Very neat question on math.stackexchange.
Cantor proved that
|X| < 2^|X|
for any set X. If we assume that X is infinite, from the axiom of choice we have that |X^n|=|X| for any positive integer n, and so
|X^n| < 2^|X|.
But if we do not assume choice, it may well be that
|X|<|X^2|<|X^3|...
Specker proved in 1954 (without choice) that if X has at least 5 elements, then 2^|X| is *not* smaller than or equal to |X^2|.
On the other hand, |X^2| is at most (2^|X|)^2 = 4^|X|.
1/
The question (by Jade Vanadium) is: in ZF (that is, without assuming choice), suppose that X is infinite. Given a positive integer n, what is the smallest k such that
|X^n| is at most k^|X|?
Easily, k is at most 2^n. More serious work gives that k is at most n+1. (!)
Call k_n the smallest such k.
From Specker's results, k_2>2. The posted question has a proof that k_n is at most n+1. In particular, k_2=3.
But we don't even seem to know k_3 at the moment.
The poster of the question, Jade Vanadium, has solved it: k_n=n+1. That it is at most n+1 was proved in the body of the question. That it is at least n+1 appear now in their answer:
https://math.stackexchange.com/a/4633716/462
Specifically, the argument shows that if X is *strictly amorphous*, then |X^n| does not inject in n^|X|.
Recall that an infinite set is amorphous if and only if it cannot be split into two infinite sets.
It is consistent with ZF that there are amorphous sets of reals.
If an amorphous set is partitioned into finite sets, then there is exactly one m such that infinitely many of the pieces have size m.
An amorphous set is strictly amorphous if and only if this number m is 1.
This variety of amorphous sets is known to be consistent, and John Truss provided an in-depth analysis of it in ""The structure of amorphous sets", Annals of Pure and Applied Logic 73 (1995) 191–233
https://www.sciencedirect.com/science/article/pii/016800729400024W
Posted a follow up question,
whether ZF proves that
|X| n^|X| ≤ (n+1)^|X|
for n natural and X infinite,
which I expect should follow similarly. I confess my naive attempts at finding more direct arguments ended up failing.
@AndresCaicedo Is this ZF, or ZFC? My knowledge is a bit rusty, but let me try....I may be wrong...
n^|X| = the cardinality of the set of all functions from |X| to n? If we speak of cardinality, we assume AC, right?
Any way, we have 2.X ~ X for infinite X, so for n>1 we have n^X ~ n^(2.X) ~ (n.n)^X >= (n+1)^X
For n=1 we have the inequality directly.
@AndresCaicedo I fear I totally misunderstood the question :) Sorry
