# Čech functions and the axiom of choice

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.)

• 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. – Asaf Karagila May 20 at 11:26
• 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). – Asaf Karagila May 20 at 13:04
• 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). – Asaf Karagila May 22 at 10:27
• 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 $f=f_0$? – YCor May 22 at 10:53
• @YCor: I got a partial answer to this in this question. – Gro-Tsen May 23 at 23:11