I am trying to show that a sequence converges if and only if each of its subsequences converges. I am sure there are other posts for this, but I am trying to write my own proof for practice.
Attempt:
We recall that for a subsequence $(a_{n_{k}})$ of a sequence $(a_{n})$ we make a strictly monotonic map $\mathbb{N} \rightarrow \mathbb{N}: k \rightarrow n_{k}$. If we consider $n_{k} = k$ then clearly the original sequence is its own subsequence therefore it converges.
For the converse statement, $\lim_{n \rightarrow \infty}a_{n} = a \Leftrightarrow \forall \epsilon > 0, \exists N \in \mathbb{N}: \forall n \geq N, |a_{n}-a| < \epsilon$. We observe that clearly $n_{k} \geq k, \forall k$ (can prove by induction if necesary). Therefore we have $|a_{n_{k}} – a| < \epsilon, \forall k \geq N \Rightarrow \lim_{k \to \infty} a_{n_{k}} = a$.
Does the above seem reasonable?
Best Answer
Yes, it is correct. The key observation is indeed that $n_k \geq k$. The same statement is even true in more general contexts (metric spaces, topological spaces) with more or less the same proof.
Here is how I would have written it:
Let $(a_{n_k})_k$ be a subsequence of $(a_n)_n$ and suppose $a_n \to a$. Given $\epsilon > 0$, choose $N$ such that $|a_n-a| <\epsilon$ when $n \geq N$. Then if $k \geq N$, from the inequality $n_k \geq k$ also $n_k \geq N$ so that we also have $|a_{n_k}-a| < \epsilon$. Hence, we have proven:
$$\forall \epsilon > 0: \exists N: \forall k \geq N: |a_{n_k}-a| < \epsilon$$ and this exactly means that $a_{n_k}\to a$.