# Prove that permutation of a sequence does not affect convergence.

Given a sequence $$\{y_k\}$$ is obtained by permuting the sequence $$\{x_n\}$$ and: $$\{\forall n\in \mathbb N\ \exists k_n \in \mathbb N:x_n = y_{k_n}, n_1 \ne n_2 \implies k_{n_1} \ne k_{n_2}\}$$ and: $$\{\forall k\in \mathbb N\ \exists n_k \in \mathbb N:y_k = x_{n_k}, k_1 \ne k_2 \implies n_{k_1} \ne n_{k_2}\}$$ Prove that: $$\lim_{n\to \infty}x_n = a \implies \lim_{k\to \infty}y_k=a$$

I've been thinking of the following. Let there exist a bijection $$P:\mathbb N\to \mathbb N$$ which converts $$k$$ to $$n_k$$. So basically we have that if $$x_n$$ is convergent to some $$a$$ then:

$$|x_n - a| < \varepsilon$$

Thus : $$\forall \varepsilon>0 \ \exists N \in \mathbb N:n \ge N \implies |x_n - a| < \varepsilon$$

This is only valid after some $$N$$. The only terms violating that condition are $$x_1, x_2, \dots, x_N$$. My guess was that we could take $$\max\{P(n)\}$$ and that would imply that $$|y_{P(n)} - a| < \varepsilon$$, so: $$\forall \varepsilon>0 \ \exists\ M = \max\{n: 1 \le P(n) \le N\}: n>M \implies |y_n-a| < \varepsilon$$

I decided to test this idea graphically and here is what I got. $$x_n = \frac{10}{n}$$

So from the graph for $$\varepsilon = 4$$ it follows that for $$N \ge 3$$ all terms of $$x_n$$ are satisfying the condition of convergence. Now if we take the maximum value for $$P(n)$$ where $$n \le 3$$ we get $$16$$. But after $$y_{16}$$ there exists a value $$y_{19}$$ which falls out of the neighborhood of $$0$$.

Where did I get it wrong and how to prove the statement in the problem section?

if $$\{a\}$$ converges then for any $$\epsilon > 0$$ there is an $$N>0$$ such than $$n>N \implies |a_n - a| < \epsilon$$
This means that there are only finitely many $$a_n$$ such that $$|a_n - a| > \epsilon$$
When we permute $$\{a\}$$ to construct $$\{b\}$$ then there are only finitely many $$b_m$$ such that $$|b_m - a| > \epsilon$$
And we can choose $$M$$ such that $$M$$ is greater than the largest index associated with $$|b_m - a| < \epsilon$$
and $$m>M \implies |b_m - a| < \epsilon$$