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This question is a follow up to that posting.

Recall the definition of super/hyper/ultra-singular set given in the linked posting.

Is there a model of $\sf ZF$ in which every uncountable set is super-singular?

Same question but in terms of the other two kinds of singularity?

We know that a Gitik model of $\sf ZF$ has every uncountable set being singular? But can we have Gitik models satisfying any of the above three conditions?

To re-iterate the definitions:

A set $A$ is super-singular, if and only if, there exists a set $x$ such that $|A|=|\bigcup x|$ and $| A| > |x|$, and for each $y \in x$ we have: $ |y| \not > |x|$ .

A set $A$ is hyper-singular, if and only if, there exists a set $x$ such that: $|A|=|\bigcup x|$ and $| A| > |x|$, and for each $y \in x$ we have: $|y| \not > |x| \land |y| \neq |x|$.

A set $A$ is ultra-singular, if and only if, there exists a set $x$ such that $|A|=|\bigcup x|$ and $| A| > |x|$, and for each $y \in x$ we have: $|y| < |x|$.

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Sketching the proof of my comment. $\aleph_\omega$ is never super-singular. Suppose that $\bigcup x=\aleph_\omega$. Then for each $y\in x$, let $\alpha_y$ be the order type of $y$, and $\alpha=\sup\{\alpha_y\mid y\in x\}$. Then we have a surjection $f\colon x\times\alpha\to\aleph_\omega$ by sending $\langle y,\beta\rangle$ to the $\beta$th element of $y$ (or $0$ if $\beta\geq\alpha_y$). [This is what I mean by a 'Specker-type argument'.]

If $|x|<\aleph_\omega$ then $|x|=\aleph_n$ for some $n<\omega$, and if $y\in x\implies|y|\not>|x|$ then $\alpha_y<\aleph_{n+1}$ for all $y\in x$. Hence $\alpha\leq\aleph_{n+1}$. So if $\aleph_\omega$ is super-singular then there is a surjection $\aleph_n\times\aleph_{n+1}\to\aleph_\omega$ for some $n<\omega$.

The same argument shows that any limit ordinal is not super-singular.

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The ordinal $\omega_1$ is uncountable, even in ZF, but it can never be hyper-singular, since otherwise we would have $\omega_1=\cup x$ where $\omega_1>|x|$ and so $x$ is countable, and the elements $y\in x$ are all not larger than or equinumerous with $x$. So $x$ would be a countable set of finite sets of countable ordinals. But every such set has a countable union, without using AC.

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  • $\begingroup$ why $\omega_1=\bigcup x$? $\endgroup$
    – Zuhair Al-Johar
    Commented Nov 7 at 17:56
  • $\begingroup$ I take same-cardinality to mean equinumerous, so we can assume $\omega_1=\cup x$ without loss. $\endgroup$
    – Joel David Hamkins
    Commented Nov 7 at 19:03
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    $\begingroup$ Also limit cardinals are never super-singular by a Specker-type '$\omega_2$ is not a countable union of countable sets' argument. $\endgroup$
    – Calliope Ryan-Smith
    Commented Nov 7 at 19:13
  • $\begingroup$ @JoelDavidHamkins, was that because $\omega_1$ is well-orderable? So, the result must be restricted to non-well orderable sets? $\endgroup$
    – Zuhair Al-Johar
    Commented Nov 7 at 20:21
  • $\begingroup$ I just meant that if $|A|=|\cup x|$, meaning they are equinumerous, then $A$ is equinumerous with $\cup x$, and so we can replace $x$ with $x'$ where $A=\cup x'$. So that part of your change didn't actually change anything. Meanwhile, the rest of my answer here does use that $\omega_1$ is well-orderable, so that all the cardinals below it are also well-orderable, and this is how I concluded that $x$ must be countable and that the elements of $x$ must be countable as well, and indeed finite for the hypersingular case. $\endgroup$
    – Joel David Hamkins
    Commented Nov 8 at 0:37

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