To prove a Cauchy sequence is convergent. If $\{x_n\}$ is Cauchy and has a convergent subsequence with limit $x$, show that $\{x_n\}$ is convergent with limit $x$.
 A: I'll try to help you unravel some of the definitions and give a hint.
Let $\epsilon >0$. 
Suppose $\{x_{n}\}$ has a subsequence $\{x_{n_{k}}\}$ that converges to $x$. Then, there is an $N_{1}\in\mathbb{N}$ such that $n_{k}\geq k\geq N_{1}$ implies $|x_{n_{k}}-x|<\epsilon/2$. 
Since $\{x_{n}\}$ is Cauchy, there is an $N_{2}\in\mathbb{N}$ such that $n,m\geq N_{2}$ imply $|x_{n}-x_{m}|<\epsilon/2$. 
Let $N=\max\{N_{1},N_{2}\}$. Can you think how you might use the triangle inequality to conclude that $|x_{n}-x|<\epsilon$ for all $n\geq N$? 
A: Let {$x_n$} be cauchy sequence in metric space $X$ with metric $d$.
By Assumption, $x_n$ has convergent subsequence $x_{n_k}$ such that $\lim_{k\rightarrow\infty}x_{n_k}=a$ for some $a\in X$.
Let $\epsilon>0$. There exists $K\in\mathbb{N}$ such that if $k\geq K$, then $d(x_{n_k},a)<\epsilon$.
Since $x_n$ is cauchy, there exists $M\in\mathbb{N}$ such that if $n,m\geq M$, then $d(x_n,x_m)<\epsilon$.
Let $N=max\{M,K\}$.
Then If $n\geq N$, then $d(x_n,a)\leq d(x_n,x_{n_N})+d(x_{n_M},a)<2\epsilon$
