Question about compact sets

compactnessmetric-spacesreal-analysis

I know what compact sets are in $R^n$:A set S of real numbers is called compact if every sequence in S has a subsequence that converges to an element again contained in S.like i know similar definition is in every metric space(please correct me if i am wrong).my question is:every set in arbitrary metric space is compact, because i can take the sequence of elements in that set, and as a subsequence i can take only first element.Hence subsequence contains only one element so it is convergent subsequence, therefore every set in arbitrary metric space is compact.What is wrong in my logic?

Best Answer

What is wrong is that, no, you cannot just take the first element. A subsequence of a sequence $(x_n)_{n\in\Bbb N}$ is a sequence $(x_{n_k})_{k\in\Bbb N}$, where $(n_k)_{k\in\Bbb N}$ is a strictly increasing sequence of natural numbers. And the sequence $1,1,1,1,\ldots$ is not strictly increasing.

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