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

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!

• Have you already proved, for example, that a non-decreasing sequence which is bounded above converges? – André Nicolas Sep 25 '14 at 6:22
• @JobinIdiculla 0,2,0,4,0,6.... is not monotone – Petite Etincelle Sep 25 '14 at 6:23
• @LiuGang: Thanks. That was a blunder indeed. – Train Heartnet 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

• Why is $x_n \leq L$ for all $n \geq N_\epsilon$? – jodag Feb 9 '18 at 7:53
• Because $x_{n}$ is monotonically increasing to the limit $L$. – Jack Moody Feb 15 '18 at 3:37
• But that's what you're trying to show. Why is it assumed? – Rafael Vergnaud Jan 28 at 0:01
• It's also not apparent to why you can post $x_n \leq L.$ – Rafael Vergnaud Jan 28 at 0:02

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.