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I want to show that $\{x_n\}_n$ converges to $x$ if $\{x_n\}_n$ is Cauchy and the subsequence $\{x_{n_k}\}_k$ converges to $x$.

I did the proof but I want to know why I can say that $|x_n-x_{n_k}| < \frac{\epsilon}{2}$.

Intuitively I know that the subsequence has to be epsilon-close to the terms in the sequence because one is just a subset of the other. I think I'm supposed to use the fact that the subsequence is increasing but I'm not sure how I get $|x_n-x_{n_k}| < \frac{\epsilon}{2}$ from that. Any help understanding that step would be greatly appreciated.

Proof:

Given: $\{x_n\}_n$ is Cauchy and the subsequence $\{x_{n_k}\}_k$ converges to $x$.

It follows that for all positive $\epsilon$, $\exists N_1 \in \mathbb{N}^+, \forall k\in \mathbb{N}^+$ such that if $k\geq N_1$, then $|x_{n_k}-x| < \frac{\epsilon}{2}$

and for all positive $\epsilon$, $\exists N_2 \in \mathbb{N}^+, \forall i,j\in \mathbb{N}^+$ such that if $i,j\geq N_2$, then $|x_i-x_j| < \frac{\epsilon}{2}$

So...

$|x_n-x| = |x_n-x_{n_k} + x_{n_k}-x| \leq |x_n-x_{n_k}| + |x_{n_k}-x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Therefore $\{x_n\}$ converges to $x$.

Is this correct? I'm not sure whether I have a solid grasp on subsequences.

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    $\begingroup$ Presumably you want $n \ge N_1 $ **and** $n \ge N_2$? $\endgroup$ – copper.hat Oct 19 '20 at 0:37
  • $\begingroup$ Naturally... It wouldn't hold otherwise. I think I'm having a little trouble with notation. $\{x_{n_k}\}$ can be thought of as $\{x_{n(k)} = x_i\}$, yes? $\endgroup$ – xhsbm Oct 19 '20 at 0:45
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you are missing something it should be written like this:

It follows that for all positive $\epsilon$, $\exists N_1 \in \mathbb{N}^+, \forall k\in \mathbb{N}^+$ such that if $k\geq N_1$, then $|x_{n_k}-x| < \frac{\epsilon}{2}$

It follows that for all positive $\epsilon$,$\exists N_1\in \Bbb{N}^+,\forall k\in\Bbb{N^+}$ with $k\geq N_1$, then $n_k\geq k\geq N_1$, Therefore, $$|x_{n_k}-x| < \frac{\epsilon}{2}.$$

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    $\begingroup$ Why do you need $N_3$? You don't. $\endgroup$ – xhsbm Oct 19 '20 at 1:09
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    $\begingroup$ @xhsbm Ahhh Sorry, You are right. $\endgroup$ – John Nash Oct 19 '20 at 2:19

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