Uniform closure of algebra of bounded functions is uniformly closed

[psie]

TL;DR

I'm reading theorem 7.29 in Rudin's PMA. I struggle with a seemingly basic conclusion Rudin makes regarding the uniform closure $\mathcal B$ of some algebra $\mathcal A$ of bounded functions, namely that it is uniformly closed.

First some definitions:

Definition A family $\mathcal A$ of complex or real functions defined on a set $E$ is said to be an algebra if it is closed under addition, multiplication and scalar multiplication.

Definition If $\mathcal A$ has the property that $f\in\mathcal A$ whenever $f_n\in \mathcal A$ ($n=1,2,\ldots$) and $f_n\to f$ uniformly on $E$, then $\mathcal A$ is said to be uniformly closed.

Definition Let $\mathcal B$ be the set of all functions which are limits of uniformly convergent sequences of members of $\mathcal A$. Then $\mathcal B$ is called the uniform closure of $\mathcal A$.

Theorem 7.29 Let $\mathcal{B}$ be the uniform closure of an algebra $\mathcal{A}$ of bounded functions. Then $\mathcal{B}$ is a uniformly closed algebra.

After Rudin has shown that $\mathcal B$ is an algebra, he invokes another theorem to claim it is uniformly closed. That theorem states that if $X$ is a metric space and $E\subset X$, then $\overline E$ is closed. I don't understand Rudin's reasoning here, and I've tried to go back to the basics.

If we denote $X=\mathbb R^S$ the set of all bounded functions from $S$ to $\mathbb R$, equipped with the uniform metric, I'm trying to show that given $\mathcal A\subset X$, the uniform closure $\mathcal B$ of $\mathcal A$ is simply $\overline{\mathcal A}$. That is, I'm trying to prove both inclusions in $\overline{\mathcal A}=\mathcal B$.

One direction is simple. Observe that if $f\in\overline{\mathcal A}$, then there exists a sequence $(f_n)\in\mathcal{A}$ converging to $f$. Since the metric is the uniform metric, the convergence is uniform. Thus by definition $f\in \mathcal B$.

In the other direction, I'm getting stuck. Let $f\in \mathcal B$. Then by definition, there is $(f_n) \in \mathcal A$ so that $(f_n)$ converges uniformly to $f$ on $S$. That is, for all $\epsilon >0$ there is $N\in \mathbb N$ so that $$|f_n(x) - f(x)| <\epsilon,$$for all $x\in S$ and $n\ge N$. The above inequality implies that $$d_\infty (f_n, f) = \sup_{x\in S} |f_n(x) - f(x)| \le \epsilon,$$ for all $n\ge N$. So $f_n \to f$ in $(X, d_\infty)$. Here's where I'd like to conclude somehow that $f$ must be a limit point of $\mathcal A$ and thus in $\overline{\mathcal A}$, but I'm unable to do this since $f$ certainly could be an isolated point, or?

 

[psie]

Ok, I think I've had some time to look at my topology book, and apparently a point $f$ is in the closure of $\mathcal A$ (which is a subset of a metric space) if and only if there's a sequence in $\mathcal A$ converging to $f$.

I used one direction of this statement already. Where I'm stuck I need to use the other direction.

 

[Euge]

Let $(X,d_\infty)$ be the space of all bounded functions $f : E \to \mathbb{C}$, equipped with the sup metric $d_\infty$. Convergence of a sequence of functions with respect to $d_\infty$ is the same as uniform convergence of the sequence. So the condition that $\mathcal{B}$ is the uniform closure of $\mathcal{A}$ means $\overline{\mathcal{A}} = \mathcal{B}$ in $X$, and the condition that $\mathcal{B}$ is uniformly closed means $\overline{\mathcal{B}} = \mathcal{B}$. So Theorem 7.29 implies that if $\mathcal{B} = \overline{\mathcal{A}}$, then $\overline{\mathcal{B}} = \mathcal{B}$, i.e., $\mathcal{B}$ is closed in $X$. This indeed follows from the fact that the closure of a subset of a metric space is closed:

Spoiler: Proof

Let $M$ be a metric space and $S\subset M$. Given $m\in M \setminus S$, there is an open ball $B\ni x$ disjoint from $S$. Thus $S$ is contained in the closed set $M\setminus B$, which implies $\overline{S} \subset M\setminus B$. Therefore $B\subset M\setminus \overline{S}$, and since $m$ was arbitrary, $M\setminus \overline{S}$ is open. In other words, $\overline{S}$ is closed.