17
$\begingroup$

Is it true that if $X$ and $Y$ are sets and there is a bijection between $\mathcal{P}(X)$ and $\mathcal{P}(Y)$ then there is a bijection from $X$ to $Y$ ?. I believe this should be obvious, but I can't see why this is so. A proof or a counter example would be highly appreciated.

Thanks.

$\endgroup$
16
$\begingroup$

This cannot be proved from the axioms of $\sf ZFC$ (and so certainly not from naive set theory) but it cannot be refuted either.

That is to say, assuming that the axioms of set theory (read: $\sf ZFC$) are consistent there are models of set theory in which $2^X\sim 2^Y\implies X\sim Y$, and there are other models in which there are $X\nsim Y$ such that $2^X\sim 2^Y$.

For example if $\sf GCH$ holds then the statement is true, because the power set is "as small as possible", but it is consistent that there is an uncountable set of real numbers whose power set is equipotent with the real numbers themselves, i.e. $X$ such that $\Bbb N<X$ but $2^X\sim 2^\Bbb N$.

The statement is weaker than $\sf GCH$, and in a related (but unrelated) post on MathOverflow I called it "Injective Continuum Function", ICF. I have seen mentioning that this was called by Tarski "Weak Power Hypothesis", WPH.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How puzzling for non set-theoricists...+1. At least for me. $\endgroup$ – Julien Apr 29 '13 at 19:22
  • 3
    $\begingroup$ @julien: I agree. One of the greatest difficulties is to understand how sets can have such different properties from one model to another. But let's put on the logical goggles and ask, can we prove from the axioms of fields that $\sqrt32$ exists? Can we prove it is unique? No, there are three fields, one in which there is no cube roots, one where there is just one, and another with three. But that doesn't surprise anyone, does it now? :-) $\endgroup$ – Asaf Karagila Apr 29 '13 at 19:26
  • $\begingroup$ Nice point. I'll keep that example in mind, thanks. $\endgroup$ – Julien Apr 29 '13 at 19:31
  • 1
    $\begingroup$ Argh, I meant $\sqrt[3]2$, not $\sqrt32$... $\endgroup$ – Asaf Karagila Apr 29 '13 at 19:41
  • 1
    $\begingroup$ @AsafKaragila I was trying to prove it for the OP's conjecture for the last 30 minutes until I saw your answer +1 $\endgroup$ – Amr Apr 29 '13 at 19:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.