Can iterating countable unions give every set? (ZF)

Question

Does ZF prove that there exists a set S such that S is not in the closure of {{s} : s in S} under at-most-countable unions?

Answer 1

Moti Gitik proved (assuming large cardinals) that all uncountable alephs can have cofinality ω. [All uncountable cardinals can be singular, Israel J. Math. 35 (1980), 61–88] I believe this may be the model you're looking for, but I don't know what happens to non-wellorderable sets in that model. According to the abstract copied below, this model is very close to what you have in mind.

Assuming the consistency of the existence of arbitrarily large strongly compact cardinals, we prove the consistency with ZF of the statement that every infinite set is a countable union of sets of smaller cardinality. Some other statements related to this one are investigated too.

Every wellorderable set $S$ in this model belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions. The proof is by induction on the cardinality of the infinite set $S$. As in the abstract, we may write $S = \bigcup_{i<\omega} S_i$ where $|S_i| < |S|$ for each $i$. By the induction hypothesis, each $S_i$ belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions, and thus $S$ belongs to this closure too.

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Actually, every set $S$ in this model belongs to the closure of $\{\{s\} : s \in S\}$ under iterated countable unions; this is essentially what Gitik's Theorem 6.3 says.

Answer 2

If we assume ZF plus the assertion that $\omega_1$ is regular (which is provable from countable choice), or that indeed there is any uncountable regular cardinal $\delta$, then such a set $S$ exists. (Note that François provided a model having no uncountable regular cardinal, where there is no well-orderable counterexample.)

Let $S$ simply contain elements of Levy rank unbounded in $\delta$. To be specific, you could let $S$ be $\delta$ itself, or $\omega_1$ if this is regular as we usually expect. If a set $X$ has Levy rank $\alpha$, so that $X\in V_{\alpha+1}$, then the union set $\bigcup X$ is also in $V_{\alpha+1}$. In particular, $V_\delta$ is closed under arbitrary unions of its elements. Also, it contains every element of $S$ and also {s} for $s\in S$. Furthermore, since $\delta$ has uncountable cofinality, $V_\delta$ contains as elements all of its countable subsets, since any such subset would have rank bounded below $\delta$. Thus, the clsoure of your set is contained within $V_\delta$, but $S$ is not in $V_\delta$.

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A simpler instance of the idea:

The union of any set of ordinals is still an ordinal. Thus, if $\delta$ is an ordinal with uncountable cofinality, then it is already closed under the process of taking countable unions of its elements, but doesn't contain $\delta$ itself as a member.

More specifically, if we take $S=\omega_1$, provided this is regular, then if we start with {s} for all $s\in S$ and close under the process of taking countable unions, we simply get all countable ordinals, and do not generate $\omega_1$ itself this way.