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Bumped by Community user
Bumped by Community user
added a comment by @Gro-Tsen as a note
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Is there a model of ${\sf ZF}$ such that there is an infinite set $X$ and a injective map $f:{\cal P}(X)\to {\cal P}(X)$ so that for $a\neq b \in {\cal P}(X)$ we have $|f(a)\cap f(b)| \leq 1$?


Note. As user Gro-Tsen in the comments below points out, if we weaken the condition $|f(a)\cap f(b)| \leq 1$ to "$f(a)\cap f(b)$ is finite", then the resulting statement is a theorem of ${\sf ZFC}$.

Is there a model of ${\sf ZF}$ such that there is an infinite set $X$ and a injective map $f:{\cal P}(X)\to {\cal P}(X)$ so that for $a\neq b \in {\cal P}(X)$ we have $|f(a)\cap f(b)| \leq 1$?

Is there a model of ${\sf ZF}$ such that there is an infinite set $X$ and a injective map $f:{\cal P}(X)\to {\cal P}(X)$ so that for $a\neq b \in {\cal P}(X)$ we have $|f(a)\cap f(b)| \leq 1$?


Note. As user Gro-Tsen in the comments below points out, if we weaken the condition $|f(a)\cap f(b)| \leq 1$ to "$f(a)\cap f(b)$ is finite", then the resulting statement is a theorem of ${\sf ZFC}$.

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Model ${\sf ZF}$ that "spreads" members of ${\cal P}(X)$

Is there a model of ${\sf ZF}$ such that there is an infinite set $X$ and a injective map $f:{\cal P}(X)\to {\cal P}(X)$ so that for $a\neq b \in {\cal P}(X)$ we have $|f(a)\cap f(b)| \leq 1$?