Proof of Stone-Wierstrass theorem

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SUMMARY

The discussion focuses on proving that the set \( U_{(a,b)} \) is open in the context of the Stone-Weierstrass theorem. The proof begins with the assumption that \( A \) is an algebra of continuous functions on a compact Hausdorff space \( X \) that separates points. The user seeks clarification on how to demonstrate that the set defined by \( U_{(a,b)} = \{ x \mid f_{(a,b)}(x) < f(x) + \epsilon \} \) is open, ultimately leading to the conclusion that \( U_{(a,b)} \) can be expressed as the pre-image of an open interval, confirming its openness.

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Homework Statement



During the proof of the Stone-Wierstrass theorem we define a particular set and claim it is open. I can't figure out how to show it is open.
I wrote out the proof until that point to show how it is constructed, I apologize for the long windedness.

Assume that X is a compact hausdorf space[tex]A\subset C(X)[/tex]is an algebra that sperates points and has a continuosu function in it. then [tex]\overline{A}=C(X).[/tex]

Proof:

Since A separates points then for any [tex]a,b\in X[/tex] there exists [tex]h\in A[/tex] such that [tex]h(a)\neq h(b)[/tex]. define [tex]g(x)=\frac{h(x)-h(a)}{h(b)-h(a)}.[/tex] Then [tex]g(a)=0[/tex]and [tex]g(b)=1[/tex].

for any function [tex]f\in A[/tex] define [tex]f_{\left(a,b\right)}(x)=\left(f(b)-f(a)\right)g(x)+f(a)\text{then }f_{\left(a,b\right)}(a)=f(a)\text{ and }f_{\left(a,b\right)}(b)=f(b)[/tex].

Set [tex]\epsilon>0[/tex] and define [tex]U_{\left(a,b\right)}=\left\{ x\quad|f_{\left(a,b\right)}(x)<f(x)+\epsilon\right\} .[/tex]

My question is how do i gaurantee that [tex]U_{\left(a,b\right)}[/tex] is open? My professor took it for granted. I'm sure it has to do with the pre-image of f but i can't write it out.

Homework Equations





The Attempt at a Solution


 
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With such a questions, you have to rewrite [tex]U_{(a,b)}[/tex] in such a way so that it is clear that it's open. In this case:

[tex]U_{(a,b)}=\{x~\vert~f_{(a,b)}(x)<f(x)+\epsilon\}=\{x~\vert~(f_{(a,b)}-f)(x)<\epsilon\}=(f_{(a,b)}-f)^{-1}(]-\infty,\epsilon[)[/tex].
 

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