Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.

I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!

  • $\begingroup$ Have you already proved, for example, that a non-decreasing sequence which is bounded above converges? $\endgroup$ Sep 25 '14 at 6:22
  • $\begingroup$ @JobinIdiculla 0,2,0,4,0,6.... is not monotone $\endgroup$ Sep 25 '14 at 6:23
  • $\begingroup$ @LiuGang: Thanks. That was a blunder indeed. $\endgroup$ Sep 25 '14 at 6:26

Suppose $\{x_n\}$ is increasing and has a subsequence $\{x_{n_k}\}$ which converges to $L$. We will prove that $\{x_n\}$ itself converges to $L$.

For any $\epsilon > 0$, we want to find an integer $N_\epsilon$ such that $|x_n - L| \leq \epsilon$ for any $n \geq N_\epsilon$.

Since $\{x_{n_k}\}$ is increasing and converges to $L$, we can find $k_\epsilon$ such that for any $k \geq k_\epsilon$, $ -\epsilon < x_{n_k} - L <0$.

Take $N_\epsilon = n_{k_\epsilon}$, then for any $n \geq N_\epsilon,$ $ x_{n_{k_\epsilon}}\leq x_n \leq L$, so $-\epsilon \leq x_{n_{k_\epsilon}} - L\leq x_n - L \leq 0$.

Similarly, we can prove when $\{x_n\}$ is decreasing

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    $\begingroup$ Why is $x_n \leq L$ for all $n \geq N_\epsilon$? $\endgroup$
    – jodag
    Feb 9 '18 at 7:53
  • $\begingroup$ Because $x_{n}$ is monotonically increasing to the limit $L$. $\endgroup$
    – Jack Moody
    Feb 15 '18 at 3:37
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    $\begingroup$ But that's what you're trying to show. Why is it assumed? $\endgroup$ Jan 28 '19 at 0:01
  • 1
    $\begingroup$ It's also not apparent to why you can post $x_n \leq L.$ $\endgroup$ Jan 28 '19 at 0:02
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    $\begingroup$ @RafaelVergnaud It's not assumed. It follows from the given. If a subsequence $x_{n_k} \to L$ then $x_n≤L, \, \forall n \in \mathbb{N}$. Assume the negation: $\exists m \in \mathbb{N} : x_m >L$ and hence $\forall n≥m \Rightarrow x_n≥x_m$, so what does that tell you about the subsequence? That the subsequence has a finite number of terms. $\endgroup$
    – Ibrahim
    Aug 9 at 12:23

Let $(y_k)$ be a convergent subsequence with a limit $A$. Since $(x_n)$ is monotone, $(y_k)$ is monotone as well. Then every $y$ is smaller than $A$. Then We know that for every $eps > 0$ there exists $K: k > K \Rightarrow y_k > A - eps$. Then for every x after $y_k$ in original sequence it's also true since sequence is monotone.Also, for every x there exists an y which stands further in $(x_n)$. Then every $x$ is $< A$. Then the sequence converges to A. (It was for increasing sequence, for decreasing anlogically).


Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$

Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b \in \mathbb{N},$ that is, $x_{n_i} \leq b,$ $\forall i \in \mathbb{N}.$

Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $\forall M \in \mathbb{N},$ $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M \in \mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).

So, $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ it follows that $x_n > b.$

Given that $(x_{n_i})$ is bounded above by $b,$ this means that $\forall i \in \mathbb{N},$ $n_i < N.$

Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < \cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).

However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.

Therefore, $(x_n)$ is convergent.


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