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.

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: $$ \lim_{k\to\infty}x_{n_k} = \lim_{n\to\infty}x_n = L $$

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?

  • $\begingroup$ I don't like the "take the limit of both sides" step. You're trying to show that the sequence has a limit, and without knowing that it has a limit beforehand, then the step is potentially meaningless. I would use the monotone convergence theorem. $\endgroup$ Jan 7, 2019 at 13:36
  • $\begingroup$ @TheoBendit You're right, but MCT is hilariously overkill for this (and might well be circular, depending on how you proved MCT). Instead, use the bound on the distance from the terms of the subsequence to its limit to establish a bound on the distance from an arbitrary term to the limit. $\endgroup$ Jan 7, 2019 at 13:38
  • $\begingroup$ @TheoBendit Thank you for the notice, i've updated the post $\endgroup$
    – roman
    Jan 7, 2019 at 13:39
  • $\begingroup$ @user3482749 I could only see it be circular when proving the MCT from local compactness of the real line. Is that a common approach? $\endgroup$ Jan 7, 2019 at 13:43
  • 1
    $\begingroup$ @user3482749 The MCT is a nail in a board at best. It's a very fundamental, close-to-the-metal tool in a real analysts toolkit. The asker is clearly looking for an elegance to the proof and avoiding epsilons and deltas. I think the MCT is (typically) a fine tool to use. Also, I just don't see Heine-Borel being proven before MCT. :-P $\endgroup$ Jan 7, 2019 at 13:50

1 Answer 1


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 $(x_n)\to L$.

  • $\begingroup$ Even though you've used some bare metal machinery this is still a nice alternative. Thank you! $\endgroup$
    – roman
    Jan 7, 2019 at 14:09

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