Limit of a Cauchy Sequence 
Let $\langle X_n \rangle$ be Cauchy with a sub sequence $\langle X_{n_k} \rangle$ such that $\lim_{n\to\infty} X_{n_k}=A$. Show $\lim_{n\to\infty} X_n=A$.

My work so far:
Choose $\epsilon$ such that there exists an $N$ such that $\|(X_{n_k},A)\| = \|(X_k, X_l)\| = \|(X_n,A)\|$ for $k,l\ge N$. Therefore $\|(X_n, A) < \epsilon$, so $\lim_{n\to\infty}  X_n=A$.
 A: Let $\varepsilon>0$. Since $\{X_n\}_{n=1}^\infty$ is a Cauchy sequence, there exists $N_1\in \Bbb N$ such that $m,n>N_1$ imply $\Vert X_n-X_m\Vert<\frac{\varepsilon}{2}.$ Now, since $X_{n_k}\to A$ as $k\to\infty$, there exists $N_2\in\Bbb N$ such that $\Vert X_{n_l}-A\Vert<\frac{\varepsilon}{2}$. Let $N=\max\{N_1,N_2\}$. Then for all $n>N$ we have $$\Vert X_n-A\Vert\leq \Vert X_n-X_{n_a}\Vert+\Vert X_{n_a}-A\Vert<\varepsilon.$$ Thus, $X_n\to A$ as $n\to\infty$.
A: Pick some $\epsilon > 0$.  Find a number $N_1$ so that if $n,m \ge N$ then $|x_n - x_m| \le \epsilon$.  Next find an $N_2 > N_1$ so that if $n_k \ge N_2$ then $|x_{n_k} - A| < \epsilon$.  
So for $n \ge N_2$, find an $n_k \ge n$; then $|A - x_n| = |A - x_{n_k} + x_{n_k} - x_n| \le |A - x_{n_k}| + |x_{n_k} - x_n| \le 2\epsilon$.
Edit: This is of course in $\mathbb{R}$.  In more general spaces, adapt the argument.
A: You say this is in a real space of dimension $n$, so in a complete metric space.
Since $(X_n)$ is Cauchy, it converges to $X$.
Every subsequence also converges to $X$.
So $X=A$.
Note: for a proof of this fact (still true), in a general metric space, see anonymous' argument.
