13.10. EXERCISES 251

Now consider B ≡ U \A. I claim this is also open. If B is not open, there exists apoint z ∈ B such that every open set containing z is not contained in B. Therefore, lettingB(z,δ ) be such that z ∈ B(z,δ ) ⊆U, there exist points of A contained in B(z,δ ) . Butthen, a repeat of the above argument shows z ∈ A also. Hence B is open and so if B ̸= /0,then U = B∪A and so U is separated by the two sets B and A contradicting the assumptionthat U is connected.

It remains to verify the connected components are open. Let z ∈ Cp where Cp is theconnected component determined by p. Then picking B(z,δ ) ⊆U, Cp ∪B(z,δ ) is con-nected and contained in U and so it must also be contained in Cp. Thus z is an interiorpoint of Cp. ■

As an application, consider the following corollary.

Corollary 13.9.15 Let f : Ω→ Z be continuous where Ω is a connected open set in Rn.Then f must be a constant.

Proof: Suppose not. Then it achieves two different values, k and l ̸= k. Then Ω =f−1 (l)∪ f−1 ({m ∈ Z : m ̸= l}) and these are disjoint nonempty open sets which separateΩ. To see they are open, note

f−1 ({m ∈ Z : m ̸= l}) = f−1(∪m ̸=l

(m− 1

6,m+

16

))which is the inverse image of an open set while f−1 (l) = f−1

((l− 1

6 , l +16

))also an open

set. ■

13.10 Exercises1. Suppose {xn} is a sequence contained in a closed set C such that limn→∞xn = x.

Show that x ∈C. Hint: Recall that a set is closed if and only if the complement ofthe set is open. That is if and only if Rn \C is open.

2. Show using Problem 1 and Theorem 13.3.8 that every closed and bounded set issequentially compact. Hint: If C is such a set, then C ⊆ I0 ≡ ∏

ni=1 [ai,bi]. Now if

{xn} is a sequence in C, it must also be a sequence in I0. Apply Problem 1 andTheorem 13.3.8.

3. Prove the extreme value theorem, a continuous function achieves its maximum andminimum on any closed and bounded set C, using the result of Problem 2. Hint:Suppose λ = sup{ f (x) : x ∈C}. Then there exists {xn} ⊆ C such that f (xn)→λ . Now select a convergent subsequence using Problem 2. Do the same for theminimum.

4. Let C be a closed and bounded set and suppose f : C→Rm is continuous. Show thatf must also be uniformly continuous. This means: For every ε > 0 there exists δ >0 such that whenever x,y ∈C and |x−y|< δ , it follows |f (x)−f (y)|< ε . Thisis a good time to review the definition of continuity so you will see the difference.Hint: Suppose it is not so. Then there exists ε > 0 and {xk} and {yk} such that|xk−yk|< 1

k but |f (xk)−f (yk)| ≥ ε . Now use Problem 2 to obtain a convergentsubsequence.