Axiom Of Choice

We have the following indirect implication of form equivalence classes:

333 \(\Rightarrow\) 211 given by the following sequence of implications, with a reference to its direct proof:

Implication	Reference
333 \(\Rightarrow\) 67	clear
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 \(\Rightarrow\) 337	Non-constructive properties of the real numbers, Howard, P. 2001, Math. Logic Quart.
337 \(\Rightarrow\) 211	Non-constructive properties of the real numbers, Howard, P. 2001, Math. Logic Quart.

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

Howard-Rubin Number	Statement
333:	\(MC(\infty,\infty,\mathrm{odd})\): For every set \(X\) of sets such that for all \(x\in X\), \(\|x\|\ge 1\), there is a function \(f\) such that for every \(x\in X\), \(f(x)\) is a finite, non-empty subset of \(x\) and \(\|f(x)\|\) is odd.
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.
337:	\(C(WO\), uniformly linearly ordered): If \(X\) is a well ordered collection of non-empty sets and there is a function \(f\) defined on \(X\) such that for every \(x\in X\), \(f(x)\) is a linear ordering of \(x\), then there is a choice function for \(X\).
211:	\(DCR\): Dependent choice for relations on \(\Bbb R\): If \(R\subseteq\Bbb R\times\Bbb R\) satisfies \((\forall x\in \Bbb R)(\exists y\in\Bbb R)(x\mathrel R y)\) then there is a sequence \(\langle x(n): n\in\omega\rangle\) of real numbers such that \((\forall n\in\omega)(x(n)\mathrel R x(n+1))\).

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