# Question on Cauchy sequence in any metric space.

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.

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 $$d(x_n,x_{i_n})<\varepsilon$$.