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Let $(X, d)$ be a (complete) metric space, and $C(X)$ be the space of continuous maps over $X$. If $X$ is compact, one often uses the topology of uniform convergence when analyzing $C(X)$. If $X$ is non-compact, there are a bunch of other topologies that one can attach to $C(X)$: the compact-open topology, the strong topology, ...

My question is the following: Why don't we just use the topology of uniform convergence on $C(X)$ regardless of the compactness of $X$? Generally, the first answer I get to this question is that the metric $d(f, g) = \sup_{x \in X} d(f(x), g(x))$ becomes unbounded when $X$ is not compact. Why is that a problem? We can always consider it an extended metric, and all the theory carries over without any problems. Even if infinite-valued metrics turn out to break the theory somehow, we can always use $d'(f, g) \equiv d(f, g)(1 + d(f, g))^{-1}$.

Therefore, I am suspecting that the topology of uniform convergence has some other disadvantage(s) when treating functions over non-compact spaces. What are these disadvantages? What concern(s) made mathematicians invent the other topologies? Why is this topology not used commonly?

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  • $\begingroup$ what do you mean by "the topology of uniform convergence"? what are the open sets? $\endgroup$ Commented Apr 12, 2014 at 21:08
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    $\begingroup$ @IttayWeiss: Even if $X$ is not compact, $d(f, g)$ defines an (extended) metric, right? I am talking about the topology whose basis consists of open balls defined through this metric. $\endgroup$
    – iheap
    Commented Apr 12, 2014 at 21:29

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A few things that come to mind:

  • The topology of uniform convergence is too fine on e.g. $C(X)$ where $X$ is an open subset of some $\mathbb{R}^n$. In many situations, you don't have uniform convergence, but locally uniform convergence, and the locally uniform convergence is sufficient for many theorems (the limit of a locally uniformly convergent sequence of continuous functions is continuous; interchange of limit and integral for integrals over compact subsets [path or surface integrals]), and when it isn't, globally uniform convergence is often also not sufficient (integrals over sets of infinite measure).

  • With the topology of uniform convergence, $C(X)$ is not a topological vector space if there are unbounded continuous functions on $X$ (scalar multiplication is not continuous), nor is it a topological ring then.

The topology of uniform convergence is a natural topology for the space $C_b(X)$ of bounded continuous functions, however. Note that when $X$ is compact, we have $C(X) = C_b(X)$.

Still, for some purposes, the topology of uniform convergence can be an/the adequate topology on $C(X)$ to consider for non-compact $X$, but there are purposes where it is inadequate, and hence other topologies have to be considered.

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  • $\begingroup$ Thanks for answering. Can you give an example for the inadequacy of the topology when $X$ is an open subset of $\mathbb{R}^n$? $\endgroup$
    – iheap
    Commented Apr 13, 2014 at 1:21
  • $\begingroup$ I think the fact that $C(X)$ isn't a topological vector space then is a pretty good example. Also that for example power series expansions of analytic functions don't converge uniformly in general, only locally uniformly. I'm not saying that the topology of uniform convergence is never adequate for such an $X$ (although I'm not aware of it being used then), but other topologies have nice properties the topology of uniform convergence hasn't. $C(X)$ is a Fréchet space in the topology of locally uniform convergence for open $X\subset\mathbb{R}^n$. $\endgroup$ Commented Apr 13, 2014 at 1:43
  • $\begingroup$ For metric spaces the answer is true, but, as a general remark, there are plenty of non-T₁ spaces (including non-compact ones) that do not admit real-valued continuous functions but constants. IMHO is it not difficult to realize which topological vector space will constitute such C (X). $\endgroup$ Commented Aug 30, 2014 at 20:32
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    $\begingroup$ @IncnisMrsi Yes, if all continuous real-valued functions on a space $X$ are constant, then $C(X)$ with the topology of uniform convergence is a (slightly boring, however) topological vector space. Then it coincides with $C_b(X)$. As mentioned, it is not a topological vector space in the topology of uniform convergence if there are unbounded continuous functions. $\endgroup$ Commented Aug 30, 2014 at 20:44

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