19.4. STONE WEIERSTRASS THEOREM 509

topology” and the following major theorem will not be true. You might go through theproof and see that this is the case. By the Alexander subbasis theorem, compactness isequivalent to saying that every open cover of subbasic sets admits a finite subcover.

Theorem 19.3.2 (Tychanoff) If (Xi,τ i) is compact, then so is (∏i∈I Xi,∏τ i).

Proof: By the Alexander subbasis theorem, the theorem will be proved if every sub-basic open cover admits a finite subcover. Therefore, let O be a subbasic open cover of∏i∈I Xi. Let

O j = {Q ∈ O : Q = Pj (A) for some A ∈ τ j}.Thus O j consists of those sets of O which have a possibly proper subset of Xi only in theslot i = j. Let

π jO j = {A : Pj (A) ∈ O j}.Thus π jO j picks out those proper open subsets of X j which occur in O j.

If no π jO j covers X j, then by the axiom of choice, there exists f ∈ ∏i∈I Xi \∪π iOi.Therefore, f ( j) /∈ ∪π jO j for each j ∈ I. Now f is a point of ∏i∈I Xi and so f ∈ Pk (A) ∈Ofor some k. However, this is a contradiction as it was shown that f (k) is not an elementof A. (A is one of the sets whose union makes up ∪πkOk.) This contradiction shows thatfor some j, π jO j covers X j. Thus X j = ∪π jO j and so by compactness of X j, there existA1, · · · ,Am, sets in τ j such that X j ⊆∪m

i=1Ai and Pj (Ai)∈O . Therefore, {Pj (Ai)}mi=1 covers

∏i∈I Xi. By the Alexander subbasis theorem this proves ∏i∈I Xi is compact. ■

19.4 Stone Weierstrass TheoremThis theorem was presented earlier in the context of a real algebra of functions on a compactset. Here this is extended to the case where the functions are defined on a locally compactHausdorff space and also extended to the case where the functions have values in C.

19.4.1 The Case of Locally Compact Sets

Definition 19.4.1 Let (X ,τ) be a locally compact Hausdorff space. C0 (X) denotesthe space of real or complex valued continuous functions defined on X with the propertythat if f ∈C0 (X) , then for each ε > 0 there exists a compact set K such that | f (x)|< ε forall x /∈ K. Define || f ||

∞= sup{| f (x)| : x ∈ X}.

This norm is well defined because | f (x)| < 1 for x not in some compact set K and| f (x)| achieves its maximum on K.

Lemma 19.4.2 For (X ,τ) a locally compact Hausdorff space with the above norm,C0 (X) is a complete space.

Proof: Let(

X̃ , τ̃)

be the one point compactification described in Lemma 19.1.27.

D≡{

f ∈C(

X̃)

: f (∞) = 0}.

Then D is a closed subspace of C(

X̃)

. For f ∈C0 (X) ,

f̃ (x)≡{

f (x) if x ∈ X0 if x = ∞