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If \(E\) is any support, then there is a \(\psi\in G\) that fixes \(E\) pointwise
and an \(n\in\omega\) such that \(\psi|P_n\) is odd. Since such a \(\psi\)
fixes \(\{C_0^n,C_1^n\}\) but moves both its elements, \(E\) is not a support
of a choice function on \(\left\{\{C_0^n,C_1^n\}: n\in\omega\right\}\).
Multiple Choice Axiom (\(MC(\infty,\infty)\), form 67) is false in \(\cal N43\): Let \(F\) be the set of all
choice functions \(f: \omega\to A\) such that \(\forall n\in\omega\), \(f(n)
\in P_n\). Suppose that \(E\) is a support of a multiple choice
function \(H\) on \(\cal P(F) - \{\emptyset\}\) in \(\cal N43\). Let \(k =
\sup\{|E\cap P_n|: n\in\omega\}\) and define \(T\subseteq F\) by
\(T=\{f\in F: (\forall i>k)(f(i)\in P_n - E)\}\). Since \(E\) is a
support of \(T\) and of \(H\), \(E\) is a support of \(H(T)\) which must be a
finite, non-empty subset of \(T\). Fix \(f\) in \(H(T)\) then for every \(g\in T\)
for which \(g(i)=f(i)\) for \(i \le k\), there is a \(\phi\in G\) which fixes \(E\)
pointwise such that \(\phi(f)=g\). Therefore \(g\in H(T)\). Since there are
infinitely many such \(g\)'s, \(H(T)\) is infinite. This is a contradiction.
Howard-Rubin number: 122
Type: proofs of results
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