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Suppose that $\{a_n\}$ has a finite range, then it has a convergent subsequence.

Here is what I have, but I don't think its enough.

Since $\{a_n\}$ has a finite range. Then we can say the sequence is bounded. Since the sequence is bounded, then it has a convergent subsequence.

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  • $\begingroup$ You can have a bounded sequence with finite range, but still being divergent. For example, $s_n = i^n$ does not converge, but it is bounded and it has a finite range. $s_n = \left\{i, -1, -i, 1, i, ... \right\}$ $\endgroup$
    – Alan
    Commented Nov 6, 2017 at 9:39

2 Answers 2

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In fact, $\{a_n\}$ must have a constant subsequence. Since the range is finite, if each element in the range appeared only a finite number of times then the sequence would only have finitely many terms. Hence one element must appear infinitely often. This is your convergent subsequence.

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HINT: Since it has a finite range, there must be some $y$ in the range such that $\{n\in\Bbb N:a_n=y\}$ is infinite. A constant sequence is convergent.

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