calculus - Proof verification. If $x_n$ is a monotone sequence and it has a convergent subsequence $x_{n_k}$, then $x_n$ is convergent to the same limit.

Let $x_n$ denote a monotone sequence where $n\in \Bbb N$. Let $x_n$ have a convergent subsequence $x_{n_k}$. Prove that $x_n$ is convergent to the same limit as $x_{n_k}$.

I've decided to consider two separate cases. $x_n$ is either increasing or decreasing. For the case of a stationary sequence the result follows immediately.

Below i'm using the fact that a bounded monotone sequence has a limit.

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Case 1. Let $x_n$ be a monotonically increasing sequence. Thus:
$$x_{n+1} \ge x_n$$

Consider a subsequence of $x_n$ namely $x_{n_k}$. Then:
$$x_{n_k} \ge x_n,\ \forall n_k \ge n$$

We are also given that:
$$\lim_{k\to\infty} x_{n_k} = L$$

Given that fact we know that $x_{n_k}$ is bounded above. Therefore:
$$x_n \le x_{n_k} \le L$$

Now by monotone convergence theorem a monotone bounded sequence must be convergent. Also by uniqueness of a limit for a convergent sequence and the fact that all its subsequences are also convergent to the same limit:
$$\lim_{k\to\infty}x_{n_k} = \lim_{n\to\infty}x_n = L$$

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Case 2. I know of two possible ways to follow for this case. First is reproduce the reasoning above for the monodically decreasing sequence, which is almost the same as case 1. Or, as mentioned in comments, consider a new sequence:

$$y_n = (-x_n)_n$$
Then the result follows immediately from case 1.

Is my proof rigorous enough to consider it complete?

Answer

Here's a more formal (and shorter) proof:

Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$, which we know exists, and let $L$ be its limit. Then, given any $\varepsilon > 0$, there is some $K \in \mathbb{N}$ such that for all $k \geq K$, $|x_{n_k} - L| < \varepsilon$ (this is the definition of the limit).

Now, for any $n \geq n_K$, we will show than $|x_n - L| < \varepsilon$, which will complete the proof. For this purpose, suppose that this is not true. Then there is some $n \geq n_K$ such that $|x_n - L| \geq \varepsilon > |x_{n_K} - L|$.

First, note that if $x_1 \leq L$ then we must have $x_i \leq L$ for all $i$: if not, then we have some $i$, such that $x_1 \leq L < x_i$, but then for all $m > i$, $x_m \geq x_i > L$, so $|x_m - L| \geq |x_i - L| > 0$, so in particular, $(x_{n_k})\not\to L$, a contradiction. Symmetrically, if $x_1 \geq L$, then $x_i \geq L$ for all $i$.

Now, we have a problem: we have either $x_n \geq L + \varepsilon > x_{n_K} \geq L$ or $x_n\leq L - \varepsilon < x_{n_K} \leq L$, but then monotonicity gives us either $x_m \geq L + \varepsilon$ or $x_m \leq L - \varepsilon$ for all $m \geq n$, in particular for all but finitely many $n_k$, so $|x_{n_k} - L| \geq \varepsilon$ for all but finitely many $n_k$, so $(x_{n_k})\not\to L$, a contradiction.

Thus, we must have $|x_n - L| <\varepsilon$ for all $n \geq n_K$, hence $(x_n)\to L$.