Can someone please critique my proof? I am a little skeptical of the very last equation I have. Thanks!
Q. Suppose $\{p_n\}$ is a Cauchy sequence in a metric space $X$, and some subsequence $\{p_{n_i}\}$ converges to a point $p \in X$. Prove that the full sequence $\{p_n\}$ converges to $p$.
Proof. Let $\epsilon > 0$. Suppose $\{p_n\}$ is a Cauchy sequence in a metric space $X$. Then, $\exists N \in \mathbb{N}$ s.t. \begin{equation} m, n > N \implies \left| p_m - p_n \right| < \epsilon/ 2 \end{equation} Now, let $\{p_{n_i}\}$ be an arbitrary subsequence of $\{p_n\}$ that converges to a point $p \in X$. Then, \begin{equation} n_i > n > N \implies \left| p_{n_i} - p \right| < \epsilon/2 \end{equation} Then, (1) and (2) imply that \begin{equation*} \begin{split} \left|p_m - p_n + p_{n_i} - p \right| \leq \left|p_m - p_n\right| + \left| p_{n_i} - p \right| &< \epsilon \\ \left|p_m + p_{n_i} - 2p_n + (p_n - p) \right| & < \epsilon \\ \left|p_m + p_{n_i} - 2p_n \right|+ \left|p_n - p \right| & < \epsilon \\ \left|p_n - p \right| &< \epsilon - \left|p_m + p_{n_i} - 2p_n \right| < \epsilon \end{split} \end{equation*} since $\left|p_m + p_{n_i} - 2p_n \right| > 0$, obviously. This completes the proof.
$$\left| p_n - p \right| = \left| p_n -p_{n_i}+p_{n_i}- p \right| \leqslant \left| p_n -p_{n_i} \right| + \left| p_{n_i}- p \right|$$ In metric notation: $$d(p_n, p) \leqslant d(p_n, p_{n_i}) + d(p_{n_i}, p)$$