In the basic Fraenkel (permutation) model, any functions from $\omega$ to $P(A)$ has finite range.

Question

Working within the Basic Fraenkel (permutation) model. As defined in Jech's axiom of choice.

Let $A$ be the set of atoms.

I want to show that any function $f: \omega \rightarrow P(A)$ has finite range. But honestly I am really stuck, any help or insights is well appreciated.

Cheers

Answer 1

The same way you show, e.g. that there is no injective function $\omega\to A.$ Assume $f:\omega\to P(A)$ has infinite range, and let $E$ be an arbitrary finite subset of $A.$ Since the range is infinite, you can find two atoms $a_1,a_2$ such that $a_1,a_2\notin E$ and there is an $n\in\omega$ such that $a_1\in f(n)$ but $a_2\notin f(n).$ So the permutation that swaps $a_1$ and $a_2$ fixes $E$ but does not fix $f.$ Thus $E$ is not a support for $f,$ and since $E$ was arbitrary, $f$ is not symmetric.

Answer 2

Here is an indirect way, which might not reveal much on the Fraenkel model itself, but it is more informative in other ways.

Theorem 1. (Kuratowski) There is a surjection from $X$ onto $\omega$ if and only if there is an injection from $\omega$ into $\mathcal P(X)$.

This one has a bit of a difficult proof, so I will skip it here.

Recall that a set $A$ is amorphous if it cannot be written as a disjoint union of two infinite sets.

Theorem 2. If $A$ is amorphous, then any function from $A$ into a set which can be well-ordered has a finite range.

Proof. Suppose not, then by the basic properties of well-ordered sets we may assume that the function is onto $\omega$. Now take the pre-image of the odd integers and the even integers, and these must be two disjoint infinite subsets.

Remark. We can extend the proof to "linearly ordered", but conceptually it is easier to prove it about well-ordered sets, and it is enough for our case.

Theorem 3. In the basic Fraenkel model the set of atoms is amorphous.

Proof. If $B\subseteq A$ is an infinite set, let $E$ be a finite set such that whenever $\pi\colon A\to A$ is a permutation satisfying $\pi\restriction E=\operatorname{id}$, $\pi(B)=B$.

If $a\in A\setminus B$, but $a\notin E$, then we may take any $b\in B\setminus E$ and consider the $2$-cycle, $\pi=(a\ b)$. Easily, $\pi\restriction E=\operatorname{id}$ and $p(B)\neq B$. Therefore $A\setminus B\subseteq E$, and thus finite.

Corollary. There is no injection from $\omega$ into $\mathcal P(A)$ in Fraenkel's basic model, where $A$ is the set of atoms.

Proof. There is no surjection from $A$, the set of atoms, in Fraenkel's model onto $\omega$, and therefore no injection from $\omega$ into $\mathcal P(A)$.