# A concern on the definition of compactness in a metric space [duplicate]

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

• Definitions are written as "if"s but it is implied that it is "if and only if." It's a definition. Dec 3, 2018 at 15:35
• @Randall thanks
– user346936
Dec 3, 2018 at 15:37

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.