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While thinking about permuting(arrange elements in order in a given set), this is available always for finite sets and this fact could be proven by mathematical induction.

However, when to think about permuting of infinite set, it is unsure of whether the permuting is available or not. For example, for given set $\Bbb N$, there's no way to exhaustively count up its available permuting methods by mathematical induction as finite case.

Is there any recommended way to go deeper about this infinite case of permutation?

Beverlie
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    Just to say, people tend not to agree on what the definition of a "permutation" of an infinite set is. This is because the group of all bijections tends to be hard to work with. Instead, people often just mean the subgroup which fixes all but finitely many elements. – lulu Sep 08 '17 at 11:25
  • This struck some curiosity. I think there could, "technically" be a permutation of an infinite set as long as we define a sequence $a_n$ containing all elements of the infinite set. For example, ${0,1,3,2,4,5,6,7,8,9,10,11,12,13,14,...}$ could be considered a permutation of the natural numbers. – J. Linne Oct 25 '21 at 07:11

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A permutation of a set $S$ is just a bijection $b : S \to S$. In the case of the natural numbers $\mathbb{N}$, there are uncountably many such permutations. See https://mathoverflow.net/a/29476/102928 for a short proof of this.

Hans Hüttel
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