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A Čech closure function on $\omega$ is a function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ such that (i) $X\subseteq\varphi(X)$ for all $X\subseteq\omega$, (ii) $\varphi(\emptyset)=\emptyset$, and (iii) $\varphi(X\cup Y)=\varphi(X)\cup\varphi(Y)$ for all $X,Y\subseteq\omega$; in other words, it obeys the Kuratowski closure axioms except possibly the idempotent law $\varphi(\varphi(X))=\varphi(X)$.

In 1947 E. Čech asked if there exists such a closure function (on any set, not necessarily $\omega$) which is also surjective and nontrivial (not the identity map). Čech's question has been answered in the affirmative under various set-theoretic assumptions, including ZFC. The ZFC example is rather complicated and makes heavy use of the axiom of choice; it seems unlikely that one could construct such a thing without choice.

Question. Is it consistent with ZF that there is no nontrivial surjective Čech closure function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$?

That is, there is no function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ satisfying the conditions (i)-(iii) above and also (iv) for every $Y\subseteq\omega$ there exists $X\subseteq\omega$ such that $\varphi(X)=Y$, and (v) $\varphi(X)\ne X$ for some $X\subseteq\omega$. (Condition (ii) is now redundant, as it follows from (i) and (iv) that $\varphi(X)=X$ if $X$ is finite.)

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  • $\begingroup$ I could imagine several candidate models: Maybe Cohen's model is a suitable example with the Dedekind-finite set of reals "spoiling things", but more likely would be Solovay's model where there are no MAD families, etc. $\endgroup$ – Asaf Karagila May 20 at 11:26
  • $\begingroup$ Also a candidate: the Feferman–Levy model where the reals are a countable union of countable sets, or the Truss model (although if this holds there, I'd imagine the Solovay model would already capture this). $\endgroup$ – Asaf Karagila May 20 at 13:04
  • $\begingroup$ I think it would be a good point in time to migrate this to MathOverflow. I don't have an obvious solution at hand, and I have quite a lot of things on my hands right now (which is why I might be missing something obvious). $\endgroup$ – Asaf Karagila May 22 at 10:27
  • $\begingroup$ Is it correct to think as "in ZF there's such a function" as "one can construct such a function"? More precisely, is it true (from some general principle) that if "there is such a function" is a theorem of ZF, then there exists $f_0\in P(\omega)^{P(\omega)}$ satisfying the given condition, and a formula of set theory $F(x)$ with only parameter $x$, such that for every $f\in P(\omega)^{P(\omega)}$, $F(f)$ holds iff $f=f_0$? $\endgroup$ – YCor May 22 at 10:53
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    $\begingroup$ @YCor: I got a partial answer to this in this question. $\endgroup$ – Gro-Tsen May 23 at 23:11

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