This is kind of a general question about establishing that two sets have the same cardinality. The definition of two sets S and T having the same cardinality is that there is a function f:S→T that is one-to-one and onto all of T.
My question is: are there any two sets for which we (and by that I mean "the mathematical community" or whatever) know that the two sets have the same cardinality by other means, but cannot find a bijection between them? I feel like the continuum hypothesis is in some vague sense related to this, but anyways I'm curious and would like to know.
Answer
This is independent from ZFC.
There is a first-order formula φ such that ZFC+V=L⊢(∀S)(∀T)((card(S)=card(T))→({x∣φ(x,S,T)} is a bijection from S to T)).
So it's consistent with ZFC that we can explicitly specify a bijection between any two sets of the same cardinality (assuming that ZFC is consistent, of course). [By the way, the formula φ can be spelled out — it's not mysterious, just long if written out in full.]
On the other hand, if, for example, you use finite forcing to make ℵL1 countable, then, in the generic extension, there is a bijection between ω and ℵL1, but there is no bijection between them that is definable without parameters.
So (again assuming the consistency of ZFC), there is a model of ZFC in which ℵL1 is countable but in which there is no definable bijection between ω and ℵL1.
Some brief remarks on how to prove the statements above:
The existence of explicit bijections in L is due to the fact that there is, if V=L, a definable well-ordering of the universe.
The non-existence of a definable bijection between the two specified countable sets in the generic extension above can be proven using the homogeneity of the partial ordering used in the forcing argument.
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