Download Topology of Rn - Will Rosenbaum

Topology of Rn
Will Rosenbaum
Updated: November 9, 2014
Department of Mathematics
University of California, Los Angeles
1 Basic Notions
1.1 Open sets In this section, we describe the standard topology on Rn .
Definition 1. Suppose x ∈ Rn and r ∈ R with r > 0. We define the open ball of radius r
centered at x to be
B(x, r) = {y ∈ Rn | |y − x| < r} .
Definition 2. We call a subset A ⊆ Rn open if for every x ∈ A, there exists ε > 0 such that
B(x, ε) ⊆ A. A subset F ⊆ Rn is called closed if its complement F c is open.
Proposition 3. Open sets in Rn have the following properties:
(a) The sets ∅ and Rn are open.
(b) Given any set I and collection {Ai | i ∈ I} where every Ai is open, the union
[
Ai
i∈I
is open.
(c) For any finite collection A1 , A2 , . . . , Ak of open sets, the intersection
k
\
Ai
i=1
is open.
Remark 4. The properties (a), (b), and (c) are called the topology axioms. Any set X with a
family of subsets τ which satisfies (a), (b), and (c) is called a topological space. Notice that
Rn and ∅ are both open and closed.
Proof. It is vacuously true that ∅ is open, and clear that Rn is open, hence (a) holds.
For (b), suppose {Ai }i∈I is a family of open sets. Suppose
x∈
[
Ai .
i∈I
By the definition of union, this implies that x ∈ Aj for some j.S
Since Aj is open and x ∈ Aj ,
there exists some ε > 0 such that B(x, ε) ⊆ Aj . Since Aj ⊆ i∈I Ai , we also have
B(x, ε) ⊆
[
i∈I
as desired.
Ai
Topology of Rn
Now suppose A1 , A2 , . . . , Ak are open in Rn , and x ∈
there exist ε1 , ε2 , . . . , εk such that
B(x, εi ) ⊆ Ai
Tk
i=1
Ai . Since x ∈ Ai for all i,
for i = 1, 2, . . . , k.
Now take ε = min {ε1 , ε2 , . . . , εk }. Then for all i,
B(x, ε) ⊆ Ai
hence
B(x, ε) ⊆
k
\
Ai .
i=1
Thus
Tk
i=1
Ai is open, as desired.
Exercise 5. (a) Show that if {Fi | i ∈ I} is a family of closed subsets in Rn then F =
is closed.
Sk
(b) Show that if F1 , F2 , . . . , Fk ⊆ Rn are closed, then F ′ = i=1 Fi is closed.
T
i∈I
Fi
Exercise 6. Give an example of a set in R which is neither open nor closed and prove that
your example works.
Exercise 7. (a) Give an example of a family of open subsets of R, {Ai | i ∈ I} such that
\
A=
Ai
i∈I
is not open.
(b) Given an example of a family of closed subsets of R, {Fi | i ∈ I} such that
[
F =
Fi
i∈I
is not closed.
Example 8. For every r > 0 and x ∈ Rn , the open ball B = B(x, r) is open. To see this,
we must show that for every other point y ∈ B, B contains a (smaller) ball centered at y.
Since y ∈ B, |x − y| < r. Let ε = r − |x − y|. Then for all z ∈ B(ε, y), we have
|x − z| ≤ |x − y| + |y − z| < r.
Therefore, z ∈ B, hence B(ε, y) ⊆ B, which is what we needed to show.
Exercise 9. Suppose I1 , I2 , . . . , In are open intervals Ii = (ai , bi ) in R. Prove that
I1 × I2 × · · · × In ⊆ Rn
is open in Rn .
Exercise 10. Let F = {x1 , x2 , . . . , xk } ⊆ Rn be a finite set. Prove that F is closed.
1.2 Sequences in Rn A sequence (xi )i∈N in Rn is an set of elements in Rn indexed by the
natural numbers. Given a sequence (xi )i∈N , a subsequence (xi )i∈I is an infinite subset of
(xi ) where I ⊆ N. We will sometimes denote a subsequence of (xi ) by (xij )j∈N where
I = {i1 < i2 < i3 < · · · }.
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Topology of Rn
Definition 11. Let (xn ) be a sequence in Rn . We say that (xi ) converges to x ∈ Rn and
write
lim xi = x
if for every ε > 0, there exists N ∈ N such that i > N implies |xi − x| < ε.
Proposition 12. Suppose (xi ) is a convergent sequence in Rn , say x = lim xi . Then every
subsequence of (xi ) also converges to x.
Proof. Suppose (xij ) is a subsequence of (xi ). In particular, we have i1 < i2 < · · · . Therefore, for every j ∈ N, ij ≥ j. Since x = lim xi , for every ε >
0, there exists N ∈ N such
that |xi − x| < ε. Since ij ≥ j, j > N implies that xij − x < ε. Therefore, lim xij = x,
as desired.
Exercise 13. Give an example of a sequence (xi ) in R which does not converge and a subsequence (xij ) of (xi ) which does converge.
Proposition 14 (Convergence is component-wise). Suppose (xi ) is a sequence in Rn where
xi = (x1,i , x2,i , . . . , xn,i )
for
i = 1, 2, . . . .
Then (xi ) converges to x = (x1 , x2 , . . . , xn ) if and only if
lim xj,i = xj
i→∞
for
j = 1, 2, . . . , n.
Exercise 15. Prove proposition 14.
Exercise 16. Give an example of a set S ⊆ R and a convergent sequence (xi ) such that
xi ∈ A for all i, but limi→∞ xi ∈
/ A.
Proposition 17 (Sequential characterization of closed sets). Suppose F ⊆ Rn . Then F is
closed if and only if every convergent sequence (xi ) with xi ∈ F for all i has its limit in F .
That is,
xi ∈ F for all i =⇒ x = lim xi ∈ F.
Proof. First, suppose F is closed and (xi ) is convergent with xi ∈ F for all i. Suppose
towards a contradiction that lim xi = x ∈
/ F . Since F c is open and x ∈ F c , there exists
c
ε > 0 such that B(x, ε) ⊆ F . Since lim xi = x, there exists some N ∈ N such that
i > N implies xi ∈ B(x, ε) ⊆ F c , which contradicts the hypothesis that xi ∈ F for all i.
Therefore, x ∈ F , as desired.
Now suppose F has the property that every convergent sequence (xi ) with xi ∈ F for
all i has x = lim xi ∈ F . Suppose that F is not closed, i.e., F c is not open. Since F c is not
open, there exists x ∈ F c such that for all ε > 0, B(x, ε) * F c . Therefore, there exists some
y ∈ F ∩ B(x, ε) for all ε. We construct a sequence (xi ) with x = lim xi in the following
way. For each i > 0, take xi to be any point in F ∩ B(x, 1/i). Then this choice of xi satisfies
|xi − x| < 1/i. Therefore lim xi = x ∈
/ F , but xi ∈ F for all i, which contradicts our
hypothesis! Therefore, F c must be open, hence F is closed.
1.3 Interior, boundary, and limit points Here, we give an alternate characterization of open
and closed sets in terms of interior, boundary and limit points of sets.
Definition 18. Suppose S ⊆ Rn is a set and x ∈ Rn .
(a) We call x an interior point of S if there exists δ > 0 such that B(x, δ) ⊆ S. We denote
the set of interior points of S by int(S).
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Topology of Rn
(b) We call x a boundary point of S if for every ε > 0, there exist y ∈ S and z ∈
/ S such
that y, z ∈ B(x, ε). We denote the set of boundary points of S by ∂S.
(c) We say that x is a limit point of S if for every ε > 0, there exists y ∈ S, y 6= x such that
y ∈ B(x, ε). We denote the set of limit points of S by S ′ .
(d) A point x ∈ S which is not a limit point is called a isolated point of S. A set which
contains only isolated points (i.e., no limit points) is called a discrete set.
(e) We define the closure of S to be S = S ∪ ∂S.
Proposition 19 (Properties interior, boundary, and limit points). Let S ⊆ Rn .
(a) int(S) ∩ ∂S = ∅.
(b) S is open if and only if S = int(S) if and only if S ∩ ∂S = ∅.
(c) S ⊆ int(S) ∪ ∂S.
(d) S is closed if and only if S ′ ⊆ S.
(e) S is closed if and only if ∂S ⊆ S.
Proof. We leave the proofs of (a), (b), and (e) as exercises. For (c), suppose x ∈ S. Suppose
x ∈
/ int(S). Then for all δ > 0, B(x, δ) * S, hence there exists z ∈
/ S such that z ∈
B(x, δ). Therefore x ∈ ∂S. Now suppose x ∈
/ ∂S. Then there exists ε > 0 such that either
B(x, ε)∩S = ∅ or B(x, ε)∩S c = ∅. The former cannot occur, as x ∈ S and x ∈ B(x, ε).
Therefore, B(x, ε) ∩ S c = ∅, hence B(x, ε) ⊆ S. Therefore, x ∈ int(S).
For (d), first suppose S is closed. Let x ∈ S ′ . We must show that x ∈ S. By Proposition
17, every convergent sequence (xi ) with xi ∈ S for all i satisfies lim xi ∈ S. Therefore, it
suffices to construct a sequence (xi ) with x = lim xi and xi ∈ S for all i. To this end, for
every i > 0, take xi to be any element
xi ∈ B(x, 1/i) ∩ S.
Such an xi exists by the definition of S ′ . Then for any ε > 0, taking i > ⌈1/i⌉ gives
|xi − x| < ε. Hence x = lim xi as desired.
On the other hand, suppose S ′ ⊆ S. Again by Proposition 17, it suffices to show that
every convergent sequence (xi ) in S has lim xi ∈ S. Suppose (xi ) satisfies xi ∈ S for all
i and x = lim xi . We claim that x ∈ S ′ . To see this, note that for every ε > 0, by the
definition of lim xi , there exist (infinitely many) xi ∈ S with xi ∈ B(x, ε). Therefore,
x ∈ S ′ ⊆ S, hence S is closed.
Exercise 20. Prove parts (a), (b), and (e) in the preceding proposition.
Problem 21. Let D ⊆ R be the set of dyadic rationals in the unit interval. That is,
o
na n
n,
a
∈
N,
a
≤
2
D=
2n
Prove that
(a) int(D) = ∅
(b) D ⊆ ∂D
(c) D = [0, 1].
(Hint: for part (c), you may find the Archimedean property useful.)
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Topology of Rn
1.4 Bounded and compact sets
Definition 22. Let S ⊆ Rn . We say that S is bounded if there exists r ∈ R, r > 0 such
that S ⊆ B(0, r). If (xi ) is a sequence in Rn we say that (xi ) is a bounded sequence if
S = {xi | i ∈ N} is bounded.
Exercise 23. Show that if T ⊆ S and S is bounded, then T is bounded. In particular, if S is
bounded and (xi ) is a sequence in S, then (xi ) is a bounded sequence.
Proposition 24. Suppose (xi ) is a convergent sequence. Then (xi ) is bounded.
Proof. Write x = lim xi . Then there exists N ∈ N such that i > N implies |x − xi | < 1.
For i = 0, 1, . . . , N define ri = |xi | and define rN +1 = |xN +1 | + 1. By the triangle
inequality, we have
|xi | ≤ rN +1 for all i ≥ N + 1.
Now choose
r = max {r1 , r2 , . . . , rN , rN +1 }
so that
xi ∈ B(0, r)
for all
i ∈ N.
Thus, (xi ) is bounded.
Theorem 25 (Bolzano-Weierstrass). A set S ⊆ Rn is bounded if and only if every sequence
(xi ) in S has a convergent subsequence.
Proof. We begin by proving the ⇒ direction. Suppose S is bounded and (xi ) is a sequence
in S. We will argue by induction on n.
Base case, n = 1. Suppose (xi ) is a bounded sequence in R. Call an element xk a max
point if for all i > k, xi ≤ xk . We consider two cases separately. If (xi ) has infinitely
many max points, then denote this subsequence of max points by xi1 , xi2 , xi3 , . . ..
Notice that this subsequence is monotonically decreasing, and bounded from below.
Therefore, it is a convergent subsequence by the completeness of R.
On the other hand, suppose (xi ) has only finitely many max points. Thus, for every
i ∈ N, there exists k > i such that xk > xi . Therefore, we can construct a subsequence xi1 , xi2 , xi3 , . . . such that xij < xik for all j < k. In particular, this subsequence is monotonically increasing and bounded from above, hence convergent by
the completeness of R.
Inductive step. Suppose the theorem holds for n − 1. Now let (xi ) be a bounded sequence
in Rn . Write the terms of this sequence in terms of their components
xi = (xi,1 , xi,2 , . . . , xi,n ).
By Proposition 14, it suffices to show there is a subsequence (xij ) of (xi ) for which
each component converges. By the case n = 1, we can find a subsequence (xij ) for
which the n-th component converges. Viewing the first n − 1 components of this
sequence as a sequence in Rn−1 , we can find yet another subsequence, (xjk ) of (xij )
for which the first n − 1 components converge, by the inductive hypothesis. Since the
n-th component of this subsequence is a subsequence of (xij ,n ) which is convergent,
the n-th component (xjk ,n ) of (xjk ) is also convergent by Proposition 12. Therefore,
(xi ) has a convergent subsequence, as desired.
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Topology of Rn
For the ⇐ direction, we argue by contraposition. Suppose S is not bounded. Then for
every i ∈ N, we can find xi ∈ S such that xi ∈
/ B(0, i). Clearly, this sequence (xi ) is not
bounded, hence is not convergent by Proposition 24.
Definition 26. We say that a set K ⊆ Rn is compact if every sequence (xi ) in K has a subsequence which converges in K. That is, if xi ∈ K for all i, then there exists a subsequence
(xij ) such that lim xij ∈ K.
Proposition 27 (Characterization of compact sets). A set K ⊆ Rn is compact if and only if
it is closed and bounded.
Proof. This proposition follows from Proposition 17 and Theorem 25. The details are left to
the reader.
Corollary 28. If K ⊆ Rn is compact and F ⊆ K is closed, then F is compact.
Proof. By Proposition 27, K is bounded. Since F ⊆ K, F is also bounded. Therefore, F is
closed and bounded, hence compact by Proposition 27.
Exercise 29. Prove that the n-dimensional unit sphere
S n = x ∈ Rn+1 |x| = 1
is compact.
Problem 30. Let S ⊆ Rn be a set and {Ai }i∈I a family of open subsets of Rn . We say that
{Ai } is an open cover of S if
[
S⊆
Ai .
i∈I
We say that {Ai } has a finite subcover of S if there exists k ∈ N and sets A1 , A2 , . . . , Ak ∈
{Ai } such that
S ⊆ A1 ∪ A2 ∪ · · · ∪ Ak .
Prove that K ⊆ Rn is compact if and only if every open cover of K has a finite subcover.
1.5 Subspace topology Given a subset S ⊆ Rn , we can think of S as being a topological space
whose topology is inherited from the topology of Rn .
Definition 31. Let S ⊆ Rn . Then a subset A ⊆ S is open in the subspace topology if there
exists an open subset A0 ⊆ Rn such that A = A0 ∩ S. A subset F ⊆ S is closed in the
subspace topology if there exists F0 ⊆ Rn such that F = F0 ∩ S.
This definition has some properties which may at first seem counter-intuitive, if not
entirely counterproductive. For example no matter what set S you choose, S is both open
and closed in the subspace topology. Indeed S = S ∩ Rn , which is both open and closed.
Example 32. Consider the half-open interval I = [0, 1) ⊆ R. Viewed as a subset of R with
the standard topology, I is neither open nor closed. However, in the subspace topology I
is both open and closed. Subsets of I of the form [0, a) for a ≤ 1 are open in the subspace
topology because we can write, for example,
[0, a) = (−1, a) ∩ I
where (−1, a) is open in the standard topology on R. In the standard topology on R, no
set of the form A = [0, a) is open as 0 is a boundary point for A. Similarly, for 0 ≤ a < 1,
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Topology of Rn
sets of the form [a, 1) are closed in I with the subspace topology. However, no such subset is
closed in R, as 1 is a limit/boundary point not contained in [a, 1) in the standard topology
on R.
Although the subspace topology on S is perhaps initially less intuitive than the standard
topology on Rn , we will see in the following section that the subspace topology is precisely
the right tool to characterize continuous functions whose domain is S.
2 Continuous functions
2.1 Three equivalent definitions The goal of this section is to give three equivalent characterizations of continuous functions defined on a subset X ⊆ Rn . We begin with the “standard”
ε, δ definition.
Definition 33. Let X ⊆ Rn and f : X → Rm a function. For x0 ∈ X, we say that the
limit as x approaches x0 of f (x) equals y ∈ Rn and write
lim f (x) = y
x→x0
if for every ε > 0, there exists δ > 0 such that |x − x0 | < δ implies |f (x) − y| < ε. We say
that f is continuous at x0 if
lim f (x) = f (x0 ).
x→x0
We say that f is continuous on X or simply continuous if for all x0 ∈ X, f is continuous at
x0 .
Proposition 34 (Equivalent characterizations of continuous functions). Let X ⊆ Rn and
f : X → Rm . Then the following are equivalent.
(a) f is continuous on X.
(b) For every x ∈ X and sequence (xi ) in X with lim xi = x, we have lim f (xi ) = f (x).
(c) For every open set A ⊆ Rm , f −1 (A) ⊆ X is open in the subspace topology.
Proof. We will prove that (a) and (b) are equivalent, and leave the equivalence of (c) to the
reader.
(a) =⇒ (b). Suppose f is continuous on X, x ∈ X and (xi ) is a sequence with lim xi = x.
We must show that lim f (xi ) = f (x). That is, for every ε > 0 there exists N ∈ N
such that i > N implies |f (xi ) − f (x)| < ε.
Suppose ε > 0 is given. Since f is continuous at x, we can find δ > 0 such that if
|y − x| < δ, then |f (y) − f (x)| < ε. Since lim xi = x, we can find an N ∈ N such
that if i > N , then |xi − x| < δ. Then for this choice of i, we have
i > N =⇒ |xi − x| < δ =⇒ |f (xi ) − f (x)| < ε.
Therefore, lim f (xi ) = f (x) as desired.
¬(a) =⇒ ¬(b). Suppose f is not continuous on X. Then there exists x ∈ X such that f
is not continuous at x. Unwrapping the negation of the definition of continuous, this
implies that there exists ε > 0 such that for all δ > 0, there exists y ∈ X such that
|x − y| < δ and |f (x) − f (y)| ≥ ε. We will use this property to construct a sequence
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Topology of Rn
(xi ) in X such that lim xi = x, but lim f (xi ) 6= f (x). Specifically, for each i ≥ 1,
take xi such that |xi − x| < 1/i and |f (x) − f (xi )| ≥ ε (for ε as prescribed above).
It is easy to verify that lim xi = x, but that lim f (xi ) 6= f (x).
Problem 35. Prove that f : X → Rm is continuous if and only if for every open subset
A ⊆ Rn , f −1 (A) is open in subspace topology on X.
Exercise 36. Consider the function f : Q → R (where Q ⊆ R has the subspace topology)
defined by
(
f (x) =
1
0
x2 > 2
x2 < 2.
Is f continuous? Is there a continuous function g : R → R such that g(x) = f (x) for all
x ∈ Q?
The power of Proposition 34 is that it gives us three different ways to prove that a function
is continuous:
(a) Analysis using the good old ε, δ definition.
(b) Sequential using the characterization in part (b) of the proposition. Continuous functions are those that map convergent sequences to convergent sequences.
(c) Topological using characterization (c) of the proposition. Continuous functions are functions such that the preimage of open sets are open.
Although these definitions are logically equivalent, using one or the other tends to admit
different styles of proof. Often one can prove theorems about continuous functions more
efficiently by judiciously choosing which characterization to use.
2.2 Composition of continuous functions Here we prove some useful properties of continuous functions.
Proposition 37 (Composition of continuous functions is continuous). Suppose X ⊆ Rn ,
Y ⊆ Rm . Let
f : X −→ Y and g : Y −→ Rk
be continuous functions. Then the composition of f and g, g ◦ f : X → Rk is continuous.
First proof. It suffices to show that for every x ∈ X and sequence (xi ) with lim xi = x,
lim g(f (xi )) = (g(f (x)). To this end, we appeal to the continuity of f and g:
lim xi = x =⇒ lim f (xi ) = f (x) =⇒ lim g(f (xi )) = g(f (x)).
The first implication holds by the continuity of f , while the second holds by the continuity
of g. Thus, g ◦ f is continuous, as desired.
Exercise 38. Suppose X ⊆ Rn , Y ⊆ Rm and f : X → Y , g : Y → Rk . Prove that for all
A ⊆ Rk
(g ◦ f )−1 (A) = f −1 (g −1 (A)).
8
Topology of Rn
Second proof. It suffices to show that for every open subset A ⊆ Rk , (g ◦ f )−1 (A) =
f −1 (g −1 (A)) is open in X with the subset topology. To this end, since g is continuous
g −1 (A) is open in Y . Therefore, by the continuity of f , f −1 (g −1 (A)) = (g ◦ f )−1 (A) is
open in X. Therefore, g ◦ f is continuous.
Both of the proofs above are simpler than a proof using the ε, δ definition of continuity.
2.3 Topological properties Suppose X ⊆ Rn . We say that X is connected if X cannot be
written X = A ∪ B where A and B are nonempty, open and A ∩ B = ∅. If X is not
connected, we say it is disconnected. Notice that the decomposition X = A ∪ B with
A ∩ B = ∅ implies that A and B are also both closed in the subspace topology. Thus we say
that A and B are clopen. In fact, this gives an alternative characterization of connectedness:
a topological space X is connected if and only if it only has trivial (that is, equal to X or ∅)
clopen sets.
Example 39. Every interval I ⊆ R is connected. To see this, suppose to the contrary that I
is not connected. Thus, there exist A, B ⊆ R such that A and B are open (in the subspace
topology for I), A ∩ B = ∅ and I = A ∪ B. Suppose a ∈ A and b ∈ B. Without loss of
generality, we assume that a < b. Since I is an interval, [a, b] ⊆ I.
Define c = sup ([a, b] ∩ A). Clearly we have a < c < b. Since A is closed with respect
to the subspace topology on I, we have c ∈ A. Thus c 6= b. On the other hand, for every
(sufficiently small) ε > 0, c + ε ∈ B from the definition of c. But this implies that c is a
limit point in B, hence c ∈ B since B is closed. This contradicts the hypothesis that I is
disconnected, hence I is connected.
Proposition 40 (Connectedness is a topological property). Suppose X ⊆ Rn is connected
and f : X → Rm is continuous. Then f (X) is connected.
Proof. Suppose to the contrary that f (X) is not connected. That is, there exist disjoint open
subsets A, B ⊆ Rn such that f (X) = A ∪ B. Therefore, we have X = f −1 (A ∪ B) =
f −1 (A) ∪ f −1 (B). Since f is continuous, f −1 (A) and f −1 (B) are open. Suppose x ∈
f −1 (A) ∩ f −1 (B). Then f (x) ∈ A and f (x) ∈ B. But B ∩ A = ∅, a contradiction!
Therefore, f −1 (A) ∩ f −1 (B) = ∅. Thus, X is a union of disjoint open sets, implying X is
not connected, contradicting our hypothesis. Therefore, f (X) must be connected.
Corollary 41 (Intermediate value theorem). Suppose I ⊆ R is an interval and f : I → R is
continuous. Suppose a, b ∈ I and f (a) < f (b). Then for all d ∈ (f (a), f (b)), there exists
c between a and b such that f (c) = d.
Proof. Without loss of generality, assume that a < b (the case where b < a is analogous).
Since f is continuous, and the sub-interval (a, b) ⊆ I is connected, f ((a, b)) is connected.
Suppose d ∈
/ f ((a, b)). Then we can write∗
f ((a, b)) = A ∪ B
where A = f ((a, b)) ∩ (−∞, d), B = f ((a, b)) ∩ (d, ∞)
so that f ((a, b)) is not connected.
Proposition 42 (Compactness is a topological property). Suppose K ⊆ Rn is a compact
set and f : K → Rm is continuous. Then f (K) is compact.
9
∗ We endow f (X) with
the subspace topology so
that f ((a, b)) is
necessarily open.
Topology of Rn
Proof. We use the “open cover” characterization of compactness. Suppose {Bi | i ∈ I} is an
open cover of f (K). Define Ai = f −1 (Bi ) for all i ∈ I. Since f is continuous, {Ai | i ∈ I}
is an open cover of K. Since K is compact, there is a finite subcover, A1 , A2 , . . . , Ak :
K=
k
[
Ai .
i=1
This implies that
f (X) = f (A1 ∪ A2 ∪ · ∪ Ak ) =
k
[
f (Ai ) =
i=1
k
[
Bi .
i=1
Therefore, {Bi | i ∈ I} has a finite subcover, implying that f (K) is compact.
Corollary 43 (Extreme value theorem). Suppose K is a compact set and f : K → R a
continuous function. Then there exist a, b ∈ K such that for all x ∈ K, f (a) ≤ f (x) ≤
f (b).
Proof. By the previous proposition, f (K) ⊆ R is compact. Therefore, it is closed and
bounded. In particular, this implies that inf f (K) ∈ f (K) and sup f (K) ∈ f (K). Thus,
we can find a and b satisfying f (a) ∈ inf f (K) and f (b) = sup f (K). These choices give
the desired result.
Problem 44. We say that a set X ⊆ Rn is path connected if for every x, y ∈ X, there exists
a continuous curve function γ : [a, b] → X such that γ(a) = x and γ(b) = y.
(a) Prove that path-connectedness is a topological property. That is, if X is path connected,
and f : X → Rm is continuous, then f (X) is path connected.
(b) Use part (a) to prove the following generalization of the intermediate value theorem: If
X ⊆ Rn is path connected, f : X → R is continuous and f (x) < f (y), then for all
c ∈ R with f (x) < c < f (y), there exists z ∈ X such that f (z) = c.
2.4 Homeomorphism Suppose X ⊆ Rn and Y ⊆ Rm , and f : X → Y . We say that f is
a homeomorphism if f a continuous bijection and f −1 is continuous as well. If there is a
homeomorphism f between X and Y , then we say that X and Y are homeomorphic and
write X ≃ Y .
Problem 45. Prove that all open intervals in R are homeomorphic. Is this true of closed
intervals?
Using the discussion of topological properties described in the previous section, we may
prove that certain topological spaces are not homeomorphic. For example, if K is compact
and X is homeomorphic to K, then X must also be compact. This allows us to prove, for
example, that the two-dimensional unit sphere is not homeomorphic to R2 .
Problem 46. Consider the two dimensional punctured unit sphere given by
X = {x ∈ Rn | |x| = 1} \ {(0, 0, 1)} .
Prove that X is homeomorphic to R2 .
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