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In Baby Rudin chapter 7 the author introduced the $\mathscr C(X)$ to denote the class of continuous functions on a metric space $X$, saying that:

If $X$ is a metric space, $\mathscr C(X)$ will denote the set of all complex-valued, continuous, bounded functions with domain $X$.

I'm wondering if the "boundedness" is required in general (although it is required for Rudin's usage here, see edit). I think I've been always seeing and probably even using notations like $1/x\in\mathscr C(0,1)$ or $x^2\in\mathscr C(\Bbb R)$ etc.

So is there a standard usage of the $\mathscr C$ notation? If not, then what is the more commonly used definition of the $\mathscr C$ notation?

Best regards.

EDIT$\quad$ In the specific context where Rudin introduced the notation, the boundedness is required, because he used $\mathscr C(X)$ to denote a linear function space equipped with the supremum norm. But my question is that is it also necessary in the general usage?

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If $X$ is any metric space, then the set $\mathscr{C}(X,\mathbf{R})$ of continuous $\mathbf{R}$-valued functions on $X$ makes perfect sense, but unless $X$ is compact, in which case boundedness is not automatic, the supremum norm will not generally be well-defined on this space, because continuous functions need not be bounded in general. If one restricts to the subset $\mathscr{C}^b(X,\mathbf{R})$ of bounded continuous functions, then one has the supremum norm. In general, there are a number of topologies one might put on $\mathscr{C}(X,\mathbf{R})$, but the most commonly-used one, at least if $X$ is locally compact, is probably the compact open topology.

But, again, to answer your question, yes, the space of continuous functions on a metric space is a perfectly reasonable object to consider. But you can't, in general put the supremum norm on it.

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  • $\begingroup$ Thank you for your time. May I ask one more question? Since I'm not familiar with topology, I don't really know what "locally compact" means (compact in a small closed neighbourhood for every point, I guess?), but do my examples, like $1/x\in\mathscr C(0,1)$ and $x^2\in\mathscr C(\Bbb R)$, belong to the "locally compact" cases, namely, the "most commonly used ones" that you mentioned? $\endgroup$
    – Vim
    Commented Aug 9, 2015 at 16:41
  • $\begingroup$ Dear @Vim, A metric space $X$ is locally compact if for each point $x\in X$, there is a compact set $C$ in $X$ containing an open neighborhood of $x$. The spaces $(0,1)$ and $\mathbf{R}$ are both locally compact, but not compact. It doesn't make sense to ask about a function being locally compact; only a space. Anyway, it is logically perfectly fine to talk about the space $\mathscr{C}(X,\mathbf{R})$ for any metric space whatsoever, locally compact or not. But if you want to put a topology on it, you might want to impose further conditions on $X$. $\endgroup$ Commented Aug 9, 2015 at 16:49
  • $\begingroup$ Thank you so much for your detailed explanation, that's very kind of you! Now I think I got the idea. $\endgroup$
    – Vim
    Commented Aug 9, 2015 at 16:54

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