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

Question

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.

Answer 1

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, $$|x_{n_k}-x| < \frac{\epsilon}{2}.$$