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

Demonstrating an alternative proof, please provide feedback :) Thank you.


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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  • $\begingroup$ Misfire in my brain. Fixed it. $\endgroup$ Jan 28 '19 at 23:14
  • $\begingroup$ Thanks for pointing that out... lol $\endgroup$ Jan 28 '19 at 23:14
  • $\begingroup$ I fixed it!!! haha. Please do not vote to close. I wanted feedback on my potential answer. $\endgroup$ Jan 28 '19 at 23:14
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    $\begingroup$ Ok. I'll change it. $\endgroup$ Jan 28 '19 at 23:15
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If $x_n$ did not converge then, by monotonicity, $x_n$ would be unbounded, in particular $x_n - 1 > x = \lim x_{n_i}$ for all $n \geq n_0,$ for some $n_0.$ But there exists a firs $i_0$ such that $n_i \geq n_0$ for all $i \geq i_0,$ and so $x_{n_i} \geq x + 1$ for all $i \geq i_0,$ a contradiction. Q.E.D.

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