# Does changing the order of a sequence change its limit?

Say $$a_n$$ is a convergent sequence, that is, that $$a:\mathbb{N}\rightarrow\mathbb{R}$$ and that there exists a number $$L$$ (the limit) such that forall $$\epsilon>0$$ exists $$n_0\in\mathbb{N}$$ such that forall $$n\geq n_0$$ $$|a_n-L|<\epsilon$$. Is there a biyection $$\sigma:\mathbb{N}\rightarrow\mathbb{N}$$ such that $$\lim_n a_{\sigma(n)}\neq L$$.

Notes: Because $$a_n$$ converges to $$L$$ that must be the only accumulation point of the image of the sequence and since the limit of a sequence is an accumulation point, $$a_{\sigma(n)}$$ can't be convergent and not have $$L$$ for its limit; then, $$a_{\sigma(n)}$$ can only be divergent.

Also, because of the ($$\epsilon$$) definition of limit there are only a finite amount of terms of the sequence such that $$|a_{\sigma(n)}-L|>\epsilon$$ for any chosen $$\epsilon>0$$, therefore there is a maximum $$m$$ such that $$|a_{\sigma(n)}-L|>\epsilon\implies n\leq m$$, therefore if we define $$n_0=m+1$$ all $$n\geq n_0$$ satisfies $$|a_n-L|<\epsilon$$ so $$L$$ should be the limit (QED?).

My problem appears when we say there is a choice function $$f$$ over $$\text{Sub}(\mathbb{N}):=\{A:A\subseteq\mathbb{N}\}$$ such that we define $$g(0)=\mathbb{N}$$ and $$g(n+1)=g(n)-f(g(n))$$ such that $$\sigma(n)=f(g(n))$$ (Notice that since all the subsets of $$\mathbb{N}$$ are numerable, we don't need the axiom of choice since it's provable).

$$\lim_{n\to \infty}a_n=L$$ iff, for every $$\epsilon>0,$$ the set $$\{n\in \Bbb N: |L-a_n|\ge \epsilon\}$$ is finite.

Let $$\lim_{n\to \infty}a_n=L$$ and let $$f:\Bbb N\to \Bbb N$$ be a bijection. Let $$b_n=a_{f(n)}$$ for each $$n\in \Bbb N.$$

Given $$\epsilon >0,$$ the set $$S(\epsilon)=\{n\in \Bbb N: |L-a_n|\ge \epsilon\}$$ is finite (and possibly empty). So take $$n_1\in \Bbb N$$ such that no member of $$S(\epsilon)$$ is greater than $$n_1.$$ Let $$n_2=1+\max \{f^{-1}(n): n\le n_1\}.$$

Now if $$n\ge n_2$$ then $$|L-b_n|<\epsilon,$$ because $$n\ge n_2\implies$$ $$f(n)>n_1\implies$$ $$f(n)\not \in S(\epsilon)\implies$$ $$\epsilon>|L-a_{f(n)}|=|L-b_n|.$$

So $$\{n:|L-b_n|\ge \epsilon\}$$ is a subset of $$\{n\in \Bbb N: n which is a finite set.

Therefore $$L=\lim_{n\to \infty}b_n=\lim_{n\to \infty}a_{f(n)}.$$

A similar argument shows that if $$\lim_{n\to \infty}a_n=L$$ and if $$f:\Bbb N \to \Bbb N$$ is finite-to-one (that is, for each $$n\in \Bbb N,$$ the set $$\{m:f(m)=n\}$$ is finite), then $$\lim_{n\to \infty}a_{f(n)}=L.$$