Let $X$ be a topological space and let $BC( X \to \mathbb R)$ be the space of bounded continuous functions from $X$ to $ \mathbb R$ equipped with the supnorm $||*||_{\infty}$. How to prove $BC( X \to \mathbb R)$ is complete?
I know the statement is true when $X$ is a metric space and I try to mimic the proof: If $f_n$ is Cauchy, then for every $x \in X$, $f_n (x)$ is Cauchy thus converges to some $f(x)$ as $\mathbb R$ is complete. Then we can show this $f$ is bounded and $f_n \to f$ in the supnorm. But it remains to show that $f$ is continuous and I have trouble with that. Any help is appreciated.
Convergence w.r.t. the sup-norm means uniform convergence on $X$. That uniform limits of continuous functions are continuous is proved exactly as in the case of metric spaces ($\varepsilon/3$-argument).
EDIT. Here is the argument: Given $x\in X$ and $\varepsilon>0$ there is $n\in\mathbb N$ such that $\|f-f_n\|_\infty<\varepsilon/3$. Since $f_n$ is continuous at $x$ there is a neighborhood $V$ of $x$ such that $|f_n(x)-f_n(y)|<\varepsilon/3$ for all $y\in V$. Therefore, for all $y\in V$ you get $$|f(x)-f(y)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon.$$