# Suppose $\{x_n\}_n$ is Cauchy and that the subsequence $\{x_{n_k}\}_k$ converges to $x$. Prove that $\{x_n\}_n$ converges to $x$.

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.

• Presumably you want $n \ge N_1$ **and** $n \ge N_2$? – copper.hat Oct 19 '20 at 0:37
• 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? – xhsbm Oct 19 '20 at 0:45

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, $$|x_{n_k}-x| < \frac{\epsilon}{2}.$$
• Why do you need $N_3$? You don't. – xhsbm Oct 19 '20 at 1:09