I learned today in class that $\mathcal C([a,b])$ is complete with the supremum norm. That is, any uniformly convergent sequence $(f_n)$ is Cauchy, and the converse is true. I asked my teacher about other subsets of $\mathbb R$ where this held, and he pointed out that it's true for any subset, as long as the supremum norm makes sense, i.e. the functions are bounded. This made sense to me.

I wonder, though, for what other kinds of spaces this is true. So I ask the question: When is $\mathcal C(X)$ complete? Say $X$ is some metric space, and we induce (if possible?) some metric on $\mathcal C(X)$ from the metric in $X$. I'm sure some extra hypotheses must be added for the question to be meaningful. In that case, so be it.

I guess what I'm asking is for a result on the completeness of $\mathcal C(X)$ that is as general as possible. If I'm wrong though, about extra hypotheses being necessary, I would much enjoy a result where $X$ is not even a metric space, and we have to talk about filters and the sort (though I'm just beginning to learn general topology). Thank you all.

Edit: I'm referring to real valued functions.

Another edit: I realized that the first result I talked about was actually on $\mathbb R^{[a,b]}$, so drop continuity. As in, replace $\mathcal C(X)$ with $\mathbb R^X$. Maybe this makes things impossible though, so my comment about extra hypotheses still holds.

  • $\begingroup$ By $\mathcal{C}(X)$, you denote the space of real- (or complex-) valued continuous functions on $X$? $\endgroup$ – Daniel Fischer Feb 12 '14 at 12:56
  • $\begingroup$ @DanielFischer Yes, I mean real valued continuous functions, I'll add that in. $\endgroup$ – GPerez Feb 12 '14 at 12:58
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    $\begingroup$ We can topologize $\mathcal{C}(X)$ with the norm $\lVert f\rVert_\infty = \sup \{ \lvert f(x)\rvert : x \in X\}$ whenever every continuous function on $X$ is bounded. That property is pseudocompactness. For the completeness, the completeness of $\mathbb{R}$ is responsible, and that the limit of a uniformly convergent sequence of continuous functions is continuous. So it works on every pseudocompact space. $\endgroup$ – Daniel Fischer Feb 12 '14 at 13:03
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    $\begingroup$ If we don't restrict ourselves to $\lVert\,\cdot\,\rVert_\infty$, we can also endow $\mathcal{C}(X)$ with other useful topologies. E.g. if $X$ is a locally compact (Hausdorff) space, the topology of uniform convergence on compact subsets of $X$ is a useful topology. If $X$ is the union of countably many compact subsets, that topology is metrizable, and it's complete (in the usual metric defining the topology). $\endgroup$ – Daniel Fischer Feb 12 '14 at 13:06

The only requirement is the existence of the $\sup$. $X$ compact is enough. $C(X)$ is a subset of ${\Bbb R}^X$, so the important structure is the structure of ${\Bbb R}$.

  • $\begingroup$ So you mean if we can use the supremum norm? I was hoping for a result on not necessarily normed spaces. This probably is too ambiguous though, I really don't know. $\endgroup$ – GPerez Feb 12 '14 at 13:05
  • $\begingroup$ Your question: "is complete with the supremum norm". Completeness depends of more than topology. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 12 '14 at 14:01
  • $\begingroup$ Oh? I thought it at least could be stated with a metric... $\endgroup$ – GPerez Feb 12 '14 at 14:36
  • $\begingroup$ If my memory serves me, if your space is metrizable, then is normable. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 12 '14 at 14:49
  • $\begingroup$ Only if it's a vector space... $\endgroup$ – GPerez Feb 12 '14 at 15:35

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