2.5. COMPACTNESS AND CONTINUOUS FUNCTIONS 47

3. (X ,d) is complete and totally bounded.

Proof: By Theorem 2.5.5, the first two conditions are equivalent.2.⇒ 3. If (X ,d) is sequentially compact, then by Lemma 2.5.7, it is totally bounded.

If {xn} is a Cauchy sequence, then there is a subsequence which converges to x ∈ X byassumption. However, from Theorem 2.3.3 this requires the original Cauchy sequence toconverge. Thus (X ,d) is complete and totally bounded.

3.⇒ 2. Suppose {xk} is a sequence in X . It suffices to show it has a Cauchy subse-quence. By assumption there are finitely many open balls of radius 1/n covering X . Thisfor each n ∈ N. Therefore, for n = 1, there is one of the balls, having radius 1 which con-tains xk for infinitely many k. Therefore, there is a subsequence with every term containedin this ball of radius 1. Now do for this subsequence what was just done for {xk} . There isa further subsequence contained in a ball of radius 1/2. Continue this way. Denote the ith

subsequence as {xki}∞

k=1. Arrange them as shown

x11,x21,x31,x41 · · ·x12,x22,x32,x42 · · ·x13,x23,x33,x43 · · ·

...

Thus all terms of {xki}∞

k=1 are contained in a ball of radius 1/i. Consider now the diagonalsequence defined as yk ≡ xkk. Given n, each yk is contained in a ball of radius 1/n wheneverk≥ n. Thus {yk} is a subsequence of the original sequence and {yk} is a Cauchy sequence.By completeness of X , this converges to some x ∈ X which shows that every sequence in Xhas a convergent subsequence. This shows 3.)⇒ 2.). ■

Lemma 2.5.9 The closed interval [a,b] in R is compact and every Cauchy sequence inR converges.

Proof: To show this, suppose it is not. Then there is an open cover C which admitsno finite subcover for [a,b] ≡ I0. Consider the two intervals

[a, a+b

2

],[ a+b

2 ,b]. One of

these, maybe both cannot be covered with finitely many sets of C since otherwise, therewould be a finite collection of sets from C covering [a,b] . Let I1 be the interval whichhas no finite subcover. Now do for it what was done for I0. Split it in half and pick thehalf which has no finite covering of sets of C . Thus there is a “nested” sequence of closedintervals I0 ⊇ I1 ⊇ I2 · · · , each being half of the preceding interval. Say In = [an,bn] . Bythe nested interval Lemma, Lemma 1.11.12, there is a point x in all these intervals. Thepoint is unique because the lengths of the intervals converge to 0. This point is in someO ∈ C . Thus for some δ > 0, [x−δ ,x+δ ] , having length 2δ , is contained in O. For klarge enough, the interval [ak,bk] has length less than δ but contains x. Therefore, it iscontained in [x−δ ,x+δ ] and so must be contained in a single set of C contrary to theconstruction. This contradiction shows that in fact [a,b] is compact.

The second claim was proved earlier, but here it is again. If {xn} is a Cauchy sequence,then it is contained in some interval [a,b] which is compact. Hence there is a subsequencewhich converges to some x ∈ [a,b]. By Theorem 2.3.3 the original Cauchy sequence con-verges to x. ■

Now the next corollary pertains more specifically to Rp.

Corollary 2.5.10 For each r > 0,Q≡ [−r,r]p ≡∏pi=1 [−r,r] is compact in Rp.