Issues with theorem 7.15 from Rudin’s PMA.

Question

Some prerequisite definitions and theorems. Given a metric space $X$, let
$$
\mathscr{C}(X) = \{f: X \rightarrow \mathbb{C} \; | \; f \text{ bounded and continuous.}\}.
$$
Next we have the familiar sup-norm for elements of $\mathscr{C}(X)$ as
$$
\|f\| = \sup_{x \in X} |f(x)|,
$$
and from this norm we can always induce a metric on $\mathscr{C}(X)$ by defining $d(f,g) = \|f-g\|$.

Theorem 7.8: A sequence of functions $(f_n)$ defined on a set $E$, converges uniformly on $E$ if and only if for every $\varepsilon>0$ there exists an integer $N$ such that $n,m \geq N$, $x \in E$ implies that
$$
|f_n(x)-f_m(x)| < \varepsilon.
$$

Theorem 7.12: If $(f_n)$ is a sequence of functions defined on $E$, and if $f_n \rightarrow f$ uniformly on $E$, then $f$ is continuous on $E$.

Theorem 7.15: The metric space $(\mathscr{C}(X),\|f-g\|)$ is complete. The proof given by Rudin is as follows.

Let $(f_n)$ be a Cauchy sequence in $\mathscr{C}(X)$. This means that to each $\varepsilon>0$ corresponds an $N$ such that $||f_n-f_m|| < \varepsilon$ if $n \geq N$ and $m \geq N$. It follows from theorem 7.8 that there is a function $f$ with domain $X$ to which $(f_n)$ converges to uniformly. By Theorem 7.12, $f$ is continuous. Moreover $f$ is bounded since there is an $n$ such that $|f(x)-f_n(x)|< 1$ for all $x \in X$, and $f_n$ is bounded. Thus $f \in \mathscr{C}(X)$, and since $f_n \rightarrow f$ uniformly on $X$, we have $||f-f_n|| \rightarrow 0$ as $n\rightarrow \infty$.

I think I have two main issues.

1. So I understand that when we're first defining a Cauchy sequence in $\mathscr{C}(X)$ we need to be working with the norm, but why is it the case that theorem 7.8 tells us that $(f_n)$ converges to some $f$ uniformly on $X$? When Rudin writes theorem 7.8, are we supposed to realize that closeness will depend on which metric space we're considering? Specifically should 7.8 really be saying that for every $\varepsilon>0$ there's an integer $N$ for such that for all $n>m\geq N$ we have
   $$
   d(f_n,f_m) < \varepsilon , \; (\forall x \in E)?
   $$
2. Looking past the first issue, I get that using theorem 7.8 tells us that $(f_n)$ actually converges uniformly on $X$ to some $f$, and then we go on to show that $f$ is bounded and continuous. In my mind isn't this enough to show completeness? What is the use of the last sentence of the proof? What are we conveying when we note that $\|f-f_n\| \rightarrow 0$ as $n \rightarrow \infty$? Isn't the point of using theorem 7.8 to show that this convergence occurs in the first place?

Sorry it's a lot and I don't know if I've included all of the prerequisites. If anything is ambiguous please let me know. Any help here is greatly appreciated!

Answer 1

1. By definition, if for each $\varepsilon > 0$, there exists an $N \in \Bbb N^*$ such that $\Vert f_n - f_m \Vert < \varepsilon$ for any $n,m > N$, then $$ \sup_{x \in E} \vert f_n(x) - f_m(x) \vert < \varepsilon, $$ so whenever $m,n > N$ and $x \in E$, $\vert f_n(x) - f_m (x) \vert < \varepsilon$. Thus by Theorem 7.8, $(f_n)$ converges uniformly on $E$. Here we don't require anything on the metric to proceed, just plain definition of $\sup$. But for Theorem 7.8, we never required functions to be continuous and bounded, so it is doubtful to use the notation $d(f_n, f_m)$.
2. To show completeness we obviously need to show $f_n \to f$ w.r.t. this norm $\Vert - \Vert$. The proof uses this sentence to summarize what we are looking for. Also note that we are only allowed to use $\Vert - \Vert$ after we proved the continuity and the boundedness.