Description:
If \(N=N(A,\cal G,S)\) is a permutationmodel, then the \(2m=m\) principle (form 3) is false in \(N\) if thefollowing holds in \(N\)...
Content:
If \(N=N(A,\cal G,S)\) is a permutation model, then the \(2m=m\) principle (form 3) is false in \(N\) if the following holds in \(N\):
\[
\begin{array}{ll}
\forall E\in S) [A-E\not=\emptyset & \\
& \wedge(\forall a\in A-E) (\forall b\in A-\{a\})(\exists\sigma\in\cal G)[((\forall c\in E) (\sigma(c)=c)\\\
(*) & \wedge(\sigma(a)\not= a\wedge\sigma(b)=b))\\
& \vee(\sigma(a)=b\wedge\sigma(b)=a\wedge(\forall c\in A-\{a,b\})\sigma(c)=c)]].
\end{array}\]
Suppose there is a \(1-1\) function \(f\) in \(N\) mapping \(2\times A\) into \(A\). Let \(E\) be a support of \(f\) and let \(a\in A-E\). Then, since \(f\) is \(1-1\), \(f(i,a)\not=a\) for\(i=0\) or \(i=1\). Suppose that \(f(i,a)=b\not=a\). It follows from (\(*\)) that there is a \(\sigma\in\cal G\) that leaves \(E\) pointwise fixed (so \(\sigma(f)=f\)) and \(\sigma(a)=c\not=a\).If \(\sigma(b)=b\), then \(f(i,a)=b\) and \(f(i,c)=b\), contradicting the fact that \(f\) is \(1-1\). If \(\sigma(a)=b\) and \(\sigma(b)=a\), then \(f(i,a)=b\) and \(f(i,b)=a\). If \(j\in\{0,1\}\) and \(j\not=i\), then \(f(j,a)=c\), where \(c\not= a\) and \(c\not= b\). Consequently, \(f(j,b)=c\) again contradicting the fact that \(f\) is \(1-1\).
Examples of models that satisfy (\(*\)) are \(\cal N1\), \(\cal N2\),\(\cal N2(\aleph_{\alpha})\), and \(\cal N12(\aleph_{\alpha})\).
Howard-Rubin number:
104
Type:
Proof
Back