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This question is from Spivak Calculus Ch.22, #4b)

Q: Prove that any subsequence of a convergent sequence converges.

Attempt:

In part a) we showed that if a subsequence of a Cauchy sequence converges, then so does the original Cauchy sequence. (I think I applied this idea indirectly, which I feel I needed to do)

Choose any subsequence $\{a_{n_{i}}\}$ of the original sequence $\{a_{n}\}$. Given that $\{a_{n}\}$ converges to a limit (call it $l$), this means that for all $\epsilon >0$, there exists $N>0$ such that for all $n > N$, $|a_{n} - l| < \frac{\epsilon}{2}$.

Now since $\{a_{n}\}$ also converges it means it is Cauchy, hence for all $\epsilon >0$, there exists $I>0$ such that for all $n_{i}, m_{i} > I$, $|a_{n_{i}} - a_{m_{i}}| < \frac{\epsilon}{2}$. (Probably should do something about this syntax since not the clearest)

Since $n_{1} < n_{2} < n_{3} < \dots$, this means there exists $I$ such that $n_{i} > N$ for some $i > I$.

Therefore:

$$|a_{n_{i}}-l| < |a_{n_{i}} - a_{n}| + |a_{n} - l| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

I know I have the flavor of the proof and the overall idea of how I should approach it, but I feel I might be messing up some of the technical details. Feedback on my attempt would be nice.

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You’re working much too hard: there’s no reason to bring in the Cauchy property at all. Given $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|a_n-\ell|<\epsilon$ for all $n\ge n_\epsilon$, and I claim that $|a_{n_k}-\ell|<\epsilon$ whenever $k\ge n_\epsilon$. This follows immediately from the fact that $n_k\ge k$ for each $k\in\Bbb N$, which is easily proved by induction on $k$: certainly $n_0\ge 0$, and if $n_k\ge k$ for some $k\in\Bbb N$. Then $n_{k+1}\ge n_k+1\ge k+1$.

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  • $\begingroup$ I professor, could I ask your assistance here, please? I edit the precedint question. Excuse me for the bother. $\endgroup$ – Antonio Maria Di Mauro Oct 1 '20 at 21:11

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