Implications

We have the following indirect implication of form equivalence classes:

407 \(\Rightarrow\) 357
given by the following sequence of implications, with a reference to its direct proof:

Implication Reference
407 \(\Rightarrow\) 43 clear
43 \(\Rightarrow\) 8 clear
8 \(\Rightarrow\) 27 clear
27 \(\Rightarrow\) 31 clear
31 \(\Rightarrow\) 32 L’axiome de M. Zermelo et son rˆole dans la th´eorie des ensembles et l’analyse, Sierpi'nski, W. 1918, Bull. Int. Acad. Sci. Cracovie Cl. Math. Nat.
32 \(\Rightarrow\) 357 clear

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

Howard-Rubin Number Statement
407:

Let \(B\) be a Boolean algebra, \(b\) a non-zero element of \(B\) and \(\{A_i: i\in\omega\}\) a sequence of subsets of \(B\) such that for each \(i\in\omega\), \(A_i\) has a supremum \(a_i\). Then there exists an ultrafilter \(D\) in \(B\) such that \(b\in D\) and, for each \(i\in\omega\), if \(a_i\in D\), then \(D\cap\ A_i\neq\emptyset\).

43:

\(DC(\omega)\) (DC), Principle of Dependent Choices: If \(S\)  is  a relation on a non-empty set \(A\) and \((\forall x\in A) (\exists y\in A)(x S y)\)  then there is a sequence \(a(0), a(1), a(2), \ldots\) of elements of \(A\) such that \((\forall n\in\omega)(a(n)\mathrel S a(n+1))\).  See Tarski [1948], p 96, Levy [1964], p. 136.

8:

\(C(\aleph_{0},\infty)\):

27:

\((\forall \alpha)( UT(\aleph_{0},\aleph_{\alpha}, \aleph_{\alpha}))\): The  union of denumerably many sets each of power \(\aleph_{\alpha }\) has power \(\aleph_{\alpha}\). Moore, G. [1982], p 36.

31:

\(UT(\aleph_{0},\aleph_{0},\aleph_{0})\): The countable union theorem:  The union of a denumerable set of denumerable sets is denumerable.

32:

\(C(\aleph_0,\le\aleph_0)\): Every denumerable set of non-empty countable sets  has a choice function.

357:

\(KW(\aleph_0,\aleph_0)\), The Kinna-Wagner Selection Principle for a denumerable family of denumerable sets: For every denumerable set \(M\) of denumerable sets there is a function \(f\) such that for all \(A\in M\), if \(|A| > 1\) then \(\emptyset\neq f(A)\subsetneq A\).

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