Notes

Description: In this note we prove that in \(\cal N51\), Form 64 (There are no amorphous sets.) and form Form 90 (Every linearly orderable set is well orderable.) are true and Form 221 (Every infinite set has a non-principal measure.) is false. The Hahn-Banach Theorem (form 52) implies Form 221, so it is also false. (See Brunner/Howard/Rubin [1997].)

Content: In this note we prove that in \(\cal N51\), Form 64 (There are no amorphous sets.) and form Form 90 (Every linearly orderable set is well orderable.) are true and Form 221 (Every infinite set has a non-principal measure.) is false. The Hahn-Banach Theorem (form 52) implies Form 221, so it is also false. (See Brunner/Howard/Rubin [1997].) We first note that for each finite ordered partition \(P = (B_1,\ldots,B_n)\) of \(A\), \(G_P \in\cal F\), where \(G_P = \{\psi\in G : \psi(P) = P\}\) and \(\cal F\) is the filter of subgroups which determine \(\cal N51\).  Further, if \(H\in \cal F\) then there is a finite ordered partition \(P\) of \(A\) such that \(G_P\subseteq H\).  Therefore for each \(x\in \)\(\cal N51\) there is a finite ordered partition \(P\) of \(A\) such that \((\forall \psi\in G_P)(\psi(x) = x)\).  We will refer to such a partition \(P\) as a support of \(x\).
    In order to prove that Form 221 is false we shall give a proof that in \(\cal N51\) there is no non-principal measure on \(A\), the set of atoms.
     Suppose, \(m\) is a non-principal measure on \(A\) supported by the partition \(P =(B_1, \ldots, B_n)\) of \(A\).  Since \(m\) is finitely additive on disjoint sets, \(m(A)=1\), and \(m(\{a\}) =0\), for all \(a\in A\).  It follows that there must be an \(i \le n\) such that \(B_i\) is infinite and \(m(B_i)\ne 0\). There exist pairwise disjoint infinite subsets, \(S_1\), \(S_2\), and \(S_3\), of \(B_i\) such that \(B_i=S_1\cup S_2\cup S_3\). Suppose \(\sigma_1, \sigma_2\in G_P\) are such that \begin{align} \sigma_1(S_1)= S_2\cup S_3\hbox{ and } \sigma(S_2\cup S_3)=S_1 \tag{1}\\ \sigma_2(S_1)=S_2,\ \sigma_2(S_2)=S_3,\hbox{ and }\sigma_2(S_3)=S_1 \tag{2}\\ \end{align} Then, we would have
(a) \(m(S_1) = m(S_2\cup S_3) = m(S_2) + m(S_3)\), by (1).
(b) \(m(S_1) = m(S_2) = m(S_3)\), by (2).
     Consequently, \(m(S_1) = 2m(S_1)\), which implies that \(m(S_1)=0\). Thus, by b., \(m(B_i)=0\), which is a contradiction.
     In order to prove that Form 90 and Form 64 are true, we need the following lemma which is a slight strengthening of Form 314 (from Degen [1988]). The proof requires the axiom of choice but we intend to use the lemma in the ground model where \(AC\) holds. If \(\psi\) is a permutation of a non-empty set \(A\) then there are two reflections \(\rho_1\) and \(\rho_2\) (a reflection on a set \(A\) is a permutation \(\rho\) for which \(\rho^2 = 1_A\).) satisfying
(i) \(\psi = \rho_2\circ\rho_1\) and
(ii) For all \(B\subseteq A\), if \(\psi(B) = B\) then \(\rho_1(B) = B\) and \(\rho_2(B) = B\).
 Assume \(\psi\) is a permutation of \(A\).  Write \(\psi = \prod_{i\in K} c_i\) as a product of disjoint cycles. Each \(c_i\) can be written as a product of reflections \(c_i = \rho_{2,i}\circ\rho_{1,i}\) as follows.  If \(c_i =(a_{-n},a_{-(n-1)},\ldots, a_{-1},a_0,a_1,\ldots,a_n)\) has odd length then \[ \rho_{1,i} = (a_{-1},a_1)(a_{-2},a_2)\cdots (a_{-n},a_n) \hbox{ and } \] \[ \rho_{2,i} =  (a_0,a_1)(a_{-1},a_2)\cdots (a_{-(n-1)},a_n). \] If \(c_i =(a_{-(n-1)},a_{-(n-2)},\ldots, a_{-1},a_0,a_1,\ldots,a_n)\) has even length then \[ \rho_{1,i} = (a_{-1},a_1)(a_{-2},a_2)\cdots (a_{-(n-1)},a_{(n-1)}) \hbox{ and } \] \[ \rho_{2,i} = (a_0,a_1)(a_{-1},a_2)\cdots (a_{-(n-1)},a_n). \] If \(c_i = (\ldots,a_{-2},a_{-1},a_0,a_1,a_2,\ldots)\) is infinite then \[ \rho_{1,i} = \prod_{n=1}^{\infty} (a_{-n},a_n) \hbox{ and } \rho_{2,i} = \prod_{n=0}^{\infty} (a_{-n},a_{n+1}). \] Then if \(\rho_1 = \prod_{i\in K} \rho_{1,i}\) and \(\rho_2 = \prod_{i\in K} \rho_{2,i}\) we have \(\rho_1^2 = 1_A\), \(\rho_2^2 = 1_A\) and \(\psi = \rho_2\circ \rho_1\). For the argument that ii) holds we first note that if \(\eta\) is a permutation whose expression as a product() of disjoint cycles is \(\eta = \prod_{j\in J} \sigma_j\) and \(B\subseteq\hbox{dom}(\eta)\) then \[ \eta(B) = B \Leftrightarrow (\forall j\in J)(\{x: \sigma_j(x) \ne x\} \subseteq B \hbox{ or } \{x : \sigma_j(x) \ne x\} \cap B = \emptyset). \tag{*} \] Now assume that \(B\subseteq A\) and that \(\psi(B) = B\).  By (\(*\)) for each \(i\in K\), either \(\{a\in A : c_i(a) \ne a\} \subseteq B\) or \(\{a\in A : c_i(a) \ne a\} \cap B = \emptyset\).  When \(\rho_1\) is written as a product of disjoint cycles each cycle is a transposition \((a,b)\) which is a cycle of \(\rho_{1,i}\) for some \(i\in K\).  Hence by the definition of \(\rho_{1,i}\), \(\{a,b\} \subseteq \{a\in A : c_i(a) \ne a\}\).  Therefore \(\{a,b\} \subseteq B\) (if \(\{a\in A : c_i(a) \ne a\} \subseteq B\)) or \(\{a,b\} \cap B = \emptyset\) (if \(\{a\in A : c_i(a) \ne a\} \cap B = \emptyset\)). It follows from (\(*\)) that \(\rho_1(B) = B\).  Similarly, \(\rho_2(B) = B\). In \(\cal N51\) every linearly orderable set is well orderable. Let \(X\) be linearly orderable in \(\cal N51\) and assume that the ordered partition \(P = (B_1,\ldots,B_k)\) is a support of a linear ordering \(<\) on \(X\).  We shall show that if \(\psi\in G_P\), then \(\psi\) fixes \(X\) pointwise and hence \(P\) is a support of a well ordering of \(X\).  Assume \(\psi\in G_P\).  By the lemma, \(\psi = \rho_2\circ\rho_1\) where \(\rho_1\) and \(\rho_2\) are reflections and (by part (ii)) \(\rho_1\) and \(\rho_2\) are in \(G_P\).  The permutations \(\rho_1\) and \(\rho_2\) are therefore order automorphisms of \((X,<)\).  Since no order automorphism of a linearly ordered set can have a finite cycle of length greater than 1, we conclude that for all \(x\in X\), \(\rho_1(x) = x\) and \(\rho_2(x) = x\) hence \(\forall x\in X\), \(\psi(x) = x\). In \(\cal N51\) there are no infinite amorphous sets. Assume that \(X \in\) \(\cal N51\) is infinite and that the ordered partition \(P = (B_1,\ldots,B_k)\) of \(A\) is a support of \(X\).  We will show that there is an infinite set \(R\in \cal N51\) such that \(R\subset X\) and \(X - R\) is infinite.
      If every element of \(X\) has support \(P\) then \(X\) is well orderable in \(\cal N51\) and the proof is easy.  We therefore assume that \(y\in X\) and there is a \(\phi\in G_P\) such that \(\phi(y) \ne y\).
      Step 1. We show that we may assume that \(y\) has a support \[ Q = (C_1,C_2,C_3,\ldots,C_{n_0},C_{n_0+1},\ldots,C_n) \] where
a. \(n_0\ge 3\) and \(C_3\) is infinite.
b. For \(1\le j\le n_0\), \(C_j \subseteq B_1\) and for \(n_0 < j\le n\), \(C_j \cap B_1 = \emptyset\).
c. \(\phi(C_1) = C_2\), \(\phi(C_2) = C_1\), and for every \(a\in A - (C_1\cup C_2)\), \(\phi(a) = a\).
      First, we may assume (by refining \(P\) if necessary) that each \(B_i\) is either infinite or a singleton.  Secondly, by writing \(\phi = \prod_{i=1}^k \phi_i\) where \(\phi_i(a) = a\) for \(a\in A-B_i\) and replacing \(\phi\) by one of the \(\phi_i\), we may assume that there is an \(i \le k\) such that \(\phi(a) = a\) for \(a\in A - B_i\). (If \(\prod_{i=1}^k \phi_i(y) \ne y\) then for some \(i\), \(\phi_i(y) \ne y\).) Further it is no loss of generality to assume that \(i = 1\).
      Let \(Q = (C_1,\ldots,C_n)\) be a support of \(y\).  By replacing each \(C_j\) with the two sets \(C_j\cap B_1\) and \(C_j - B_1\) (if necessary) and reordering the \(C_j\)'s we may assume that there is an \(n_0\le n\) such that \(C_j \subseteq B_1\) for \(j\le n_0\) and \(C_j \cap B_1 = \emptyset\) for \(j > n_0\).  As in the proof of the previous theorem (using the lemma) we may assume that \(\phi\) is a reflection.  By replacing \(Q\) by a partition whose cells are \(\{C_i \cap \phi(C_j) : 1\le i,j\le n\} - \{\emptyset\}\) we may assume (since \(\phi\) is a reflection) that \(\forall i\), \(1\le i\le n\), \(\exists j\), \(1\le j\le n\) such that \(\phi(C_i) = C_j\) and \(\phi(C_j) = C_i\).  (By our simplifying assumptions so far, \(\phi(C_j) = C_j\) for \(j > n_0\).) By decomposing \(\phi\) into a product \(\phi = \prod_{r=1}^m \gamma_r\) where for each \(r\), \(1\le r\le m\) there are \(i\) and \(j\) such that \(\gamma_r(a) =a\) for \(a\in A - (C_i\cup C_j)\), \(\gamma_r(C_i) = C_j\) and \(\gamma)r(C_j) = C_i\) and replacing \(\phi\) by one of the \(\gamma_r\), we may assume that \(\phi(a) = a\) for \(a\in A - (C_i\cup C_j)\), \(\phi(C_i) = C_j\) and \(\phi(C_j) = C_i\).  It is no loss of generality to assume that \(i = 1\) and \(j = 2\).
      Finally, we may assume that for some \(i\), \(3\le i\le n_0\), \(C_i\) is infinite.  Assuming that this is not the case and using the facts that \(C_1 \cup C_2 \subseteq B_1\) and that \(B_1\) is infinite we conclude that \(C_1\) and \(C_2\) are infinite.  Let \(C'\) be any infinite subset of \(C_1\) whose complement in \(C_1\) is infinite.  Note that \(C_2\) is the disjoint union \(C_2 = \phi(C')\cup \phi(C_1 - C')\). Therefore, \(Q' = (C',\phi(C'),C_1 -C',\phi(C_1 -C'), C_3,\ldots,C_n)\) is a partition of \(Q\) which supports \(y\).  Define \(\phi'\) and \(\phi''\) (both in \(G_P\)) as follows: \[ \phi'(a) = \begin{cases} \phi(a) & a\in C'\cup \phi(C')\\                    a & \hbox{otherwise} \end{cases} \] and \[ \phi''(a) = \begin{cases} \phi(a) & a\in (C_1 - C')\cup \phi(C_1 - C')\\                     a & \hbox{otherwise} \end{cases} \] Clearly \(\phi = \phi'\circ\phi''\) and therefore one of \(\phi'\) or \(\phi''\) moves \(y\).  Assume without loss of generality that \(\phi'(y) \ne y\) then replace \(Q\) with \(Q'\) and \(\phi\) by \(\phi'\).
This completes step 1.
      Step 2.  We use \(y\) to obtain an infinite subset \(R\) of \(X\) whose complement in \(X\) is infinite. Our primary tool will be the following lemma. Assume that \(\delta,\lambda \in G_P\), that \(\lambda(a) = a\) for all \(a\in A - (\delta(C_1) \cup \delta(C_2))\), that \(\lambda(\delta(C_1)) = \delta(C_2)\), and that \(\lambda(\delta(C_2)) = \delta(C_1)\).  Then \(\lambda(\delta(y)) \ne \delta(y)\). By the hypotheses, \(\delta^{-1}\lambda \delta(C_1) = C_2\) and \(\delta^{-1}\lambda \delta(C_2) = C_1\) and for all \(a\in A - (C_1 \cup C_2)\), \(\delta^{-1}\lambda \delta(a)= a\). This means that \(\delta^{-1}\lambda \delta(Q) = \phi(Q)\) and since \(Q\) is a support of \(y\), \(\delta^{-1}\lambda \delta(y) = \phi(y) \ne y\).  It follows that \(\lambda(\delta(y)) \ne \delta(y)\).      Now by using permutations in \(G_P\) which move \(C_1\) and \(C_2\) we can obtain infinitely many "copies" of \(y\) all of which are in \(X\).
     Partition \(C_3\) into sets \(\{S_j, T_j, U_j, V_j : j\in \omega\}\) so that \(|S_j| = |T_j| = |U_j| = |V_j| = |C_1|\) (\(=|C_2|\)). Let \(C_3'\) and \(C_3''\) denote the sets \(\bigcup_{j\in\omega}(S_j\cup T_j)\) and \(\bigcup_{j\in\omega}(U_j\cup V_j)\) respectively.  Let \(\tau\in G_P\) satisfy \(\tau(a) =a\) for all \(a\in A - (C_1\cup C_2\cup S_0\cup T_0)\), \(\tau(C_1) = S_0\), \(\tau(S_0) = C_1\), \(\tau(C_2) = T_0\) and \(\tau(T_0) = C_2\).  Let \(\nu\in G_P\) satisfy \(\nu(a) =a\) for all \(a\in A - (C_1\cup C_2 \cup U_0 \cup V_0)\), \(\nu(C_1) = U_0\), \(\tau(U_0) = C_1\), \(\tau(C_2) = V_0\) and \(\tau(V_0) = C_2\).  For \(j\in\omega\), \(j\ge 1\), let \(\sigma_j\in G_P\) satisfy \(\sigma_j(a) =a\) for \(a\in A - (S_0\cup T_0\cup S_j\cup T_j)\), \(\sigma_j(S_0) = S_j\), \(\sigma_j(S_j) = S_0\), \(\sigma_j(T_0) = T_j\) and \(\sigma_j(T_j) = T_0\).  Similarly, for \(j\ge 1\), let \(\gamma_j \in G_P\) satisfy \(\gamma_j(a) = a\) for all \(a\in A - (U_0\cup V_0\cup U_j\cup V_j)\), \(\gamma_j(U_0) = U_j\), \(\gamma_j(U_j) = U_0\), \(\gamma_j(V_0) = V_j\), and \(\gamma_j(V_j) = V_0\). The partition \(P' = (C_1,C_2,C_3',C_3'',\ldots,C_{n_0},B_2,\ldots, B_k)\) is a refinement of \(P\) and therefore is a support of \(X\). Hence the two sets \(R_1 = \{\psi(\tau(y)) : \psi\in G_{P'}\}\) and \(R_2 = \{\psi(\nu(y)) : \psi\in G_{P'}\}\) are subsets of \(X\). Further, \(P'\) is a support of \(R_1\) and \(R_2\) so \(R_1\) and \(R_2\) are in \(\cal N51\).  We shall also prove the following:
(i) \((\forall j\in\omega, j\ge 1)(\sigma_j(\tau(y))\in R_1 \land \gamma_j(\nu(y))\in R_2)\).
(ii) \((\forall j\in\omega, j\ge 1)(\sigma_j(\tau(y)) \notin R_2\land \gamma_j(\nu(y))\notin R_1)\).
(iii) \((\forall j, r\in\omega, j,r\ge 1)\left(j\ne r \to [\sigma_j(\tau(y))\ne \sigma_r(\tau(y)) \land (\gamma_j(\nu(y))\ne \gamma_r(\nu(y))]\right)\).
      It will follow that \(R = R_1\) is an infinite subset of \(X\) with infinite complement in \(X\) and the proof will be complete. To prove (i) we observe that since \(\sigma_j\) and \(\gamma_j\) agree with the identity permutation outside of \(C_3'\) and \(C_3''\) respectively for \(j\ge 1\), they are both in \(G_{P'}\).
      For (ii) we first note that since \(Q\) is a support of \(y\), a support of \(\nu(y)\) is given by \[ \begin{align} \nu(Q) & = (\nu(C_1),\nu(C_2),\nu(C_3),\ldots,\nu(C_n)) \\   & = (U_0,V_0,[C_3 - (U_0\cup V_0)]\cup(C_1\cup C_2),C_4,\ldots,C_n). \end{align} \] Therefore if \(\psi\) fixes \(P'\) then \(\psi(\nu(y))\) has support \[ \begin{align} \psi(\nu(Q)) & = & (\psi(U_0),\psi(V_0),[\psi(C_3) - \psi(U_0\cup V_0)] \cup\psi(C_1 \cup C_2), \psi(C_4),\ldots,\psi(C_n)) \\ & = & (\psi(U_0),\psi(V_0),[C_3 - \psi(U_0\cup V_0)] \cup (C_1 \cup C_2), \psi(C_4),\ldots,\psi(C_n)).\\ \end{align} \]   Since \(C_3' \subseteq C_3 - \psi(U_0\cup V_0)\) we may conclude that for any \(\lambda\in G_P\), if \(\lambda(a) = a\) for all \(a\in A - C_3'\) then \(\lambda(\psi(\nu(Q)) = \psi(\nu(Q))\) and therefore \(\lambda(\psi(\nu(y))) = \psi(\nu(y))\).  Hence such an \(\lambda\) fixes \(R_2\) pointwise.  In particular, if we choose \(\lambda\) so that \(\lambda(a) = a\) for all \(a\in A - (S_j\cup T_j)\), \(\lambda(S_j) =T_j\) and \(\lambda(T_j) = S_j\) then \(\lambda\) fixes \(R_2\) pointwise and by the lemma (with \(\delta = \sigma_j \circ \tau\)) \(\lambda(\sigma_j(\tau(y))) \ne \sigma_j(\tau(y))\).  It follows that \(\sigma_j(\tau(y)) \notin R_2\).  A similar argument shows that \(\gamma_j(\nu(y))\notin R_1\).
      The argument for (iii) is similar.  We shall show \[ (\forall j,r\in \omega, j,r\ge 1)\left(j\ne r \to \sigma_j(\tau(y))\ne \sigma_r(\tau(y))\right) \] and leave the similar argument that \(\gamma_j(\nu(y)) \ne \gamma_r(\nu(y))\) to the reader.  First note that \(\sigma_r(\tau(y))\) has support \[ \sigma_r(\tau(Q)) = (S_r,T_r,[C_3 -(S_r\cup T_r)]\cup(C_1\cup C_2),C_4,\ldots,C_n). \] Call this support \(Q'\).  Since \(S_j\cup T_j\subseteq C_3 - (S_r\cup T_r)\) any \(\lambda\) satisfying \(\lambda(a) = a\) for all \(a\in A-(S_j\cup T_j)\) fixes \(Q'\) and therefore fixes \(\sigma_r(\tau(y))\).  In particular if we choose such a \(\lambda\) for which \(\lambda(S_j) = T_j\) and \(\lambda(T_j) = S_j\) then \(\lambda\) fixes \(\sigma_r(\tau(y))\) and by the lemma (with \(\delta = \sigma_j\circ\tau\)) \(\lambda(\sigma_j(\tau(y))) \ne \sigma_j(\tau(y))\).  We can therefore conclude that \(\sigma_r(\tau(y)) \ne \sigma_j(\tau(y))\). This completes the proof of the theorem.

Howard-Rubin number: 128

Type: proofs of results

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