The following is the question:
What sequences have a permutation, which is monotone increasing or monotone decreasing?
So first of all I believe that if a sequence is convergent, then there definitely exists such a permutation. I haven't proved it yet, but I think it shouldn't be that hard.
Now there are divergent sequences which doesn't have such permutations ($0,1,0,1,0,1,\dots$ as an example), but the trivial divergent sequence $(1,2,3,\dots)$ clearly works.
So in general, is it true that it has such a permutation if and only if it has a limit (perhaps $\pm \infty$)?
Best Answer
Clearly such a sequence must have a limit in the extended reals, but as was noted in the comments, that is not sufficient. Suppose that $\sigma=\langle x_n:n\in\Bbb N\rangle$ converges to $x$, and suppose further that there are $k,\ell\in\Bbb N$ such that $x_k<x<x_\ell$. Then there is an $m\in\Bbb N$ such that $x_k<x_n<x_\ell$ whenever $n\ge m$; clearly, however, there is no permutation of $\sigma$ that puts all of the terms $x_n$ with $n\ge m$ between $x_k$ and $x_\ell$, so $\sigma$ has no monotone permutation. Thus, if $\sigma$ does have a monotone permutation, we must have either $x_n\le x$ for all $n\in\Bbb N$, or $x_n\ge x$ for all $n\in\Bbb N$. Now I’ll show that this additional condition is also sufficient.
Suppose that $\sigma$ converges to $x$, and $x_n\le x$ for all $n\in\Bbb N$; the other case is similar. Let
$$A_0=\{n\in\Bbb N:x_n\le x-1\}\,,$$
and for $k\in\Bbb Z^+$ let
$$A_k=\left\{n\in\Bbb N:\frac1n<x\le\frac1{n+1}\right\}\,.$$
Since $\sigma$ converges to $x$, $A_k$ is finite for each $k\in\Bbb N$, and it follows easily that $\sigma$ has a monotone increasing permutation.
Thus, $\sigma$ has a monotone permutation if and only if it converges to some $x$ in the extended reals, and either $x_n\le x$ for all $n\in\Bbb N$, or $x_n\ge x$ for all $n\in\Bbb N$.