Y. (Use the “open cover” definition of compactness.) So compactness is
a topological property: if Y is homeomorphic to X (denoted Y ≈ X),
and X is compact, then so is Y.
We proved this in class.
6. Suppose X is a topological space. Show that a subset C of X is closed
iff C contains all of the points that are arbitrarily close to C. (Recall
that the point x is arbitrarily close to C if every open neighborhood of
x contains a point of C.)
We proved this in class.
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7. Suppose that X and Y are topological spaces and f : X → Y. Show
that f is continuous iff for each closed set C ⊂ Y, f −1 (C) is closed a
closed subset of X.
Suppose first that f is continuous — i.e., for each open set V in Y,
f −1 (V ) is open in X. Let C be a closed subset of Y. Then Y − C is
open in Y. Furthermore,
f −1 (Y − C) = X − f −1 (C),
by Theorem 2.6 on page 21 of the set theory handout. But, by assumption f is continuous. So f −1 (Y − C) is open in X. Thus,
X − [X − f −1 (C)] = f −1 (C)
is closed in X.
Conversely, suppose that for each closed set C in Y, f −1 (C) is closed in
in X. Also suppose that V is an open subset of Y. Then, by assumption,
f −1 (Y − V ) = X − f −1 (V )
is closed, and hence
X − [X − f −1 (V )] = f −1 (V )
is open. That is, f is continuous.
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