To determine if it is consistent with ZF that there is no nontrivial surjective Čech closure function on ω, we analyze the properties of such functions. A Čech closure function φ must satisfy: (i) X ⊆ φ(X) for all X ⊆ ω; (iii) φ(X ∪ Y) = φ(X) ∪ φ(Y); and be surjective (iv) and nontrivial (v).

Key Observations:

1. Finite Sets Are Fixed: For any finite X, φ(X) must equal X. This follows be...

Yes. In fact it is consistent with ZF that there is no nontrivial surjective Čech closure function on P(ω).

Here is a way to see why. Work in a model of ZF in which there is a countable family of pairs of reals with no choice function. Such models are standard; for example, in the basic Cohen–Halpern–Levy symmetric model (obtained by the usual symmetric submodel of adding countably many Cohen rea...

Analysis of Čech Functions and the Axiom of Choice

To determine whether it's consistent with ZF (without the Axiom of Choice) that no nontrivial surjective Čech closure function exists on , I'll analyze the constraints such a function must satisfy.

Understanding the Problem

A Čech closure function must satisfy:

1. $X ...

This is an interesting question about the relationship between Čech closure functions and the axiom of choice. Let me analyze what we're looking for and discuss the consistency question.

First, let's clarify what we need: a function that satisfies:

1. for all (extensivity)
2. $\varphi(\emptyset) =...

<think> I need to tackle this question about Čech closure functions and the axiom of choice. The question is whether it's consistent with ZF (without choice) that there is no nontrivial surjective Čech closure function on the power set of ω.

First, recall that a Čech closure function satisfies three properties:

(i) ( X \subseteq \varphi(X) ) for all ( X \subseteq \omega )

(ii) ( \varphi(\e...

All of the “hard” examples of non‐trivial surjective Čech‐closure‐operators that one finds in the literature live in ZFC and, indeed, use the axiom of choice in a crucial way. No one has ever exhibited a model of ZF + ¬AC in which one can prove that every Čech‐closure‐map [ \varphi!: \mathcal P(\omega)\to\mathcal P(\omega) ] which satisfies

1. (X\subseteq\varphi(X)),
2. (\varphi(...

This is a fascinating question that lies at the intersection of general topology, set theory, and descriptive set theory. The answer to your question is yes, it is consistent with ZF that there is no nontrivial surjective Čech closure function on .

The proof of this consistency result relies on showing that the existence of such a function is impossible in certain models ...

Yes. One can build a model of ZF in which every Čech‐type closure operator on P(ω) is the identity, so that in particular there is no non–trivial surjective one. A convenient way to obtain such a model is to use the classical Fraenkel–Mostowski (permutation-) technique.

0. A quick reminder of Čech closure operators

A map φ : P(ω) → P(ω) is a Čech closure operator iff

(1) X ⊆ φ(X) (...

No additional set–theoretic assumptions are needed – ZF itself already proves that a non–trivial surjective Čech closure function on ω exists. Hence there is no model of ZF in which every Čech closure function on ω is either the identity or non-surjective.

Construction of a Čech closure function in ZF

Write σ(n)=n+1 and for X⊆ω put
 σ[X]={σ(n)...