Finding explicit bijections between sets of the same cardinality 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 $f: S \to T$ 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. 
 A: I cannot fully answer your question, but I will leave my (rather humble) two cents on this.
You can also prove that two sets have the same cardinality by either showing that there is an injection from each to the other, a surjection from each on the other, or an injection and a surjection from one to the other. Having that said, it is sometimes much easier to follow any of the strategies above instead of creating an explicit bijection.
On top of my head I can give you this example: "show that $\Bbb {N}$ and $\Bbb {N}^k$ have the same cardinality". I don't know any explicit bijections from one to the other, but I can easily build 2 injections and then the result will follow! For the injection $\Bbb {N} \rightarrow \Bbb {N}^k$ I choose $f(n) = (n, 0, \cdots, 0) $. For the injection $\Bbb {N} \leftarrow \Bbb {N}^k$ I do $g(n_1, \cdots, n_k) = 2^{x_1}3^{x_2}\cdots p_k^{x_k} $ where $p_k $ is the $k$th prime number.
A: This is independent from ZFC.
There is a first-order formula $\varphi$ such that $$\mathrm{ZFC} + V=L \vdash(\forall S)(\forall T)\Big(\big(\operatorname{card}(S)=\operatorname{card}(T)\big)\rightarrow\big(\{x \mid \varphi(x, S, T)\}\text{ is a bijection from }S\text{ to }T\big)\Big).$$
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 $\varphi$ 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 $\aleph_1^{\,L}$ countable, then, in the generic extension, there is a bijection between $\omega$ and $\aleph_1^{\,L},$ 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 $\aleph_1^{\,L}$ is countable but in which there is no definable bijection between $\omega$ and $\aleph_1^{\,L}.$
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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.
