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Any sequence $(a_n)$ of real numbers has a subsequence $(a_{n_k})$ such that ($\sin(a_{n_k})$) converges.

Is the above statement true?

My thoughts: I think so. Suppose Not. Then for every $\epsilon > 0$, there exists a K such that if $n_k \geq K$, then $\sin a_{n_k} > \epsilon$. Choose $\epsilon := 2$. Then, $\sin a_{n_k} \geq 2$, which is absurd. Is my thoughts correct?

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Every compact set including $[-1, 1]$ is sequentially compact, meaning that any sequence inside it has a convergent subsequence. The sequence $\sin(a_n)$ belongs to the compact set $[-1, 1]$ and the result follows.

Your argument doesn't make sense. In the proof by contradiction you can only say that for any subsequence $a_{n_k}$ and any real $L$ there exists a positive $\epsilon$ such that $|\sin a_{n_k}-L|>\epsilon$.

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