Question on Cauchy sequence in any metric space.

Question

In a metric space $\mathit(X,d)$, If $\langle x_n\rangle$ is a Cauchy sequence and $\langle x_{i_n}\rangle$ its subsequence, show that $\mathit d(x_n,x_{i_n})$ $\to$ 0 as $\mathit n$ $\to$ $\infty$.

My attempt:

Here, $\langle x_n\rangle$ is a Cauchy sequence. Therefore for every $\epsilon$$>0$, there exist $n(\epsilon)∈\Bbb N$ such that $$d(x_n,x_{i_n})< \epsilon,\forall n,i_n \geq n(\epsilon)$$ Here we can clearly see that $$x_n \to \infty$$ and $$x_{i_n} \to \infty$$as$$n \to \infty$$ Thus $\mathit d(x_n,x_{i_n})$ $\to$ 0 as $\mathit n$ $\to$ $\infty$.

But I am not sure my answer is correct or not..so please point out where I make mistake. Any help is appreciated.

Answer 1

Proving that $\lim_{n\to\infty}d(x_n,x_{i_n})=0$ means proving that, for every $\varepsilon>0$, there is some $N\in\Bbb N$ such that$$n\geqslant N\implies d(x_n,x_{i_n})<\varepsilon.$$I don't see that proved in your answer.

Note that if $N\in\Bbb N$ is such that $m,n\geqslant N\implies d(x_m,x_n)<\varepsilon$, then, since we always have $i_n\geqslant n$ (since $(i_n)_{n\in\Bbb N}$ is a strictly increasing sequence of natural numbers), if $n\geqslant N$, then $i_n\geqslant N$ too, and therefore $d(x_n,x_{i_n})<\varepsilon$.