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.

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    $\begingroup$ This is what you are missing: Which topology on $B_C(\Bbb R)$ is considered? $\endgroup$ Jul 10, 2018 at 10:42
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    $\begingroup$ Are you sure that the sequence converges uniformly? $\endgroup$
    – egreg
    Jul 10, 2018 at 10:43
  • $\begingroup$ @HagenvonEitzen Gee... $B_C$ with the sup norm! I completely forgot about that. Thanks! $\endgroup$ Jul 10, 2018 at 10:49
  • $\begingroup$ The first sentence in the question is not generally true. You need that the functions map to a complete metric space. $\endgroup$ Jul 11, 2018 at 2:58

1 Answer 1


The space of bounded continuous functions is complete with respect to the metric induced by the following {supremum} norm:

$$\|f\|=\sup_{x \in E}|f(x)|$$

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)

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    $\begingroup$ Thanks. Would have voted, but I'm a newbie (in math too) so can't... $\endgroup$ Jul 10, 2018 at 10:51

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