# Why is the space of bounded continuous functions complete?

The space of bounded continuous functions from some space to a complete metric space is complete. Now, if we consider the sequence $\lbrace f_n(x)=x^n \mid f_n \colon [0,1]\to [0,1] \rbrace_n^{\infty}$, then $f_n$ is continuous and bounded for all $n$, and converges to a discontinuous function. Since $\lbrace f_n \rbrace$ converges, it is Cauchy. But $\{f_n\}\subset B_C(\mathbb{R})$ (where $B_C (X)$ is the space of bounded continuous functions from $X$ to $X$). How does that comply with the fact that $B_C(\mathbb{R})$ is complete?

Clearly, I'm missing something here. I'd appreciate an explanation. Thanks.

• This is what you are missing: Which topology on $B_C(\Bbb R)$ is considered? – Hagen von Eitzen Jul 10 '18 at 10:42
• Are you sure that the sequence converges uniformly? – egreg Jul 10 '18 at 10:43
• @HagenvonEitzen Gee... $B_C$ with the sup norm! I completely forgot about that. Thanks! – Bob the Builder Jul 10 '18 at 10:49
• The first sentence in the question is not generally true. You need that the functions map to a complete metric space. – Andreas Blass Jul 11 '18 at 2:58

$$\|f\|=\sup_{x \in E}|f(x)|$$
where $f: E \to \mathbb{R}$ or $\mathbb{C}$. You can check that this indeed gives us a norm. Now define $d(f,g)=\|f-g\|$ to get a metric on the space of bounded, continuous functions.
When you talk about a metric space, you must know what metric you are working with. Surely, if $f_n \to f$ in the norm given above, $f$ will be continuous. (this is just another way to state the standard theorem that the limit of a uniformly convergent sequence of continuous functions is continuous)