We have the following indirect implication of form equivalence classes:

107 \(\Rightarrow\) 46-K
given by the following sequence of implications, with a reference to its direct proof:

Implication Reference
107 \(\Rightarrow\) 62 clear
62 \(\Rightarrow\) 61 clear
61 \(\Rightarrow\) 46-K clear

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

Howard-Rubin Number Statement
107:  

M. Hall's Theorem: Let \(\{S(\alpha): \alpha\in A\}\) be a collection of finite subsets (of a set \(X\)) then if

(*) for each finite \(F \subseteq  A\) there is an injective choice function on \(F\)
then there is an injective choice function on \(A\). (That is, a 1-1 function \(f\) such that \((\forall\alpha\in A)(f(\alpha)\in S(\alpha))\).) (According to a theorem of P. Hall (\(*\)) is equivalent to \(\left |\bigcup_{\alpha\in F} S(\alpha)\right|\ge |F|\). P. Hall's theorem does not require the axiom of choice.)

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.

46-K:

If \(K\) is a finite subset of \(\omega-\{0,1\}\), \(C(\infty,K)\): For every \(n\in K\), every set of \(n\)-element sets has a choice function.

Comment:

Back