1. ## Cauchy Sequence

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

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

2. it should look more like this

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

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

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 $\displaystyle |p_n-p_m| < \varepsilon$ but i don't see why $\displaystyle |p_{n_r}-p_{n_s}| < \frac{\varepsilon}{2}$.
I always have trouble seeing these things, but i don't see why i can't just take $\displaystyle |p_{n_r}-p_{n_s}| < \varepsilon$ for all $\displaystyle \varepsilon > 0$ as $\displaystyle p_{n_i} \rightarrow p$ and since $\displaystyle p_{n_s},p_{n_r} \in$ {$\displaystyle p_n$}, then $\displaystyle p_n$ must converge to $\displaystyle p$ as well.

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

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

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