It's well known that for any set $A$, $A < P(A)$.
But now, I have some question that, WITHOUT AC, can we guarantee that $A \leq X$ or $X \leq A$ whenever $X \subseteq P(A)$?
Thank you.
2026-04-03 20:55:11.1775249711
Comparability of a set and a subset of power set.
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No, we can't quite prove this.
Cohen's first model for the failure of the axiom of choice has a set of real numbers which is infinite but Dedekind-finite. This means exactly that this set has cardinality incomparable with that of $\Bbb N$.
But we know that $\Bbb R$ and $\mathcal P(\Bbb N)$ have the same cardinality, even without assuming the axiom of choice. So in this universe of sets there is a subset of $\mathcal P(\Bbb N)$ which is incomparable with $\Bbb N$.