I know this question has been asked many times on this site. I just wanted to make sure how I wrote this proof out is clear enough.
Theorem
$ \{ p_n \} $ converges to $p$ if and only if every subsequence of$ \{ p_n \} $ converges to $p$
Proof. Every sequence is a subsequence of itself so I won't bother with that direction.
For the other, let $ \{ p_n \} $ be a sequence in a metric space $X$ that converges to $p$. By definition this means for every $\epsilon >0$ there is an integer $N$ such that for all $n > N$ implies that $$d(p,p_n) < \epsilon.$$ Let $ \{ p_{n_i} \} $ be a subsequence of $ \{ p_n \} $. Any point $p_{n_k}$ of $ \{ p_{n_i} \} $ is also a point of $ \{ p_n \}. $ Hence $ p_{n_k} = p_n$ for some $k$. Therefore, $$d(p,p_{n_k}) =d(p,p_n) < \epsilon$$ $$d(p,p_{n_k}) < \epsilon.$$$$\tag*{$\blacksquare$}$$
This is in contrast with many solutions that states the position of point $p_{n_k}$ (in different notation):
- Link to the question is here. Is the way I give the proof "correct enough." If not, can you explain what I miss out on by stating it as such. Thanks a lot all.
It is your aim to prove that every subsequence $(p_{n_k})_k$ of a sequence $(p_n)_n$ will converge to $p$ whenever $(p_n)_n$ converges to $p$.
Unfortunately your effort makes no sense.
This is the way to do it:
Assume that $(p_n)_n$ converges to $p$ and that $(p_{n_k})_k$ is an arbitrary subsequence of $(p_n)_n$.
Then it is enough to prove that in this situation $(p_{n_k})_k$ converges to $p$.
For this let $\epsilon>0$.
It is enough now to prove that some index $K$ exists with $k>K\implies d(p,p_{n_k})<\epsilon$.
Note that some index $N$ exists with $n>N\implies d(p,p_n)<\epsilon$ because $(p_n)_n$ converges to $p$.
We have $n_1<n_2<n_3<\cdots$ so some $K$ will exist with $n_{K}>N$.
This $K$ does the job, since $k>K\implies n_k>n_{K}>N\implies d(p,p_{n_k})<\epsilon$.
We are ready.