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Puzzling

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    $\begingroup$ Now it's been solved, here is the background: A few years ago I taught Cantor's diagonal argument, that any map $f\colon X\to \mathcal{P}(X)$ is not surjective. For the midterm test I thought I would set the dual task of proving that every map $g\colon \mathcal{P}(X)\to X$ is not injective. In particular I asked: $${}$$Given $g\colon \mathcal{P}(X)\to X$, find $x\in X$ and $A,B\subseteq X$, such that $x\in A, x\notin B$ and $g(A)=x=g(B)$. $${}$$ This is of course the same as the question above, if you unpack the mathematical notation. $\endgroup$
    – tkf
    Commented Jun 6 at 0:45
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    $\begingroup$ -Who do you rule? -No one -Liar! I am the only ruler of no one! in memory of Raymond Smullyan. $\endgroup$
    – Alexey Birukov
    Commented Jun 6 at 9:21
  • $\begingroup$ Not true if the population is empty. SCNR. $\endgroup$
    – Torsten Schoeneberg
    Commented Jun 7 at 16:13
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    $\begingroup$ @Torsten Schoeneberg Population is not empty, because there is at least a person in charge of empty group. $\endgroup$
    – Alexey Birukov
    Commented Jun 7 at 22:31