Permuting the terms of a sequence does not affect its convergence

Question

Let $x_n$ be a sequence such that $x_n \rightarrow 0$. Let $\sigma\colon\mathbb N \rightarrow \mathbb N$ be a bijection. Define a new sequence $y_n:= x_ {\sigma (n)} $. Show that $ y_n \rightarrow 0 $.

ATTEMPT Since $x_n$ converges to 0 implies after a certain $n>n_{0}$ all its terms will lie between $(0-\epsilon,0+\epsilon$). As $\sigma$ is a bijection so it is increasing function. If I take set of subscripts of original sequence, I take all $n$ after $n>n_0$ such that set $\{ n_1 ,n_2,n_3,n_4,\ldots\}$ where all $n_{i} ,i=1,2,3,\ldots$ are bigger than $n_0$. So applying $\sigma$ function which is an increasing function I get new increasing sequence of subscripts. Now I need to claim that $x_i = \sigma (n_i)$ in order to make furthure progress. But how? Thanks

Answer 1

Intuitively: The property that there exists $n_0$ such that $|x_n|<\epsilon$ for all $n>n_0$ can be reformulated as: The number of terms outside the interval $(-\epsilon,\epsilon)$ is finite. In this reformulation we see that the order of $\mathbb N$ doesn't matter.

Explicitly, one can proceed as follows: Given $\epsilon>0$, there exists $n_0$ such that $|x_n|<\epsilon$ for all $n>n_0$. Let $n_1=\max\{\sigma^{-1}(1),\ldots,\sigma^{-1}(n_0)\}$. Then for all $n>n_1$, $\sigma(n)$ is not in $\{1,\ldots,n_0\}$, hence $|y_n|=|x_{\sigma(n)}|<\epsilon$.