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I just learned about metric spaces and their basic properties and definitions, such as the definition of a continuous function that maps two metric spaces. However, when I say "continuity" on a metric space, I am not talking about continuos functions, but rather the continuity of the space itself. Like, how do you characterize the property of a metric space being "continuous", kind of like the real numbers, that is to say, having no gaps. In the real numbers, we usually characterize this with the completeness axiom, however I don't think that can be directly translated into a metric space since there is no least or bigger than. I remember reading that an equivalent formulation of the completeness axiom is the nested interval theorem, so I figured maybe it is possible to create an analogous formulation with metric spaces, instead of nested intervals, nested closed sets. Would that work? I also read about the property of completeness in a metric space, and that is something about cauchy sequences always converging to a point on the space, but I am not sure if that completeness is the same completeness we talk about in the real numbers aka having no gaps. So in conclusion I want to know, what are some characterizations of the "continuity" of a metric space? and also I would like to know if my idea of generalizing the nested interval theorem would work in the desired way. Thanks.

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    $\begingroup$ Why do you stick to the term "continuity" when you want to ask about completeness? Unless your goal is to confuse people I recommend sticking with the latter. As for completeness, "complete metric space" means exactly one thing, and that is Cauchy completeness. It is what we mean when we talk about completeness of $\mathbb{R}$, yes. If you visit Wikipedia, you'll find an equivalent characterization via nested sets, too. $\endgroup$ Commented Aug 10 at 0:01
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    $\begingroup$ I think that what you are finding out is that your notion of "continuity" becomes naive for abstract metric spaces. When we imagine the real line it is easy to characterize intuitively what subsets of it are continuous and those which aren't (that is, the ones that have "a gap"). But an abstract metric space is not generally embedded in any completely ordered and complete space such as the reals. I believe the best we can do is to refer to Cauchy completeness. $\endgroup$ Commented Aug 10 at 0:10
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    $\begingroup$ I'd also like to point out that Cauchy completeness is a satisfying notion of "having no gaps", informally, of course. The fact that a sequence that has its terms getting closer and closer together is guaranteed to converge means, pictorically, that you can tread your space with smaller and smaller steps, and you will always arrive arbitrarily close of somewhere; that is, you will never find a "gap" (granted that in this abstract space there isn't an "outside", so the notion of "gap" must be thought about carefully). $\endgroup$ Commented Aug 10 at 0:17
  • $\begingroup$ @BenSteffan I use the term "continuity" in quotes for lack of a better term, since I wasn't sure if completeness was what I was referring to. But it seems like it is. $\endgroup$
    – zlaaemi
    Commented Aug 10 at 0:22
  • $\begingroup$ @Lourenco: really? Do you consider the Cantor set to "have no gaps"? $\endgroup$ Commented Aug 10 at 0:46

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I do not agree with Ben's comment that completeness is the right formalization of what you're asking about. For example the Cantor set is also complete but I think it's hard to argue that it has "no gaps"!

I think a better way to formalize "no gaps" is that $\mathbb{R}$ is connected, which informally means that it can't be separated into two pieces. This neatly distinguishes $\mathbb{R}$ from the Cantor set, which is totally disconnected. The connectedness of $\mathbb{R}$ is related to but not the same thing as its completeness and is one of its most important basic properties; for example it immediately implies the intermediate value theorem.

In mathematics we only use the term "continuous" to refer to functions, not to spaces; for spaces we have more precise terms like connectedness and we use those terms instead. The naive distinction between "continuous" and "discrete" breaks down badly; for example already $\mathbb{Q}$ is neither (it is also totally disconnected). So we need more precise terminology.

Even connectedness may not be quite right, since for example $\mathbb{R}^2 \setminus \{ 0 \}$ is also connected and one could argue it has a "gap," namely the hole in the middle. This example is not complete wrt the induced metric, so one could ask for both complete + connected. Unfortunately $\mathbb{R}^2 \setminus \{ 0 \}$ is homeomorphic to the cylinder $S^1 \times \mathbb{R}$, which has a metric making it complete! This metric makes the hole "infinitely far away." One wonders whether the cylinder qualifies as having a "gap" or not; at this point we really need more precise terminology to say what exactly we mean. Spaces can vary in many different ways!

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  • $\begingroup$ Thanks for your reply! So I think a better way to express the "no gap" property for a metric space would be that for every point in the metric and every distance greater than zero (and lesser than the diameter of the entire space), the closed ball of that distance set minus the open ball on the same point same distance is non empty. In other words, there are always "boundary" points in the closed balls of any distance lesser than the diameter. So what do you think? Would a metric like that be connected? Complete? both? Or maybe neither? $\endgroup$
    – zlaaemi
    Commented Aug 10 at 1:10
  • $\begingroup$ Maybe something like "simply connected" works. $\endgroup$ Commented Aug 10 at 1:21
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    $\begingroup$ @lucenaposition Well, $\mathbb{R^3}\setminus\{0\}$ is simply connected, but one might say it also contains a hole. Here we have to move to higher homotopy/homology groups. I guess there is just no real definition of "no gaps". There are different types, and we just have to work with the precise definitions. $\endgroup$
    – Mark
    Commented Aug 10 at 1:26
  • $\begingroup$ @zlaaemi: well, $\mathbb{R} \setminus \{ 0 \}$ satisfies that condition, and is neither connected nor complete. $\endgroup$ Commented Aug 10 at 2:58
  • $\begingroup$ What about for any point $x$ and any distance $y$, the set $\{k|\text{dist}(k,x)\le y\}$ is compact? $\endgroup$ Commented Aug 10 at 11:11

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