I have a question concerning cardinal arithmetic in $\sf{ZF}$.
Write $ X \cong Y$ if there is a bijection between $X$ and $Y$.


Let $M$ be a set such that $2M\cong M$.
Can one show, in $\sf{ZF}$, that for any infinite subset $X$ of $M$, we also have $2X \cong X$ ?


Thank you :)



Subsets of $M$ such that $2M \cong M$
No.


Simply because if $A$ is a set of any cardinality, there is $B$ such that:


1. $|A|\leq|B$, and
2. $|B|+|B|=|B|$, and in fact we can require $|B\times B|=|B|$.


Just take $B=A^\omega$, all the functions from $\omega$ to $A$. Then we have that $|B|=|A|^{\aleph\_0}$ and therefore $|B\times B|=|A|^{\aleph\_0+\aleph\_0}=|A|^{\aleph\_0}=|B|$.


Now take $A$ to be some set such that $|A|<2|A|$, e.g. an infinite Dedekind-finite set, and you get a counterexample.


(As Joel remarks, taking $B=\omega\times A$ is sufficient for $|B|+|B|=|B|$. However, it might not be enough if we want $|B|^2=|B|$.)




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Another fun way of proving this is to prove that for every $\alpha$, $V\_\alpha$ satisfies that $|V\_\alpha|=2|V\_\alpha|=|V\_\alpha|^2$. So if the additivity would be hereditary (for infinite subsets), it would be provable in $\sf ZF$ that for every infinite cardinal, $|A|+|A|=|A|$. But this is not true, of course.


