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In one of my books about metric spaces, it is stated that "a not null set, equiped with a metric, is a metric space". On the other hand, the null set as a subset of $R$ is a compact subspace as it is complete and totally bounded.... Why is there such a contradiction?Thanks.

joriki
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dmtri
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    This is probably something that is going to vary from text to text. The discussion here might help. – Jair Taylor Jun 23 '18 at 16:30
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    You have an unbalanced quotation mark in there; it's hard to tell which part is the quote. – joriki Jun 23 '18 at 16:44
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    What is the "contradiction" you're talking about? – Kat Jun 23 '18 at 17:41
  • @joriki sorry for the quotation...it is fixed now. – dmtri Jun 23 '18 at 17:55
  • @Ron the contradiction is whether a metric space can be null or not – dmtri Jun 23 '18 at 17:57
  • Is $\varnothing$ a metric space? One book rules it out, another may not rule it out. Why do you need to know? That is, for what purpose? – GEdgar Jun 23 '18 at 18:05
  • @JairTaylor , nice link! I think I am ok now. thanks. – dmtri Jun 23 '18 at 18:07
  • @GEdgar cause....I am a Vigro and I need to clarify everyhting in maths! Sorry I am kidding..The real reason is that I had to solve an exercise determining all the compact set in a discreat metric space – dmtri Jun 23 '18 at 18:15
  • Compactness it a topological property, and from the definition it follows that the empty set is compact. I don't get the part where you compare this with whether or not it is a metric space. – Kat Jun 23 '18 at 20:20
  • What book are you using? I would suggest finding a better book, if your book claims a metric space cannot be empty. See also https://math.stackexchange.com/questions/45145/why-are-metric-spaces-non-empty – Eric Wofsey Jun 23 '18 at 22:08

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The author decided to require metric spaces to be non-empty. This is not a definition I would adopt (see Why are metric spaces non-empty? for example) but it does not contradict anything.

If we follow this definition, then $\varnothing$ is still a compact subset of $\mathbb{R}$, but it is not recognized as a metric space of its own. Not a contradiction.