Let $Frege(X)$ be the set of all equivalence classes of subsets of $X$ under equivalence relation bijection. formally:
$Frege(X)= \{\{Y \subseteq X: |Y|=|Z|\}: Z \subseteq X \}$
Where $||$ is cardinality function defined after Scott's.
My question is about the following statement:
$\forall X: |Frege(X)| \leq |X|+1$
Is it equivalent to the axiom of choice over the rest of axioms of $\sf ZF$?
More precisely can $\sf ZF$ plus the above statement prove the axiom of choice?
Afternote: To spell the above without invoking Scott's cardinality just replace $|Y|=|Z|$ by $\exists f(f: Y \to Z \land f \text{ is a bijection})$, and replace $|Frege(X)| \leq |X| + 1$ by $\exists g (g: Frege(X) \to X \cup \{X\} \land g \text{ is an injection})$
from Hot Weekly Questions - Mathematics Stack Exchange
Zuhair
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