Return to Question

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?

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?

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

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

Finite examples include $\bigcup \{\{0\},\{1,2\}\}$ being super-singular, and $\bigcup\{ \{0 \} , \{1\}, \{2,3\} \}$ being ultra-singular. Infinite examples of ultra-singular includes the set union of 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?

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?

Let's 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 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?

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

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

Finite examples include $\bigcup \{\{0\},\{1,2\}\}$ being super-singular, and $\bigcup\{ \{0 \} , \{1\}, \{2,3\} \}$ being ultra-singular. Infinite examples of ultra-singular includes the set union of 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?