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?
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Yes, it's not the question in the title of the duplicate. But the accepted answer gives a solution which covers your question as well. – Asaf Karagila Aug 07 '19 at 12:44