Why are metric spaces non-empty?

Question

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...

Must... resist... saying... because there's no point to the empty metric space. — Gunnar Þór Magnússon, Jun 13 '11 at 17:51
The metric space with one point isn't interesting, either, so what's $\frac{the}{a}$ point? :-) — Tim van Beek, Jun 13 '11 at 18:11
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. — t.b., Jun 13 '11 at 18:43
@Theo: I think you're missing the point. There is a sensible reason why 1 is not a prime number (if it were then things would not be uniquely the product of primes). My question is whether there is an analogous sensible reason as to whether the empty space is not a metric space. The noise about the diameter of the empty set is irrelevant. — Kevin Buzzard, Jun 13 '11 at 19:53
@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. — t.b., Jun 13 '11 at 20:09
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. — t.b., Jun 13 '11 at 20:36
Theo: thanks for the clarification! And also thanks a lot for checking in some books. When I posted the question the score was 2-0 to the non-empty set, but the tables appear to have turned now and the empty set is winning by a hearty margin. — Kevin Buzzard, Jun 13 '11 at 21:09
I wonder about that $-\infty.$ One could say $\sup\varnothing=-\infty,$ but within the set of possible distances, $[0,+\infty),$ one should say $\sup\varnothing=0$ and $\inf\varnothing$ doesn't exist, but within $[0,+\infty]$ one would have $\inf\varnothing=+\infty.$ That last is useful if the metric space is a manifold that is not connected: the distance between two points is the infimum of all path length, so if there is no path, the distance would be $+\infty.$ Maybe that disqualifies it from being a metric space by some definitions, but maybe the definitions should be adjusted. $\qquad$ — Michael Hardy, Jun 10 '18 at 21:55
Berberian also defines metric spaces to be nonempty; But I, for one, am in the other camp. — ashpool, Jan 18 '22 at 00:31

score 28 · Accepted Answer · answered Jun 13 '11 at 17:53

28

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.

answered Jun 13 '11 at 17:53

Qiaochu Yuan

419,620

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Qiaochu: you have restored my faith in humanity! – Kevin Buzzard Jun 13 '11 at 17:54
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So...shall I edit the wikipedia page?? – Kevin Buzzard Jun 13 '11 at 17:54
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I don't see why not. – Qiaochu Yuan Jun 13 '11 at 17:56
I beat Kevin to it. (Would you really take off points for not treating the case of the empty metric space?) – Pete L. Clark Jun 13 '11 at 18:00
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Excluding the empty metric space would deprive us of such interesting debates as «what is the diameter of the empty metric space?»! – Mariano Suárez-Álvarez Jun 13 '11 at 18:02
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@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. – Qiaochu Yuan Jun 13 '11 at 18:03
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Allowing the empty metric space would also invalidate the statement that every real-valued function on a compact metric space attains a maximum. – Mark Jun 13 '11 at 18:04
Because I've never edited a wikipedia page in my life -- that's one good reason ;-) Still, perhaps now's the time to start! I might leave it for a while though -- perhaps someone can name 5 standard references where the things are non-empty... – Kevin Buzzard Jun 13 '11 at 18:05
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@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. – Qiaochu Yuan Jun 13 '11 at 18:06
@Mariano: how would you argue for anything other than $0$ as the diameter of the empty metric space? (@Qiaochu: it's an empty supremum of non-negative real numbers.) – Pete L. Clark Jun 13 '11 at 18:06
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@Qiaochu: that has the charm of trading one silly convention (no empty metric spaces) for a kludge (the diameter of a bounded metric space (the empty space is surely bounded! :) ) is a real number, except in one case. – Mariano Suárez-Álvarez Jun 13 '11 at 18:07
@Pete: I now know that the lecturer put as his definition of metric space that it be non-empty, so I'll definitely not take a mark off. But I definitely have taken marks off in the past for this sort of thing -- it's the classic way to lose a mark, no? You forget one stupid special case and you've divided by 0 in your proof, so you're dead. – Kevin Buzzard Jun 13 '11 at 18:08
@Kevin: well, maybe your students are better than mine. In the courses in which I give exams, if the students can totally nail the nontrivial cases that's usually worth full credit. I might make a comment, but I wouldn't take points off. – Pete L. Clark Jun 13 '11 at 18:10
@Pete: now perhaps we'll find the answer to my question in a few days' time at wikipedia.org! – Kevin Buzzard Jun 13 '11 at 18:12
Whoops, I was double confused. Incorrect comments deleted, and the correct answer appears in my original comment. – Qiaochu Yuan Jun 13 '11 at 18:13
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@Qiaochu: This diverged into off-topicness, as was expected: but the least upper bound of the empty subset of $\mathbb R$ (say), is surely not $\infty$ but $-infty$! – Mariano Suárez-Álvarez Jun 13 '11 at 18:13
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@Mariano: yes, but for metric spaces we are talking about $\mathbb{R}_{\ge 0}$, not $\mathbb{R}$. – Qiaochu Yuan Jun 13 '11 at 18:14
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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 $0$. – Jonas Meyer Jun 13 '11 at 18:15
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@Mariano: take the sup not in $\mathbf{R}$ but in $\mathbf{R}{\geq0}$ -- this is a sensible convention when looking at functions taking values in $\mathbf{R}{\geq0}$. For example if you define the norm of a linear map between normed spaces to be the sup of $f(x)$ for $|x|=1$ then you are in danger of getting the norm of the zero map from the 0-space to another space being $-\infty$, unless you follow this convention -- which is sensible, because a norm can never be negative. – Kevin Buzzard Jun 13 '11 at 18:16
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@everyone: we are actually a bit late to this debate: see en.wikipedia.org/wiki/Diameter. – Pete L. Clark Jun 13 '11 at 18:18
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To me, it seems rather arbitrary to define the diameter of an empty set to be 0 instead of $-\infty$. Negative infinity diameter may be counterintuitive, but that's the most natural definition in my view. – ashpool Jan 17 '22 at 13:51

Why are metric spaces non-empty?

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