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How does one prove that a set is sequentially compact? I'm having trouble picturing what it means for every sequence to have a subsequence that converges.

Examples would be greatly appreciated too.

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You prove it by showing that every sequence of points in the set has a subsequence that converges to a point in the set. What more do you expect to be able to say at this level of total generality? –  Chris Eagle Oct 21 '12 at 9:53
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I would picture it as in the first topological examples: compact sets in $\Bbb R^n$ (here sequentially compact means the same), and these are the bounded and closed sets.

If a subset $S$ is bounded, you can split it into $2^n$ pieces (midplanes for every basic direction in $n$ dim), so if we have a sequence $a_n\in S$, there is a part which still contains infintely many elements of the sequence, this part should be again split, and so on.. so one will find a sequence within $S$ which converges somewhere in the space. To insure that the limitpoint is also in $S$, the closedness is required.

And, in practice, proving sequentially compactness is simplest to be done based on the definition: let's assume we are given a sequence, and then we will have to select a convergent subsequence..

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