TITLE: Do bounded permutations of N leave an initial segment invariant?
QUESTION [5 upvotes]: Let $p$ be a permutation of $\mathbb{N}$. We say that $p$ is bounded if there exists $k$ so that $|p(i)-i| \le k$ for all $i$. If $p$ is bounded, must there exist $M>0$ such that
$p(\{1,2,\ldots, M\}) = \{1,2,\ldots, M\}$? If so, can we bound $M$ by a function of $k$?

REPLY [3 votes]: Qiaochu's example, as noted, has no finite cycles. At the other extreme, here's an involution, that is, a product of disjoint transpositions: $$(1,5)(2,4)(3,7)\prod_{k=1}^{\infty}(8k-2,8k+2)(8k,8k+4)(8k+1,8k+5)(8k+3,8k+7)$$ Nothing gets moved by more than 4, and no initial segment is fixed.