Notes

Description: In \(\cal N23\), \(C(\infty,2)\) (Form 88) is false and \(C(\aleph_0,\infty)\) (Form 126) is true.  The proof that \(C(\aleph_0,\infty)\) is true is due to A. Rubin.

Content:

In \(\cal N23\), \(C(\infty,2)\) (Form 88) is false and \(C(\aleph_0,\infty)\) (Form 126) is true.  The proof that \(C(\aleph_0,\infty)\) is true is due to A. Rubin.

First we shall show that \(C(\infty,2)\) is false. Let \(B\) be the set of subsets \(x\) of \(A\) such that the order type of \(x\) with respect to \(<\) is \(\omega\).  Let \(D\) be the set of equivalence classes of elements of \(B\) under the relation \(x R y\) iff the symmetric difference, \((x-y) \cup (y-x)\) is finite.  (Denote the equivalence class of \(x\) by \([x]\).)  Let \(P\) be the set of pairs of elements of \(D\).  \(P\) has no choice function in the model because if \(f\) is such a choice function and \(E \subseteq A\) is a support of \(f\) it follows that for all order preserving permutations \(\phi\) of \(A\) such that \(\phi\) fixes \(E\) pointwise and for all \([x]\), \([y]\) in \(D\), if \(\phi\) fixes \(\{ [x], [y] \}\) then \(\phi\) fixes both \([x]\) and \([y]\).  We can construct a \(\phi\), \(x\) and \(y\) for which this is false as follows:  Since \(E\) is well ordered there are elements \(a\), \(b \in A\) such that the interval \((a,b)\) contains no points of \(E\).  Let \(\phi\) be an order preserving permutation of \(A\) that fixes \(E\) pointwise, fixes \(a\) and \(b\) but moves at least one element \(t\) of \((a,b)\).  We may assume (by replacing \(\phi\) with \(\phi^{-1}\) if necessary) that \(t < \phi(t)\). Let \(x = \{ \phi^{2n}(t) : n \in \omega \}\) and \(y =\{ \phi^{2n+1}(t) : n \in \omega\}\).  Then \(\phi([x]) = [y]\) and \(\phi([y]) = [x]\) (and \([x] \not= [y])\).

Next, we shall show that \(C(\aleph_0,\infty)\) is true in \(\cal N23\). Let \(X=\{Y_i: i\in\omega\}\) be a denumerable set of non-empty sets in the model. Let \(B\) be a common support for \(X\) and the \(Y_i\)'s, \(i\in \omega\). Let the open intervals of \(A-B\) be denoted by \(I_{\beta}\), for \(\beta <\alpha\). For each \(i\in\omega\), choose (in the outer model) \(z_i\in Y_i\) and let \(B_i\) be a support of \(z_i\). Let \(J_{\beta i}= I_{\beta}\cap B_i\). (Note: Whenever we choose a sequence of elements in \(A\), we choose them in increasing order. Also, \(\beta\) will range over \(\alpha\), and \(i\) and \(j\) will range over \(\omega\).)

We then define \(K_{\beta j i}\) as follows: For each \(\beta\) and \(i\), choose \(E_{\beta i}= \{e_{\beta j i}: j\in\omega\}\) cofinal in \(I_{\beta}\) such that each \(e_{\beta o i} \lt J_{\beta i}\) and \(E_{\beta i}\cap \hbox{cl}(B_i)=\emptyset\). Then define \(K_{\beta j i}=B_i\cap (e_{\beta j i}, e_{\beta j+1 i})\). (If \(J_{\beta i}\) is not cofinal in \(I_{\beta}\), choose \(E_{\beta i}\) such that \(e_{\beta o i} \lt J_{\beta i} \lt e_{\beta 1 i}\) so there is only one non-empty \(K_{\beta j i}\).)

For each \(\beta\), choose \(\{c_{\beta i}: i\in\omega\}\) cofinal in \(I_{\beta}\) and choose \(\{d_{\beta j i}: i\in\omega\}\) in the interval \((c_{\beta j}, c_{\beta j+1})\). For each \(\beta\), \(j\), and \(i\), let \(f_{\beta j i}\) be a function mapping  \(K_{\beta j i}\) into the interval \((d_{\beta j i}, d_{\beta j i+1})\) and let \(L_{\beta j i}\) be the range of \(f_{\beta j i}\). Next, let \(f_i\) be an extension of \(\bigcup_{\beta, j}f_{\beta j i}\) to a complete isomorphism of \(A\) fixing \(B\). We claim that \(F=\{\lt Y_i, f_i(z_i)\gt: i\in\omega\}\) is a choice function for \(X\) that is in the model. Since \(f_i\) preserves \(B\), we must have that \(f_i(z_i)\in f(Y_i)=Y_i\). A support of \(F\) is \(\bigcup_{\beta,j,i}L_{\beta j i}\) which is a well ordered union of the well ordered sets \(L_{\beta j i}\)  and, therefore, can be well ordered.

Howard-Rubin number: 100

Type: proof of result

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