Implications

We have the following indirect implication of form equivalence classes:

95-F \(\Rightarrow\) 316
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 \(\Rightarrow\) 89 On cardinals and their successors, Jech, T. 1966a, Bull. Acad. Polon. Sci. S'er. Sci. Math. Astronom. Phys.
89 \(\Rightarrow\) 90 The Axiom of Choice, Jech, 1973b, page 133
90 \(\Rightarrow\) 51 Variations of Zorn's lemma, principles of cofinality, and Hausdorff's maximal principle, Part I and II, Harper, J. 1976, Notre Dame J. Formal Logic
51 \(\Rightarrow\) 316 Well ordered subsets of linearly ordered sets, Howard, P. 1994, Notre Dame J. Formal Logic

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).

89:

Antichain Principle:  Every partially ordered set has a maximal antichain. Jech [1973b], p 133.

90:

\(LW\):  Every linearly ordered set can be well ordered. Jech [1973b], p 133.

51:

Cofinality Principle: Every linear ordering has a cofinal sub well ordering.  Sierpi\'nski [1918], p 117.

316:

If a linearly ordered set \((A,\le)\) has the fixed point property then \((A,\le)\) is complete. (\((A,\le)\)  has the fixed point property if every function \(f:A\to A\) satisfying \((x\le y \Rightarrow f(x)\le f(y))\) has a fixed point, and (\((A,\le)\) is complete if every subset of \(A\) has a least upper bound.)

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