FOM: New member/ Stone-Weierstrass theorem

[Klaas Pieter Hart]

On Thu, 10 May 2001, Karim Zahidi wrote:

> What prompted me to subscribe now is a question by one of my fellow
> mathematicians here in Dublin. This guy is writing a book on analysis and at
> some point he introduces the Stone-Weierstrass theorem. The proof he gives
> in his book uses very heavily the Axiom of Choice, but he read somewhere that
> it is possible to give a proof avoiding the Axiom. I wonder if somebody on
> this list might know about this or would be able to suply some source for
> this.

It suffices to prove the Real version.
Let $g$ and $\epsilon>0$ be given.

Using the properties of $A$ it is possible to find, given $a$ and $b$ in $X$,
a function $f\in A$ such that $f(a)=g(a)$ and $f(b)=g(b)$.
In the familiar proof one first fixes $a$ and chooses for each $b$ one
such function $f_{a,b}$ -- this is blatant choice -- and forms 
$U_b=\{x:f_{a,b}(x)<g(x)+\epsilon\}$.
Via a finite subcover one finds $b_1$, ..., $b_n$ and forms
$f_a=f_{a,b_1}\meet\cdots\meet f_{a,b_n}$, a continuous function which
is everywhere less than $g+\epsilon$ and satisfies $f_a(a)=g(a)$.
Next one does this for all $a$ -- choice again --, sets
$V_a=\{x:f_a(x)>g(x)-\epsilon\}$; as above, via a finite subcover, one
finds $f$ with $g-\epsilon<f<g+\epsilon$.

Note, the functions $f_a$ and $f$ belong to the closure of $A$ (no choice
needed).

To get rid of choice:
in the first step look at
$\{f\in A:f(a)=g(a)\}$ and observe that, apparently, the open sets
$U_f=\{x:f(x)<g(x)+\epsilon\}$ cover $X$.
In the second step look at
$\{f\in\cl A: f<g+\epsilon\}$ and observe that the open sets
$V_f=\{x:f(x)>g(x)-\epsilon\}$ cover $X$.

I don't know if this in Johnstone's book but it is a fairly
straightforward way to eliminate choice from the usual proof.

KP Hart

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