Axiom Of Choice

We have the following indirect implication of form equivalence classes:

109 \(\Rightarrow\) 350 given by the following sequence of implications, with a reference to its direct proof:

Implication	Reference
109 \(\Rightarrow\) 66	clear
66 \(\Rightarrow\) 67	Existence of a basis implies the axiom of choice, Blass, A. 1984a, Contemporary Mathematics
67 \(\Rightarrow\) 126	clear
126 \(\Rightarrow\) 350	clear

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

Howard-Rubin Number	Statement
109:	Every field \(F\) and every vector space \(V\) over \(F\) has the property that each linearly independent set \(A\subseteq V\) can be extended to a basis. H.Rubin/J.~Rubin [1985], pp 119ff.
66:	Every vector space over a field has a basis.
67:	\(MC(\infty,\infty)\) \((MC)\), The Axiom of Multiple Choice: For every set \(M\) of non-empty sets there is a function \(f\) such that \((\forall x\in M)(\emptyset\neq f(x)\subseteq x\) and \(f(x)\) is finite).
126:	\(MC(\aleph_0,\infty)\), Countable axiom of multiple choice: For every denumerable set \(X\) of non-empty sets there is a function \(f\) such that for all \(y\in X\), \(f(y)\) is a non-empty finite subset of \(y\).
350:	\(MC(\aleph_0,\aleph_0)\): For every denumerable set \(X\) of non-empty denumerable sets there is a function \(f\) such that for all \(x\in X\), \(f(x)\) is a finite, non-empty subset of \(x\).

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