suppose we have the metric space $(S_M,d_\infty)$, where ;
$$S_M:=\{f:\Bbb R \rightarrow\Bbb R \mid f \text{ is continuous and } |f(x)|<M \,\,\forall x \in \Bbb R \}$$
$$d_\infty:=\sup\{|f(x)-g(x)|x \in \Bbb R\}.$$
I want to prove this space is complete.
Here is what I said;
If we have a set of functions $f_n \in S_M$ which are cauchy wrt $d_{\infty}$( henceforth to be referred to as simply d), Then for all $x \in \Bbb R$ $f_n(x)$ is cauchy wrt $d$ and so convergent.
Next we define $f$ to be the pointwise limit of $f_n$. We must show that $f$ is bounded , continuous and that $d(f_n,f)\rightarrow 0,$ as $n \rightarrow \infty$. However we already have that any sequence in this space is bounded and continous so we need only show that $d(f_n,f)\rightarrow 0,$ as $n \rightarrow \infty$.
Take $\varepsilon>0$ and find N s.t. $d(f_n,f_m)<\varepsilon/ 2$ whenever $n,m \geq N$
Since $f_m(x)$ converges to $f(x)$ then there is some $N'$ st $|f_m(x)-f(x)|<\varepsilon/2$ when m$\geq N$
for $m \geq max \{N,N'\}$ we have
$$|f_n(x)-f(x)| \leq |f_n(x)-f_m(x)|+|f_m(x)-f(x)|<\varepsilon/2 + \varepsilon/2 = \varepsilon$$
so as $|f_n(x)-f(x)| < \varepsilon$, we have that $\sup|f_n(x)-f(x)|\leq \varepsilon$
since this holds for all $n \geq N$ we have $lim_{n \rightarrow \infty} \sup_{x \in X} |f_n(x)-f(x)|=0$.
which means that the limit point is contained within the space and so the space is complete.
Am I missing anything important that I should show , any major holes in logic ? I saw an example of a similar question where someone went into proving that $f$ was bounded and continuous but I think that just seems like a lot of extra work given the fact that $f$ must be both of those things by definition if it belongs to $S_M$
