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I have two questions.

  1. Can the empty set be formed into a metric space?

  2. If it exists, is it complete?

I have thought that the empty set is a complete metric space, since we can let $d:\emptyset\times\emptyset\to\mathbb{R}$ to be the empty metric on the empty set, and since there does not exists a Cauchy sequence such that it does not converge on the emptyset.

But, as I was reading a book, I found that a complete metric space cannot be written as a countable union of nowhere dense subsets... By this theorem, I found that the empty set cannot be complete, since the empty set is a nowhere dense subset of itself.

So, do I have to conclude that the empty set cannot be formed into a metric space?

Guy Fsone
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    Perhaps the book assumes metric spaces are not empty. – lhf Nov 17 '14 at 00:20
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    Or maybe the theorem wasn't meant for empty spaces. I think both answers should be yes, but I don't know if there are more drawbacks in doing this. – user2345215 Nov 17 '14 at 00:21
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    Then, the book is wrong? it doesn't assume for non-empty set. – User Nov 17 '14 at 00:23
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    Ah. I found in the wikipedia, that the Theorem (Baire Category Theorem) I mentioned in the question is for non-empty metric spaces... Thank you both Ihf and user2345215. – User Nov 17 '14 at 00:29
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    It is tedious to continuously put the disclaimer on every theorem/proposition that $|X|\notin{0,1}$ so it is often mentioned early in the first chapter or the foreword that they omit that as a condition for most of the following proofs. It is often uninteresting to work with the empty space and trivial space, so they prefer to save ink, space, time, and sanity in writing that detail. – JMoravitz Nov 17 '14 at 01:04
  • One more question: Is this space a subspace for every metric space? – SARTHAK GUPTA Jul 02 '20 at 15:35
  • Empty metric space is compact, and so complete. – Infinite Jul 26 '21 at 08:35

1 Answers1

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  1. It's a matter of convention, but I agree with Qiaochu that the "right" answer is yes. For instance, we want any subset of a metric space to be a metric space. As you noticed, we can just take the metric to be $d: \varnothing \times \varnothing \to \varnothing$, and it certainly satisfies the required axioms: $\forall x \forall y \; d(x,y) = d(y,x)$; $\forall x \; d(x,x) = 0$; $\forall x \forall y \forall z \; d(x,z) \le d(x,y) + d(y,z)$.

    There are of course some technicalities, e.g. the diameter of this space. Here is a related question with no answer yet.

  2. The empty metric space is complete.

    Firstly, as you noticed, every Cauchy sequence converges (since there are no Cauchy sequences).

    Secondly, while it's true that the empty metric space can be written as a countable union of nowhere dense sets ($\varnothing$ is nowhere dense, or more easily we can just take the empty union), it looks like this is a special case. Look at the statement of the Baire category theorem here. From the definition $\varnothing$ checks out as a Baire space, so BCT1 holds. BCT3 is stated as

    BCT3. A non-empty complete metric space is NOT the countable union of nowhere-dense closed sets.

    Notice that the space is stipulated to be non-empty, so BCT3 holds vacuously as well. (There must be something in the proof that BCT1 and BCT3 are equivalent that causes the non-empty condition to arise.)

Caleb Stanford
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