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Let $S$ be a set of complex numbers. I would like to prove the following are equivalent:

1) every sequence of elements of $S$ has a point of accumulation in $S$;

2) every infinite subset of $S$ has a point of accumulation in S

3) every sequence of elements of $S$ has a convergent subsequence whose limit is in $S$

PROOF. 1)$\rightarrow$ 2) and 3)$\rightarrow$ 1) are trivial. How to prove 2)$\rightarrow$ 3)?

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For 2 $\to$ 3, Consider a sequence $x_(n)$of element in $S$ .\

Case 1: If range of the sequence is finite then you can choose a sub sequence which converge to a element of $S$ .(namely the constant sub-sequence {a} where $a$ is the element in the range of the first sequence which occurs infinitely many times.) \

Case 2: If range is infinite set then you will have a limit point $p\in S$ of the range set.From that you will have a sub-sequence of $x_(n)$ converging to $p$. Take $1/n$ neighborhood around p and pick element from {$x_(n)$} to construct the sub-sequence converging to p.

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