# Math Help - Cauchy Sequence

1. ## Cauchy Sequence

Problem Statement:
Suppose that { $p_n$} is a Cauchy sequence and that there is a subsquence { $p_{n_i}$} and a number $p$ such that $p_{n_i} \rightarrow p$. Show that the full sequence converges, too; that is $p_n \rightarrow p$.

My Solution:
It will suffice to prove that { $p_n$} converges to $p$ if and only if every subsequence of { $p_n$} converges to $p$.
( $\Rightarrow$)Suppose $p_n \rightarrow p$. Consider { $p_{n_i}$} $\subseteq$ { $p_n$}.
Let $\varepsilon > 0$. For some $n \geq N$, $d(p_n,p) < \varepsilon$. Then if $i \geq N, n_i \geq i \geq N$. so $d(p_{n_i},p) < \varepsilon$.
( $\Leftarrow$){ $p_n$} is a subset of itself, so $p_n \rightarrow p$.

2. it should look more like this

given $\epsilon>0$, integers $n,m$ and $N$ we have
$\|p_n-p_m\|<\epsilon$ whenever $n,m\geq N$

we know that $\|p_{n_i}\|$ converges to $p$ and convergent sequences are also Cauchy sequences, so we have the inequality
$\|p_{n_r}-p_{n_s}\|<\frac{\epsilon}{2}$ whenever $r,s\geq N$

then play around with the inequality, you should find the desired result

3. Is this solved then? I realize I made the mistake of thinking that showing my iff statement is true proves the whole thing. But it's only ONE subsequence, not all of them...

4. OK, sorry. Looking at this again, I do not understand. I apologize, the sequences are particularly confusing to me.
I understand why $|p_n-p_m| < \varepsilon$ but i don't see why $|p_{n_r}-p_{n_s}| < \frac{\varepsilon}{2}$.
I always have trouble seeing these things, but i don't see why i can't just take $|p_{n_r}-p_{n_s}| < \varepsilon$ for all $\varepsilon > 0$ as $p_{n_i} \rightarrow p$ and since $p_{n_s},p_{n_r} \in$ { $p_n$}, then $p_n$ must converge to $p$ as well.

5. Originally Posted by DontKnoMaff
i don't see why $|p_{n_r}-p_{n_s}| < \frac{\varepsilon}{2}$.
That's just the definition of Cauchy sequence (and all convergent seq. are Cauchy), although I think the one that works is $|p-p_{n_k}|<\varepsilon/2$. Now think triangle inequality applied to $|p-p_n|$ with those bounds.

6. ok i think i get it. Could someone check me to make sure though?

Take $\varepsilon > 0$. take $N$ s.t. $n_k,n > N$ implies that $d(p_{n_k},p),d(p_n,p_{n_k}) < \frac{\varepsilon}{2}$. Hence $d(p_n,p) \leq d(p_{n_k},p)+d(p_{n_k},p_n) \leq \varepsilon$ Thus { $p_n$} converges to p.