My version of the Stone-Weierstrass theorem is:(adsbygoogle = window.adsbygoogle || []).push({});

Let [itex]X[/itex] be a compact space, and suppose [itex]A[/itex] is a subalgebra of [itex]C(X,\mathbb{R})[/itex] which separates points of [itex]X[/itex] and contains the constant functions. Then [itex]A[/itex] is dense in [itex]C(X,\mathbb{R})[/itex].

I can see why the subalgebra would have to separate points (this is how A is dense, no?) but I don't understand HOW it separates points. I mean, isn't [itex]A[/itex] a subalgebra? Which means it is a subset of [itex]X[/itex] and has the same algebraic structure as [itex]X[/itex].

If this subset, [itex]A[/itex] were to separate points of [itex]X[/itex], then if I take any pair of distinct points [itex]x,y \in X[/itex] then there must exist a function [itex]f \in A[/itex] such that [itex]f(x) \neq f(y)[/itex].

Oh, I think I just answered this myself! So if the subalgebra did not separate points, then [itex]A[/itex] is never dense in [itex]X[/itex].

My gripe is, [itex]A[/itex] is a subalgebra of [itex]C(X,\mathbb{R})[/itex], what is [itex]C(X,\mathbb{R})[/itex]? And why must the subalgebra contain the constant function? What if it didn't contain the constant function?

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# Homework Help: The Stone-Weierstrass Theorem

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