Let $X$ be an infinite set. Then it is provable (I believe) that $$X = \{(x_1, x_2, \ldots, x_n) \mid n ∈ \mathbb{N}, x_i ∈ X\}.$$ Without resorting to abstract results in set theory, can we find a bijection that makes this true?
1 Answers
This is an inherently vague question, but there is a precise result which indicates that the answer is no. Specifically, it is consistent with ZF (= set theory with the Axiom of Choice) that there are infinite sets for which that equality fails. Indeed, the mere statement "For all infinite sets $X$, we have $X\equiv X^2$" is equivalent to the axiom of choice.
So without knowing anything else about $X$, we can't be sure the result holds unless we accept the axiom of choice, and if that's not "abstract set theory" I don't know what is.
That said, for some infinite sets we can do this. For example, taking $X=\mathbb{N}$ we can use prime factorization: consider the map $$\langle x_1,...,x_n\rangle\mapsto p_1^{x_1+1}\cdot ...\cdot p_n^{x_n+1},$$ where $p_i$ is the $i$th prime (the "$+1$" superscript is to handle the case when $x_i=0$  if you don't consider $0$ a natural number, you don't need it). It's easy to show that this is an injection. With this in mind it's not hard to exhibit and verify a concrete bijection if you don't want to use CantorBernstein. (And there are plenty of other ways to do this too.)
It's a good exercise at this point to come up with an explicit injection from the set of all finite sequences of real numbers into $\mathbb{R}$.
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You're absolutely right. I was too quick. In fact my $X$ is an infinite _ring_, but I thought I could generalise it. I accept AC, and can work with it somewhat. I suppose a similar thing could be done with the prime elements of the ring then? Or can infinite rings contain only finitely many prime elements? Then I suppose it doesn't work anymore. – Jos van Nieuwman Sep 29 '19 at 20:02