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This question already has an answer here:

Let $(X,d)$ be a metric space. This space is compact if any sequence $x_n \subset X$ has a convergent subsequence.

This is how I'm given the definition of a compact metric space and it confuses me. How come a definition is not an "if and only if" statement and instead an "if" statement. This seems more of a theorem to me and seems like there could be some metric spaces which are compact but do not have a convergent subsequence.

Also, if say, $E \subset X$ is compact,then would the values of those convergent subsequences be a member of $E$?

Thanks

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marked as duplicate by Brahadeesh, Xander Henderson, KReiser, user10354138, Alexander Gruber Dec 4 '18 at 4:02

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    $\begingroup$ Definitions are written as "if"s but it is implied that it is "if and only if." It's a definition. $\endgroup$ – Randall Dec 3 '18 at 15:35
  • $\begingroup$ @Randall thanks $\endgroup$ – Xenidia Dec 3 '18 at 15:37
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First of all, when you are defining something, you can use the expression “if and only if” or you can just use the word “if”; in the context of definitions, they are the same thing.

And if this was a theorem, you would have to have the concept of compact space defined in some other way. Sometimes, yes, compactness is defined in a different way (through open covers, for instance) and, in that case, yes, the sentence that you quoted would then be a theorem.

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of course they need to converge in $E$, otherwise you run immediately into problems (just consider the open ball in $\mathbb{R}^n$, which is a subset of a compact sphere, hence every sequence in the open ball has a convergent subsequence, but not convergent in the open ball) also, the thing with the if was already mentioned. However, you will hopefully soon see the most abstract and in my opinion best definition for compactness (every open covering admits a finite subcovering) and then you might think of it as a theorem.

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