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The Stone-Weierstrass Theorem

  1. Nov 12, 2005 #1
    My version of the Stone-Weierstrass theorem is:

    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?
     
  2. jcsd
  3. Nov 12, 2005 #2

    HallsofIvy

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    Looks to me like you need to review some basic definitions! [itex]C(X,\mathbb{R})[/itex] is the set of all continuous functions from [itex]X[/itex] to [itex]\mathbb{R}[/itex]. As for " has the same algebraic structure as [itex]X[/itex]", X doesn't have any algebraic structure- it's just a compact topological space.
     
  4. Nov 12, 2005 #3
    It's always the definitions isn't it Halls? ;)
    That's what I thought!


    Definition of a subalgebra:

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


    Now if you have a compact space [itex]X[/itex], and suppose [itex]A[/itex] is a subalgebra (which means that multiplication defined
    above is closed in [itex]A[/itex]) which separates points of [itex]X[/itex] and contains the constant functions. Then the subalgebra [itex]A[/itex] is dense in [itex]C(X,\mathbb{R})[/itex].
    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?
     
    Last edited: Nov 12, 2005
  5. Nov 12, 2005 #4

    Hurkyl

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    Try considering special cases! What if X is a one-point set? A two-point set?
     
  6. Nov 12, 2005 #5

    mathwonk

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    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?]
     
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