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I'm just second-marking some exam scripts, and I wanted to leap on a question and made the following pedantic remark concerning the model answers: "if the metric space is empty then this proof doesn't work because something which is supposed to be finite is $-\infty$. Hence this proof is incomplete -- it's missing the line "If the space is empty then the result is trivial".

But then another question made me wonder whether in fact the lecturer of the course had actually put as part of the definition of metric space, that it be non-empty. A quick trip to Wikipedia revealed that there also the definition required the space to be non-empty.

Why?

I certainly don't want to require that a topological space be non-empty, for example. There is presumably some sensible reason why the general convention for topological spaces has been to allow the empty set (this I understand!) but the general convention for metric spaces appears to be not to allow it...

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    $\begingroup$ Must... resist... saying... because there's no point to the empty metric space. $\endgroup$
    – Gunnar Þór Magnússon
    Jun 13, 2011 at 17:51
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    $\begingroup$ The metric space with one point isn't interesting, either, so what's $\frac{the}{a}$ point? :-) $\endgroup$
    – Tim van Beek
    Jun 13, 2011 at 18:11
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    $\begingroup$ From the comment thread to Qiaochu's answer I'd assume that it is precisely to avoid such pointless debates when teaching to first year students. Let's focus on the interesting things not on the occasional pathological special case. See also too simple to be simple on the nlab. $\endgroup$
    – t.b.
    Jun 13, 2011 at 18:43
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    $\begingroup$ @Kevin: I don't think I'm missing it. I just wanted to direct your attention to that page. My second sentence wasn't at all directed at you or at your question but rather a comment on why I would not insist on the empty metric space in a first year course and a comment on the diameter noise. Sorry if that was phrased in a bit an unfortunate way. No offense intended. $\endgroup$
    – t.b.
    Jun 13, 2011 at 20:09
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    $\begingroup$ For what it's worth: I just checked in more than a dozen books whether metric spaces are explicitly required to be non-empty. Among them Dugundji, Kelley, Munkres and the like. The only book I found in which metric spaces were explicitly required to be non-empty was Royden's Real Analysis, third edition and two very basic analysis texts in German. $\endgroup$
    – t.b.
    Jun 13, 2011 at 20:36

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Rudin doesn't require that a metric space be non-empty. I agree that there is no good reason for a convention which says otherwise. For example, for those of you who are convinced by such reasons, we want subspaces of metrizable spaces to be metrizable.

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    $\begingroup$ Qiaochu: you have restored my faith in humanity! $\endgroup$
    – Kevin Buzzard
    Jun 13, 2011 at 17:54
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    $\begingroup$ Excluding the empty metric space would deprive us of such interesting debates as «what is the diameter of the empty metric space?»! $\endgroup$
    – Mariano Suárez-Álvarez
    Jun 13, 2011 at 18:02
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    $\begingroup$ @Mariano: the diameter of the empty metric space is the supremum of the empty subset of $\mathbb{R}_{\ge 0}$, so it's $0$. @Pete: I think it's a good habit to instill in students a healthy respect for the empty case. For example, it prevents them from trying to do things like applying Zorn's lemma to an empty poset. $\endgroup$
    – Qiaochu Yuan
    Jun 13, 2011 at 18:03
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    $\begingroup$ @Mark: that's the wrong statement. The correct statement is that the image of a compact space is compact, and that's true either way. A corollary is that for non-empty compact spaces, real-valued functions to $\mathbb{R}$ have maxima. I think it is completely sensible to require the adjective "non-empty" here since we are trying to show the existence of something. $\endgroup$
    – Qiaochu Yuan
    Jun 13, 2011 at 18:06
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    $\begingroup$ If the diameter of $X$ is defined to be $\inf\{r\geq 0: d(x,y)\leq r\text{ for all }x,y\in X\}$, then the diameter of the empty metric space is $0$. $\endgroup$
    – Jonas Meyer
    Jun 13, 2011 at 18:15

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