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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!

Jason
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3 Answers3

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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

Jack Moody
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Petite Etincelle
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    Why is $x_n \leq L$ for all $n \geq N_\epsilon$? – jodag Feb 09 '18 at 07:53
  • Because $x_{n}$ is monotonically increasing to the limit $L$. – Jack Moody Feb 15 '18 at 03:37
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    But that's what you're trying to show. Why is it assumed? – Rafael Vergnaud Jan 28 '19 at 00:01
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    It's also not apparent to why you can post $x_n \leq L.$ – Rafael Vergnaud Jan 28 '19 at 00:02
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    @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. – Ibrahim Aug 09 '21 at 12:23
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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 $\varepsilon > 0$ there exists $K: k > K \Rightarrow y_k > A - \varepsilon$. 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 analogically).

bob
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Petr Naryshkin
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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.

Rafael Vergnaud
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