Suppose $|X|=2^{2^{\aleph_0}}$, $A \subset X$ and $|A|=2^{\aleph_0}$. Is it possible to prove that $|X \setminus A|=2^{2^{\aleph_0}}$ without using the axiom of choice and its consequences?
2026-03-27 22:03:27.1774649007
An elementary question about cardinal arithmetic without AC
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