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Examine the compactness of $M:=\left\{\tan(1/n): n\in\mathbb{N}\right\}\cap [0,1]\subset\mathbb{R}$ with respect to the absolute value.


I came to the conclusion that $M$ is not compact, since for compactness each sequence in $M$ needs to have a convergent subsequence with limit in $M$.

I think that $(a_n)_{n\geq 2}$ with $a_n:=\tan(1/n)$ is a counter example since each subsequence of it should converge to $0\notin M$.

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You are correct, being compact (in a metric space) implies sequentially compact. However, you can use an even weaker result, compact (in a metric space) implies closed. The sequence you've given converges in $[0,1]$, but its limit is not in the given set, so your set isn't closed, hence not compact.

You may want to justify what $\tan(1/n) \to 0$ as $n\to \infty$, but that's the only additional thing I could think to do for your solution.

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  • $\begingroup$ By the way: What if we replace the $\cap$ by $\cup$? Then each sequence in $M$ should have a convergent subsequence converging in M, right, hence $M$ would be compact? $\endgroup$ – John_Doe Nov 2 '16 at 15:04
  • $\begingroup$ In that case you'd get $[0,1] \cup \{\tan(1)\}$, which is the union of closed and bounded sets, so closed and bounded. Heine-Borel Theorem says that this is compact. $\endgroup$ – Hayden Nov 2 '16 at 15:16
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Your intuition is correct. More rigorously, if $a_{n_k}$, considered as a sequence of $M$ would converge to some $p \in M$, then $a_{n_k}$ would also converge to $p$ in $\mathbb{R}$ but since $a_n \to 0$ in $\mathbb{R}$, we would have $p = 0 \in M$, a contradiction.

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