What is the Stone-Weierstrass Theorem?

[Oxymoron]

My version of the Stone-Weierstrass theorem is:

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

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 A a subalgebra? Which means it is a subset of X and has the same algebraic structure as X.

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

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

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

 

[HallsofIvy]

Looks to me like you need to review some basic definitions! C(X,\mathbb{R}) is the set of all continuous functions from X to \mathbb{R}. As for " has the same algebraic structure as X", X doesn't have any algebraic structure- it's just a compact topological space.

 

[Oxymoron]

Looks to me like you need to review some basic definitions!

It's always the definitions isn't it Halls? ;)

C(X,\mathbb{R}) is the set of all continuous functions from X to \mathbb{R}. As for " has the same algebraic structure as X", X doesn't have any algebraic structure- it's just a compact topological space.

That's what I thought!


Definition of a subalgebra:

A subalgebra of the set of all continuous functions from X to it's underlying field \mathbb{F} is a subspace A such that for f,g \in A, fg \in A, that is, multiplication of elements of a subalgebra is closed.


Now if you have a compact space X, and suppose A is a subalgebra (which means that multiplication defined
above is closed in A) which separates points of X and contains the constant functions. Then the subalgebra A is dense in C(X,\mathbb{R}).
Im trying to understand why the part in red needs to be included in this theorem. Why does the subalgebra have to contain the constant functions?

 

[Hurkyl]

Why does the subalgebra have to contain the constant functions?

Try considering special cases! What if X is a one-point set? A two-point set?

 

[mathwonk]

why don't you try proving the polynomials are dense in the space of all continuous functions on an interval? maybe you will see what is needed for the proof.[and what if you consider the set of all those continuous functions vanishing at a given point?]