Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
Can you explain what you mean by "closure"? Under my interpretation, any uncountable set will do. –  François G. Dorais Jul 23 '10 at 0:29
    
Are you wondering if ZF proves the existence of a regular cardinal? –  François G. Dorais Jul 23 '10 at 0:32
    
By "closure", I mean the minimal superset closed under it. I already know ZF proves Aleph_0 is regular. –  Ricky Demer Jul 23 '10 at 0:37
    
OK, I had missed "iterated" in the title. I meant to say uncountable regular cardinal in my second comment. –  François G. Dorais Jul 23 '10 at 0:42
    
I deleted my answer since I completely misunderstood the question. –  Carlo Von Schnitzel Jul 23 '10 at 0:59

2 Answers 2

up vote 10 down vote accepted

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.


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.

share|improve this answer
    
What a great model! But is there any hope for us to have all cardinalities (including the non-well-founded ones) having cofinality $\omega$? That is, is it consistent with ZF that every set is a countable union of smaller sets? –  Joel David Hamkins Jul 23 '10 at 11:57
    
If you did have that every set was a countable union of smaller sets, then you could get the desired answer to the question by inducing on rank (which is well-founded in ZF) rather than cardinality. –  Joel David Hamkins Jul 23 '10 at 21:49
    
Some of the subsets might have the same rank as S. More generally, the argument works for the wellfounded part of the cardinal partial order, this might include all cardinals even if choice fails but I'm not sure about this particular model. –  François G. Dorais Jul 23 '10 at 22:16
    
In my first comment above, I meant to ask about non-well-orderable cardinalities (since ZF has Foundation axiom, we have well-founded $\in$). But I guess you are right that even if every set is a countable union of strictly smaller sets, then for the reason you say one cannot seem to induct on rank to show every set is reachable. –  Joel David Hamkins Jul 24 '10 at 10:46
    
Great! I don't have access to the paper tonight and was going to check tomorrow, but you saved me the trouble. Thanks again! –  Clinton Conley Jul 12 '11 at 20:46

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 $V_\delta$.


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.

share|improve this answer
    
Okay, so how about the converse? Given such a set S, can we find an ordinal with uncountable cofinality? Perhaps just take its rank? If that is correct, then, combining that with François' answer, I would conclude the answer to the original question is «no», ZF cannot prove the existence of such an S. –  Daniel Mehkeri Jul 24 '10 at 2:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.