4.8. COMPACTNESS 69

Proof:⇒ Suppose K is sequentially compact. Why is it closed? Let kn→ k where eachkn ∈ K. Why is k ∈ K? Since K is sequentially compact, there is a subsequence

{kn j

}such

that lim j→∞ kn j = k̂ ∈ K. However, the subsequence converges to k and so k = k̂ ∈ K. ByTheorem 4.8.11, K is closed. Why is K bounded? If it were not, there would exist ∥kn∥> nwhere kn ∈K and n∈N which means this sequence could have no convergent subsequencebecause the subsequence would not even be bounded. See Theorems 4.5.3 and 4.5.2.⇐Suppose that K is closed and bounded. Since S is bounded, there exists

R =p

∏k=1

[ak,bk]

containing K. If {kn} ⊆ K, then from Corollary 4.8.3, there exists a subsequence{

kn j

}such that lim j→∞ kn j = k ∈ R. However, K is closed and so in fact, k ∈ K.

The last claim follows from a repeat of the preceding argument. Just use K in placeof R and H in place of K. Alternatively, if K is closed and bounded, then so is H, being aclosed subset of K.

What about the sequentially compact sets in Cp? This is actually a special case ofTheorem 4.8.13. For z ∈ Cp, z = (z1, · · · ,zp) where zk = xk + iyk. Thus(

x1,y1,x2,y2, · · · ,xp,yp)≡ θz ∈ R2p

A set K is bounded in Cp if and only if {θz : z ∈ K} is bounded in R2p. Also, zn→ z in Cp

if and only if θzn→ θz in R2p. Now K is closed and bounded in Cp if and only if θK ≡{θz : z ∈ K} is closed and bounded in R2pand so K is closed and bounded in Cp if andonly if θK is sequentially compact in R2p. Thus if {zn} is a sequence in K, there exists asubsequence, still denoted with n such that θzn converges in R2p if and only if zn convergesto some z ∈ Cp. However, z ∈ K because K is closed. Thus K is sequentially compact inCp.

Conversely, if K is sequentially compact, then it must be bounded since otherwise therewould be a sequence {kn} ⊆ K with ∥kn∥ > n and so no subsequence can be Cauchy sono subsequence can converge. K must also be closed because if not, there would be x /∈ Kand a sequence {kn} ⊆ K with kn → x. However, by sequential compactness, there is asubsequence

{knk

}∞

k=1 , knk → k ∈ K and so k = x ∈ K after all. This proves most of thefollowing theorem.

Theorem 4.8.14 Let H ⊆ Fp. Then H is closed and bounded if and only if H is se-quentially compact. A sequence {xn} is a Cauchy sequence in Fp if and only if it converges.In particular, Fp is complete, p≥ 1.

Proof: Consider the last claim. If {zn} converges, then it is a Cauchy sequence byTheorem 4.5.3. Conversely, if {zn} is a Cauchy sequence, then it is bounded by Theorem4.5.2 so it is contained in some closed and bounded subset of Fp. Therefore, a subsequenceconverges to a point of this closed and bounded set. However, by Theorem 4.5.4, theoriginal Cauchy sequence converges to this point.

4.8.3 Compactness and Open CoveringsIn Theorem 4.8.13 it was shown that sequential compactness in Fp is the same as closedand bounded. Here we give the traditional definition of compactness and show that this isalso equivalent to closed and bounded.