## Stone-Weierstrass Theorem (Real deal)

In this section we extend the idea of the last post to arbitrary compact Hausdorff spaces. Admittedly a lot of information will be assumed known. Including facts about lattices and subalgebras of .

We begin by recalling that if are continuous then so is and

It thus makes sense to talk about sublattices of .

Our first theorem is a potent one, it gives sufficient conditions for a closed sublattice of to be the full space.

**Theorem:** Let be compact Hausdorff with more than one point and be a closed sublattice of with the property that given any distinct and any real numbers there exists some such that . Then,

**Proof: **Since is closed it suffices to show that given any there exists some such that for any .

We start by considering a fixed but arbitrary . For each there exists some such that and . Define . Clearly this is equivalent to and since is continuous it follows that is open. Also, since and and so . Thus, is an open cover for and thus by it’s compactness it admits a finite subcover . Let be the associated functions. Clearly then since we have (by virtue of the fact that it’s a lattice) that . Also, given any we have that for some and so . Since was arbitrary it follows that for every .

Now, considering the same procedure as above for each we consider the sets for each . Since

we see that . Noticing once again that we see that is open for each . Thus, combining these two comments we see that is an open cover for . Appealing to compactness again we see that it has a finite subcover . Thus, let be the associated functions. Since each are in we have that . Now, let be arbitrary. Then, for some and so . Since was arbitrary it follows that for all . Also noting though that

for every We see that

.

Thus, by previous comment the conclusion follows.

Now, it is a problem in elementary analysis to prove that

and

It follows that every linear subspace of which contains is a sublattice. From this we come to the following theorem

**Theorem:** Let be a closed subalgebra of then it is also a closed sublattice.

**Proof:** It follows from prior comment that we must merely show that . Since is closed we must merely show that is an adherent point of . Thus, let be given.

Now, the function where given by is continuous. And from the previous post there exists some such that . In particular, . Thus, is a polynomial without a constant such that

.

Now, since and is a subalgebra we have that . From there it is fairly easy to see that . By previous comment the conclusion follows.

We are now prepared to prove the main theorem in this post

**Theorem:** Let be a compact Hausdorff space and a closed subalgebra of which contains a non-zero constant function and separates points. Then, .

**Proof:** If has one point then consists of all the constant functions on and since contains a non-zero constant function it follows that .

Thus, we may assume WLOG that . Thus, it follows from our first theorem that given any distinct and there exists some such that . Now, since separates points there exists some such that . So define

Clearly is a linear combination of elements of and so . Also, . The conclusion follows. .

There is an analgous theorem dealing with be it is omitted here.

We discuss only one application of the above theorem. But before it we need a theorem.

**Theorem:** Let be a subalgebra of then is also a subalgebra.

**Proof:** We prove that it is closed under vector addition, for scalar and vector multiplication follow similarly.

Let then there exists such that . It follows from the fact that is a subalgebra that and

.

It follows that is an adherent point of and so .

The other follows similarly.

**Theorem:** Let then is dense in .

**Proof:** Clearly . Also, let be distinct, in other words for some . Define by

Clearly but . It follows that separates points. Thus, is a closed subalgebra of which contains a non-zero constant function and separates points. The conclusion follows from the previous theorem since is a compact Hausdorff space. .

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