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Zuhair Al-Johar
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What is the relationship between non-existence of those kinds of singular sets and AC?

Lets say a set $x$ is super-singular, if and only if, $| \bigcup x| > |x|$, and for each $y \in x$ we have: $ |y| \not > |x|$ .

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

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

finite examples include $\{\{0\},\{1,2\}\}$ being super-singular, and $\{ \{0 \} , \{1\}, \{2,3\} \}$ being ultra-singular. Infinite examples of ultra-singular includes Russell's sock.

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

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

Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47