Cfrgtkky

let $A$ be a set of continuous functions over the closed interval $[0,1]$, which applies to the following conditions:

$ 1: forall f in A , forall x in [0,1] , f(x) geq 0 $

$2: forall f, g in A , f + g in A $

$3: forall x in [0,1], textbf{there is} f in A quad textbf{so that} f(x) > 0 $

By Stone -Weierstrass theorem , How can I Prove "that there is $ h in A$ so that $ forall x in [0,1], h(x) > 0$" ?.

Should I use the below theorem?

THEOREM:

Let A $subset C(K)$ such that

1) A is a subalgebra with unity 1

2) For each $ x_1, x_2 in K $ with $ x_1 neq x_2 $, exist $f in A$ such that f($ x_1$) $ neq $ f($ x_2$).

Then $ overline A = C(K)$, where C(K) is the space of continuous functions over a compact space

I would use compactness: find for every $x$ a function $f_x$ as in 3), also take $delta_x>0$ such that $f_x$ is positive inside the $delta_x$-nbd of $x$. Now apply compactness to reduce to finitely many function and add them.

Addendum: there are finitely many $x_1$, ..., $x_n$ such that $[0,1]$ is covered by the corresponding $delta_{x_i}$-nbds. Now take $f=f_{x_1}+cdots+f_{x_n}$.