238 CHAPTER 13. SOME FUNDAMENTALS∗

Suppose first that f is continuous at x. Then there exists δ > 0 such that if |x−y| < δ ,then |f (x)−f (y)| < ε . The first part of the above inequality then shows that for eachk = 1, · · · ,q, | fk (x)− fk (y)|< ε . This shows the only if part. Now suppose each functionfk is continuous. Then if ε > 0 is given, there exists δ k > 0 such that whenever |x−y|< δ k

| fk (x)− fk (y)|< ε/q.

Now let 0 < δ ≤min(δ 1, · · · ,δ q). For |x−y|< δ , the above inequality holds for all k andso the last part of (13.1) implies

|f (x)−f (y)| ≤q

∑i=1| fi (x)− fi (y)|<

q

∑i=1

ε

q= ε.

This proves part (4).To verify part (5), let ε > 0 be given and let δ = ε . Then if |x−y| < δ , the triangle

inequality implies

| f (x)− f (y)|= ||x|− |y|| ≤ |x−y|< δ = ε.

This proves part (5) and completes the proof of the theorem. ■

13.2 The Nested Interval LemmaFirst, here is the one dimensional nested interval lemma.

Lemma 13.2.1 Let Ik = [ak,bk] be closed intervals, ak ≤ bk, such that Ik ⊇ Ik+1 for all k.Then there exists a point c which is contained in all these intervals. If limk→∞ (bk−ak) = 0,then there is exactly one such point.

Proof: Note that the {ak} are an increasing sequence and that {bk} is a decreasingsequence. Now note that if m < n, then

am ≤ an ≤ bn

while if m > n,bn ≥ bm ≥ am.

It follows that am ≤ bn for any pair m,n. Therefore, each bn is an upper bound for all theam and so if c≡ sup{ak}, then for each n, it follows that c≤ bn and so for all, an ≤ c≤ bnwhich shows that c is in all of these intervals.

If the condition on the lengths of the intervals holds, then if c,c′ are in all the intervals,then if they are not equal, then eventually, for large enough k, they cannot both be containedin [ak,bk] since eventually bk−ak < |c− c′|. This would be a contradiction. Hence c = c′.■

Definition 13.2.2 The diameter of a set S, is defined as

diam(S)≡ sup{|x−y| : x,y ∈ S} .

Thus diam(S) is just a careful description of what you would think of as the diameter.It measures how stretched out the set is.

Here is a multidimensional version of the nested interval lemma.