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Let's say 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: $|y| < |x|$.

Finite examples include $3=\bigcup \{\{0\},\{1,2\}\}$ being super-singular, and $4=\bigcup\{ \{0 \} , \{1\}, \{2,3\} \}$ being ultra-singular. Infinite examples of ultra-singular includes the set union of a Russell's sock.

Is it equivalent to AC to say that no infinite super-singlular set exists?

Which of the known choice principles non-existence of an infinite set of each of the other two kinds of sets is equivalent to?

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    $\begingroup$ To match the familiar terminology, one should have the adjectives "supersingular", "ultrasingular" etc. apply to the set $\cup x$, not $x$. After all, $\aleph_\omega$ is singular, because it is the union of $x=\{\aleph_0,\aleph_1,\aleph_2,\ldots\}$, but we don't say that that set is singular. A cardinal is singular, when it is the size of a set that is the union of a small number of small sets. $\endgroup$
    – Joel David Hamkins
    Commented Nov 5 at 22:56
  • $\begingroup$ @JoelDavidHamkins, thanks! I've edited. $\endgroup$
    – Zuhair Al-Johar
    Commented Nov 6 at 4:47
  • $\begingroup$ But with your edit, the definition is now not well defined, since the property depends on $x$ and not just on $\cup x$. What I had meant was for you to say that a set $A$ is supersingular, if there is an $x$ for which $A=\cup x$ and so forth. $\endgroup$
    – Joel David Hamkins
    Commented Nov 6 at 15:10
  • $\begingroup$ @JoelDavidHamkins, I've further refined this definition to include $|A|=|\bigcup x|$ instead of $A=\bigcup x$, because singularity is a cardinality issue. $\endgroup$
    – Zuhair Al-Johar
    Commented Nov 7 at 16:09

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Yes, it is equivalent to axiom of choice to say that there is no supersingular set.

It is easy to see with AC that there is no supersingular set.

Conversely, suppose that there is no supersingular set. I claim this implies that $\kappa^2=\kappa$ for every infinite cardinal. This is because if there is a cardinal $\kappa$ with $\kappa^2>\kappa$, then we may regard $\kappa^2$ as the union of a size $\kappa$ family of size $\kappa$ sets, which would mean it is a supersingular.

So if there is no supersingular set, then $\kappa^2=\kappa$, which implies AC.

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