Axiom Of Choice

We have the following indirect implication of form equivalence classes:

95-F \(\Rightarrow\) 425 given by the following sequence of implications, with a reference to its direct proof:

Implication	Reference
95-F \(\Rightarrow\) 67	Some theorems on vector spaces and the axiom of choice, Bleicher, M. 1964, Fund. Math. The Axiom of Choice, Jech, 1973b, page 148 problem 10.4
67 \(\Rightarrow\) 76	clear
76 \(\Rightarrow\) 425	On first and second countable spaces and the axiom of choice, Gutierres, G 2004, Topology and its Applications.

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

Howard-Rubin Number	Statement
95-F:	Existence of Complementary Subspaces over a Field \(F\): If \(F\) is a field, then every vector space \(V\) over \(F\) has the property that if \(S\subseteq V\) is a subspace of \(V\), then there is a subspace \(S'\subseteq V\) such that \(S\cap S'= \{0\}\) and \(S\cup S'\) generates \(V\). H. Rubin/J. Rubin [1985], pp 119ff, and Jech [1973b], p 148 prob 10.4.
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).
76:	\(MC_\omega(\infty,\infty)\) (\(\omega\)-MC): For every family \(X\) of pairwise disjoint non-empty sets, there is a function \(f\) such that for each \(x\in X\), f(x) is a non-empty countable subset of \(x\).
425:	For every first countable topological space \((X, \Cal T)\) there is a family \((\Cal D(x))_{x \in X}\) such that \(\forall x \in X\), \(D(x)\) countable local base at \(x\). \ac{Gutierres} \cite{2004} and note 159.

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