How to prove $(x_n)$ converges to $a$ under these conditions Suppose $x_n$ is an increasing sequence and of real numbers such that there exists a subsequence $(x_{n_k})$ which converges to $a \in \mathbb{R}$. Prove that $(x_n)$ converges to $a$.
My attempt:
Suppose $(x_n)$ does not converge to $a$.
Then for any $n\in \mathbb{N}$, we have $\epsilon_0\leq|x_n-a|$ for some $\epsilon_0>0$.
In particular, we can choose $n=n_k$,then $\epsilon_0\leq|x_{n_k}-a|$ which is a contradiction. 
It is because a subsequence $(x_{n_k})$ converges to $a$ means for any $\epsilon>0$, there exists a natural number $K(\epsilon)\in\mathbb{N}$ such that for all $K(\epsilon)\leq n_k$, $|x_n-x|<\epsilon$.
Do my answer correct? If not, could you give me an answer or a hint? Thank you.
 A: Suppose $x_{n_k} \rightarrow a$ but $x_n \rightarrow a'$, $a \neq a'$. Then for some $N $ sufficiently large, $|x_{n_k} - a| < \varepsilon$  and $|x_n - a'| < \varepsilon$ for all $n_k, n > N$.  Take $\varepsilon < |a-a'|/3$.  This implies that all the terms of $x_{n_k}$ for $n_k>N$ are at least $|a-a'|/3$ away from all the terms of $x_n$ for $n>N$, despite one being a sub-sequence of the other.  This is a contradiction.
Try drawing a graph with $n$ on the axis, $a$ and $a'$ with non-intersecting balls of size $|a-a'|/3$ around them, and the two sequences $x_n$ and $x_{n_k}$ converging to $a$ or $a'$ so that they are eventually bounded away from each other.  But $\{x_{n_k}\} \subset \{x_n\}$, so that cannot be true.
A: Let's say $$(\forall \varepsilon>0,\exists k_0\in \mathbb{N},\forall k>k_0) :\\ a-\varepsilon<x_{n_k}<a+\varepsilon$$
Then $$(\forall n>k_0,  \exists n_{k_1},n_{k_2}>k_0):\\ a-\varepsilon<x_{n_{k_1}}<x_n<x_{n_{{k_2}}}<a+\varepsilon$$
where $n_{k_1}<n<n_{{k_1}}$ (because $x_n$ is an increasing sequence). Hence, $$a-\varepsilon<x_{n}<a+\varepsilon$$
$(\forall n>k_0).$
