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 $\{n\in \Bbb N: n<n_2\},$ 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 $\lim_{n\to \infty}a_{f(n)}=L.$


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