We know that whether $|P(x)|=|P(y)|$ implies $|X|=|Y|$ is dependent on CH. Let $W(X)$ be the set of all well orders over $X$. Does $|W(X)|=|W(Y)|$ imply $|X|=|Y|$? Is the answer dependent on CH? More generally, what axioms do we need to show $|W(X)|=|P(X)|$? Is a bijection between them constructible?
6
-
$\begingroup$ What do you mean by 'constructible'? $\endgroup$– James E HansonCommented Nov 2 at 8:01
-
$\begingroup$ Assuming $2^{\aleph_0}=2^{\aleph_1}$ we get that $|W(\omega)|=|W(\omega_1)|$. $\endgroup$– Asaf Karagila ♦Commented Nov 2 at 8:13
-
1$\begingroup$ $|W(0)|=|W(1)|$ $\endgroup$– bofCommented Nov 2 at 8:32
-
4$\begingroup$ Isn't $|P(X)|=|W(X)|$ just true in ZFC for all infinite $X$? $\endgroup$– Alessandro CodenottiCommented Nov 2 at 13:30
-
6$\begingroup$ @AlessandroCodenotti It is, and the OP should know this from the answer to their previous question: mathoverflow.net/a/481412 . $\endgroup$– Emil JeřábekCommented Nov 2 at 13:47
|
Show 1 more comment