Axiom Of Choice

We have the following indirect implication of form equivalence classes:

408 \(\Rightarrow\) 61 given by the following sequence of implications, with a reference to its direct proof:

Implication	Reference
408 \(\Rightarrow\) 62	clear
62 \(\Rightarrow\) 61	clear

Here are the links and statements of the form equivalence classes referenced above:

Howard-Rubin Number	Statement
408:	If \(\{f_i: i\in I\}\) is a family of functions such that for each \(i\in I\), \(f_i\subseteq E\times W\), where \(E\) and \(W\) are non-empty sets, and \(\cal B\) is a filter base on \(I\) such that For all \(B\in\cal B\) and all finite \(F\subseteq E\) there is an \(i\in I\) such that \(f_i\) is defined on \(F\), and For all \(B \in\cal B\) and all finite \(F\subseteq E\) there exist at most finitely many functions on \(F\) which are restrictions of the functions \(f_i\) with \(i\in I\), then there is a function \(f\) with domain \(E\) such that for each finite \(F\subseteq E\) and each \(B\in\cal B\) there is an \(i\in I\) such that \(f\|F = f_i\|F\).
62:	\(C(\infty,< \aleph_{0})\): Every set of non-empty finite sets has a choice function.
61:	\((\forall n\in\omega, n\ge 2\))\((C(\infty,n))\): For each \(n\in\omega\), \(n\ge 2\), every set of \(n\) element sets has a choice function.

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