Why uniform closure $\mathscr{B}$ of an algebra $\mathscr{A}$ of bounded complex functions is uniformly closed? Let $\mathscr{A}$ be an algebra of bounded complex functions. (Or if necessary, continuous and domain of functions is compact)
Definition:
$\mathscr{B}$ is uniformly closed iff $f\in\mathscr{B}$ whenever $f_n\in \mathscr{B} (n=1,2,\cdot)$ and $f_n\rightarrow f$ uniformly.
$\mathscr{B}$ is the uniform closure of $\mathscr{A}$ iff $\mathscr{B}$ is the set of all functions which are limits of uniformly convergent sequences of members of $\mathscr{A}$.
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Let $\mathscr{B}$ be a uniform closure of $\mathscr{A}$.
How do i prove that $\mathscr{B}$ is uniformly closed in ZF?
Does Stone-Weierstrass theorem require choice since it is critical to prove Stone-Weierstrass Theorem?
 A: I would answer for the first question.
Theorem: Let $\mathscr{B}$ be a uniform closure of $\mathscr{A}$. (Where $\mathscr{A}$ - algebra consisting of bounded functions). Then $\mathscr{B}$- uniformly closed algebra.
Proof: If $f\in\mathscr{B}$ and $g\in\mathscr{B}$, then there are uniformly convergent sequences $f_n\in\mathscr{A}$ and $g_n\in\mathscr{A}$ such that $f_n\to f$ and $g_n\to g$. Since the functions are bounded, we can write:
$$f_n+g_n\to f+g$$
$$f_ng_n\to fg$$
$$cg_n\to cg$$
Where $c$ is constant from the field.
So $f+g\in\mathscr{B} $,$fg\in\mathscr{B} $,$cg\in\mathscr{B}$. So $\mathscr{B}$ is algebra.
Let $f_n$ is uniformly convergent sequence of elements from $\mathscr{B}$. There are functions $g_n$ such that $|f_n(x)-g_n(x)|<\frac{1}{n}$. If $f_n\to f$ then it is clear that $g_n\to f$, so (by definition of $\mathscr{B} $) $f\in\mathscr{B}$, so $\mathscr{B} $ is uniformly closed. 
$\blacksquare$
A: Since $\mathscr{B}$ consists of the limits of uniformly convergent sequences in $\mathscr{A}$ and since the limits must be distinct from each other,
I think every point of $\mathscr{B}$ is isolated in metric sense.
So the cloure of $\mathscr{B}$ is $\mathscr{B}$ itself.
So, by Theorem 2.27 of Rudin's PMA, $\mathscr{B}$ is closed.
Hence, $\mathscr{B}$ is uniformly closed.
A: I think that $\mathscr{B}$ is the uniform closure of $\mathscr{A}$, for any $x$, $\mathscr{B}$ is the uniform closure of $\mathscr{A}$.
Recall the closure $\overline{E}$ of the set $E$ is closed. Here, $f_n(x) \rightarrow f(x)$ for any $x$ is holded. So $\mathscr{B}$ is uniform closed.
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