# Prove that any subsequence of a convergent sequence converges - Proof Verification

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.

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 $$n_{k+1}\ge n_k+1\ge k+1$$.