I was reading through a proof of the Stone - Weierstrass approximation theorem:
Let $X$ be a compact metric space. Let $A\subseteq C^\mathbb{R}_b(X)$ an algebra which contains the constant functions and separates points of $X$. Then $A$ is dense with respect to the metric induced by $\lVert . \rVert_u$ in $C^\mathbb{R}_b$
Notation:
$\lVert . \rVert_u: \text{sup norm}$.
$C^\mathbb{R}_b(X): \text{Bounded and continous real valued functions defined on X}$
In one part of the proof one wants to show that if $f\in A$, then $|f|\in \overline A$ (the uniform closure of $A$ in $C^\mathbb{R}_b(X)$).
The proof is as follows:
From previous steps one has a sequence $(p_n)_n$ of real polynomials, pointwise increasing in $[0,1]$, that converges uniformly to the function $\sqrt x$ for $x\in[0,1]$. If we consider the sequence $$ \left(p_n\left(\frac{f^2}{\lVert f \rVert_u}\right)\right) $$ this converges uniformly to $\sqrt\frac{f^2}{\lVert f \rVert_u} = \frac{|f|}{\lVert f \rVert_u}$, hence $|f|\in \overline A$.
What I understand is that using Dini's theorem for uniform convergence one can say that convergence is uniform.
What I don't understand is what makes $|f|\in \overline A$? How I understand it is that the approximation is being done by the polynomials. But my understading of uniform closure is that of wikipedia: 'The uniform closure of a set of functions A is the space of all functions that can be approximated by a sequence of uniformly-converging functions on A.
However I'm not sure if $p_n\in A$ for every $n$. (I'm aware that it is possible that they could be contained in $A$, but what if $A$ is some algebra that excludes them?).
I'll appreciate if someone gives me a hand on my confussion.
If $p$ is any polynomial with real coefficients and $g\in A$, then $p(g)\in A$. This is just because $A$ is an algebra, so it is closed under multiplication and addition. We have $p(g)=\sum_{i=0}^n c_i g^i$ for some constants $c_i\in\mathbb{R}$. This expression is just a big sum of products of elements of $A$, so it is in $A$ as well.
So in particular, since $f^2/\|f\|_u\in A$, $p_n(f^2/\|f\|_u)\in A$ for each $n$. Thus their uniform limit $\sqrt{f^2/\|f\|_u}$ is in $\overline{A}$.