The classical Cantor-Schroder-Bernstein Theorem says that there exists a bijective function $X\leftrightarrow Y$ if and only if there exist injective functions $X\hookrightarrow Y$ and $Y\hookrightarrow X$.


It is known that without Axiom of Choice the injectivity cannot be replaced by the surjectivity in this result.


Nonetheless, without AC, we have the following simple



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> **Fact.** If for two sets $X,Y$ there are surjective functions $X\twoheadrightarrow Y$ and $Y\twoheadrightarrow X$, then there exists a bijective function $\mathcal P(X)\leftrightarrow \mathcal P(Y)$ between the corresponding power-sets.
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Can the power-sets in this result be replaced by some simpler sets built over $X$, for example, $X^\omega$ or $X\times\omega$?


More precisely:



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> **Question.** Assume that for two sets $X,Y$ there exist surjective functions $X\twoheadrightarrow Y$ and $Y\twoheadrightarrow X$.
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> Is it true (in ZF) that the sets
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> (i) $X^\omega$ and $Y^\omega$ have the same cardinality?
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> (ii) $X\times\omega$ and $Y\times\omega$ have the same cardinality?
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Implications of the existence of a pair of surjective functions, without Axiom of Choice
No, and here is a counterexample.


Suppose that $|\Bbb R|<|[\Bbb R]^\omega|$, that is, there are more countable *subsets* of reals than reals. This is indeed possible, e.g. if all sets of Lebesgue measurable.


Since $\sf ZF$ proves there are bi-surjections (in fact, an injection from $\Bbb R$ into $[\Bbb R]^\omega$), this would be a counterexample.


Now, $|\Bbb R^\omega|=|\Bbb R\times\omega|=|\Bbb R|$, and even without carefully checking what is $([\Bbb R]^\omega)^\omega$ and $[\Bbb R]^\omega\times\omega$, the cardinality cannot decrease.


