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 $\displaystyle p_n \rightarrow p$.