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?]