Description: Proofs for the models given in 2B3-2B6 of Pincus [1972a]
Content:
Proofs for the models given in 2B3-2B6 of Pincus [1972a]
Theorem 1: \(\cal N4 \models\) Form 64, but Form 83 is false. (2B3)
(Form 64 is ``There are no amorphous sets. (Every infinite set is the union of two disjoint infinite sets.)'' and Form 83 is ``\(T\)-finite is equivalent to finite.'' (A set \(S\) is \(T\)-finite if every non-empty, monotone \(X \subseteq {\cal P}(S)\) has a \(\subseteq \)-maximal element.).)Proof: First we shall show that in \(\cal N4\) there is no infinite \(\subseteq\)-chain in \(\cal P(A)\). Suppose \(C=\{S_i: i\in I\}\) is such a chain with support E. Then for all \(\phi\in\) fix\((E) (=\{\psi\in\cal G : (\forall a\in A)\psi(a)=a\}\)), for all \(i\in I\), \(\phi(S_i)\in C\), and for all \(i,j\in I\), \(S_i\subseteq S_j\) iff \(\phi(S_i) \subseteq\phi(S_j)\). There must be an \(S_j\in C\) and a \(\phi\in\) fix\((E)\) such that \(\phi(S_j)\not= S_j\). Otherwise, \(C\) can be well ordered, which contradicts the fact that in \(\cal N4\), \(\cal P(A)\) is Dedekind finite. Suppose \(\phi(S_j)=S_k\). We may assume without loss of generality that \(S_j\subset S_k\). It follows that \(S_k\) is Dedekind infinite, contradicting the fact that \(A\) is Dedekind finite.
Next, we shall show that every infinite set is the union of two disjoint infinite sets (Form 64 is true). Let \(x\) be an infinite set with support \(E\). Suppose there is a \(t\in x\) and a \(\phi\in\mathrm{\; fix\; }(E)\) such that \()\phi(t)\not=t\). (If no such \(t\) exists, \(x\) can be well ordered and is therefore the union of two disjoint infinite sets.) Let \(F\) be a support of \(t\), then \(\phi(F)\) is a support of \(\phi(t)\) and \(F-E\not= \emptyset\). Let \(a\in F-E\). Since \(\precsim\) is a homogeneous universal partial ordering there are infinitely many elements \(b\in A-(E\cup F)\) such that \(a\prec b\) and infinitely many \(c\in A-(E\cup F)\) such that \(c\prec a\). Let \(B= \{b\in A-(E\cup F): a\prec b\}\) and \(C=\{c\in A-(E\cup F): c\prec a\}\). Since \(a\in F-E\), for each \(b\in B\) and \(c\in C\) there are \(\phi, \psi\in\mathrm{\;fix}(E)\) such that \(\phi(a)=b\) and \(\psi(a)=c\). It follows that both of the following subsets of \(x\) are infinite: \(\{\phi(t): \phi\in\mathrm{\;fix}(E) \wedge (\exists b\in B)(\phi(a)=b)\}\) and \(\{\psi(t): \psi\in\mathrm{\;fix}(E)\wedge(\exists c\in C)(\psi(a)=c)\}\). Thus, \(x\) can be partitioned into two infinite sets.
Theorem 2 \(\cal N2\models\) Form 83, but Form 215 is false (2B4).
(Form 83 is given in theorem 1 and Form 215 is ``If \()(\forall y \subseteq X)(y\) can be linearly ordered implies \(y\) is finite), then \(X\) is finite.'')
Proof: First we note that the set of atoms, \(A\), has the property that no infinite \(x\subseteq A\) can be linearly ordered so Form 215 is false. We shall show that for all infinite \(x\) in \(\cal N2\), \(\cal P(x)\) has an infinite \(\subseteq\)-chain. Suppose \(x\) is an infinite set with support \(E\). Suppose \(t\in x\) with support \(F\), then \(y=\{\phi(F): \phi\in\) fix\((E)\}\) is finite. Consequently, \(z_t=\{\phi(t): \phi\in\) fix\((E)\}\) is finite. Hence, since \(x\) is infinite, \(Z=\{z_t: t\in x\}\) is infinite and it can be well ordered because each element has support \(E\). Let \(u_0, u_1, u_2, \cdots\) be an infinite sequence of elements of \(Z\). We can construct an infinite \(\subseteq\)-chain \(\{U_n: n\in\omega\}\) in \()\cal P(x)\) as follows: \(U_0=u_0\) and \(U_{n+1}=\bigcup_{i=0}^n u_i\).
Theorem 3: \(\cal N34 \models\) Form 216, but Form 217 is false (2B5).
(Form 216 is ``Every infinite tree has either an infinite chain or an infinite antichain.'' and Form 217 is ``Every infinite partially ordered set has either an infinite chain or an infinite antichain.''.)Proof: First we note that the relation \(R\) on the atoms defined by \(c\,R\, d\) if and only if \((\exists q, r \in {\Bbb Q})(q < r \land c \in C_q \land d \in C_r ) \lor c = d\) is a partial ordering in the model with no infinite chain and no infinite antichain in the model (Form 217 is false). We shall prove that Every infinite tree \((X,\precsim)\) in the model has either an infinite chain (in the model) or an infinite antichain (in the model).
If \(H\) is a subgroup of \(\cal G\) and \(E \subset A\), we will use the notation fix\(_H(E)\) to denote the subgroup \(\{\phi\in H: (\forall x \in E)(\phi(x) = x)\}\) of \(H\). \(H\) will usually be \(\cal G\) or \(G_1\). One useful fact is that if \(\phi \in G_1\) and \(\psi \in G_2\) then there is a \(\psi' \in G_2\) such that \(\phi \circ \psi = \psi' \circ \phi\).
We will call a finite subset \(D\) of \(\cal C\) a \(*\)-support of \(x\) if \(\bigcup D\) is a support of \(x\). The following lemma about dense linear orderings and their automorphisms will be useful.
Lemma A: Assume \(a, b \in \Bbb Q\), \(a < b\) and that \(E\) and \(F\) are finite subsets of the interval \((a,b)\). Also assume that \(\phi\) is an order automorphism of \((a,b)\) which fixes \(E\) pointwise. Then there is a finite set \(\{ \phi_1, \ldots , \phi_n\}\) of automorphisms of \((a,b)\) each of which fixes \(E\) pointwise and such that
(Although we won't prove this lemma, the first step would be the case where \(E = \emptyset\), \(F = \{y,z\}\) with \( y < z \) and \(z < \phi(y)\). In this case we would choose \(\phi_2\) so that \(\phi_2(z) = z\) and \(y < \phi_2(y) < z\). Then \(\phi_1\) could be chosen so that \(\phi_1(y) = y\), \(\phi_1 \circ \phi_2(y) = \phi(y)\) and \(\phi_1(z) = \phi(z)\).)
Assume \((X,\precsim)\) is a tree in \(\cal N34\) with \(*\)-support \(E\). If every element of \(X\) has \(*\)-support \(E\) then \(X\) is well ordered in the model and therefore applying the axiom of choice in the ground model will give either an infinite chain or infinite antichain in the model. We therefore may assume that \()(\exists t \in X)(\exists \eta \in\) fix\(_{\cal G}(\bigcup E))(\eta(t) \ne t)\). Assume that \(t\) has \(*\)-support \(E \cup F\) where \(E\cap F = \emptyset\). By a previous remark \(\eta\) can be written \(\psi\circ \phi\) where \(\phi \in G_1\) and \(\psi \in G_2\) and by Lemma A there are permutations \(\phi_1, \ldots , \phi_n\) in fix\(_{G_1}(E)\) such that each \(\phi_i\) moves at most one element of \(F\) and \(\forall z \in F\), \(\phi_1 \circ \cdots \circ \phi_n(z) = \phi(z)\).
Claim: There is a \(\gamma\) in fix\(_{G_1}(E)\) that moves at most one element of \(F\) such that \(\gamma(t) \ne t\).
Proof: If \(\phi(t) \ne t\) then for at least one \(i\), \(1 \le i \le n\), \(\phi_i(t) \ne t\) and the proof is finished by letting \(\gamma = \phi_i\). Otherwise \(\phi(t) = t\) and therefore \()\psi(t) = \psi \circ \phi(t) = \eta(t) \ne t\). The permutation \(\psi\) is a product of transpositions one of which, say \(\tau_q = (a_q,b_q) \) moves \(t\). Further \(C_q = \{a_q, b_q \} \in F\) otherwise \(\tau_q\) fixes \(\bigcup F\) pointwise and hence \(\tau_q(t) = t\). Choose \(\gamma \in \hbox{fix}_{G_1} (\bigcup[(E\cup F)-\{C_q \}])\( so that \)\gamma(C_q) \ne C_q\). Then \(\gamma(t)\) has \(*\)-support \([(E \cup F)-\{ C_q\}]\cup \{\gamma(C_q)\}\) hence \(\tau_q(\gamma(t)) = \gamma(t)\) since \(\tau_q\) fixes a support of \(\gamma(t)\) pointwise. But \(\tau_q(t) \ne t\) therefore \(\gamma(t) \ne t\). This completes the proof of the claim.
Using the notation of the claim, let \(D = (E \cup F) - \{C_q\}\). And assume, as in the claim, that \(\gamma \in G_1\) fixes \(D\) pointwise and that \(\gamma(t) \ne t\). We will show that the set \(W =\{\beta(t): \beta \in\) fix\(_G(\bigcup D)\}\) (which is clearly in the model) is an infinite antichain in \((X,\precsim)\).
The first step is to argue by contradiction that \(\{ t, \gamma(t) \}\) is not a (two element) chain in \((X,\precsim)\). Assume \(\{ t, \gamma(t) \}\) is a chain and assume that \(\gamma(t) \prec t\). (The argument is similar is \(t \prec \gamma(t)\).) We also assume that \(\gamma(C_q) \prec C_q\) in the ordered set \((\cal C, \preceq)\) (Again, an argument similar to the one we will give here will work if \(C_q \prec \gamma(C_q)\).) The set \(S =\{\phi(C_q): \phi \in G_1\) and \(\phi\) fixes \(D\) pointwise\(\}\) is an interval in \((\cal C, \preceq)\). We will argue that, under the above assumptions, that \(T = \{\phi(t): \phi\in G_1\) and \(\phi\) fixes \(D\) pointwise\(\}\) under the ordering \(\precsim\) is order isomorphic to \(S\). The isomorphism \(f\) is defined by \(f(\phi(t)) = \phi(C_q)\). (The details of the proof that \(f\) is an isomorphism will be left to the reader.) This is a contradiction since a tree cannot have dense linearly ordered subset.
Therefore \(\{ t, \gamma(t) \}\) is an antichain. There is only one element of fix\(_{G_2}(\bigcup D)\) that could move \(t\), namely the transposition \(\tau_q = (a_q,b_q)\). If \(\tau_q(t) \ne t\) then \(\{ t, \tau_q(t) \}\) is also an antichain since \(\tau_q^2\) is the identity permutation. In addition \(\{\tau_q(t),\gamma(t)\}=\tau(\{t,\gamma(t)\}\) is an antichain because \(\tau\) fixes \(\precsim\). Similarly \(\{ t, \gamma(\tau_q(t)) \}\) is an antichain.
Now assume \(\beta_1(t)\) and \(\beta_2(t)\) are in \(W\). We first argue that if \(\beta_1(t) \ne \beta_2(t)\) then \(\{ \beta_1(t), \beta_2(t) \}\) is an antichain. Since \(\beta_1 \in G\), \(\beta_1 = \phi_1 \circ \psi_1\) where \(\phi_1 \in\) fix\(_{G_1}(\bigcup D)\) and \(\psi_1 \in\) fix\(_{G_2} (\bigcup D)\). Since the only transposition in fix\(_{G_2}\) that can possibly move \(t\) is the transposition \(\tau_q = (a_q,b_q)\) we can assume that \(\psi_1\) is either \(\tau_q\) or the identity. Therefore \(\beta_1(t) = \phi_1(t)\circ \psi_1(t)\) where \(\phi_1 \in G_1\) and fixes \(\bigcup D\) pointwise and \(\psi_1\) is either \(\tau_q\) or the identity. Similarly \(\beta_2(t) = \phi_2(t)\circ \psi_2(t)\) where \(\phi_2 \in G_1\) and fixes \(\bigcup D\) pointwise and \(\psi_2\) is either \(\tau_q\) or the identity. The proof now proceeds by cases.
Case 1: \(\beta_1(t) = \phi_1(t)\) and \(\beta_2(t) = \phi_2(t)\).
Since \(\beta_1(t) \ne \beta_2(t)\) we have \(\phi_1(C_q) \ne \phi_2(C_q)\). Therefore there is an \(\alpha \in G_1\) which fixes \(D\) pointwise and for which \(\alpha(\{C_q,\gamma(C_q)\})=\{\phi_1(C_q),\phi_2(C_q)\}\). Hence \(\alpha(\{ t, \gamma(t) \}) = \{ \phi_1(t), \phi_2(t) \}\) and therefore, since \(\alpha\) fixes \(\precsim\), \(\{ \beta_1(t), \beta_2(t) \}\) is an antichain.
Case 2: \(\beta_1(t) = \phi_1 \circ \tau_q (t)\) and \(\beta_2(t) = \phi_2 (t)\) where \(\tau_q(t) \ne t\).
Assume that \(\phi_1(C_q) \prec \phi_2(C_q)\) and that \(C_q \prec \gamma(C_q)\).
(The arguments in the other possible cases are similar.) Then there
is an \(\alpha \in G_1\) which fixes \(D\) pointwise and for which
\(\alpha(C_q) = \phi_1(C_q)\) and \(\alpha(\gamma(C_q)) = \phi_2(C_q)\).
Hence \(\alpha(\tau_q(t)) = \phi_1(\tau_q(t)\) and \(\alpha(\gamma(t)) =
\phi_2(t)\). Therefore \(\alpha(\{ \tau_q(t), \gamma(t) \}) =
\{ \beta_1(t), \beta_2(t) \}\). which must therefore be an antichain.
The other cases are similar. Since every two element subset of \(W\) is
an antichain, \(W\) is an antichain.
Theorem 4: \(\cal N1\models\) Form 217, but Form 9 is false (2B6).
(Form 217 is given in theorem 3 and Form 9 is ``Every Dedekind finite set is finite.''.)
Proof: The set of atoms is an infinite Dedekind finite set so Form 9 is false. We shall show that every infinite partially ordered set (poset) in \(\cal N1\) either has an infinite chain or an infinite antichain. Let \((X,\le)\) be a poset with support \(E\). If every element of \(X\) has support \(E\), then \(X\) can be well ordered and we are done. Assume there is a \(t\in X\) and a \(\psi\in\) fix\((E)\) such that \(\psi(t)\not=t\). We may assume that \(\psi\) moves only finitely many atoms. Let \( F=E\cup K\) be a support of \(t\), where \(E\cap K=\emptyset\).
We first argue that there is a \(k\in K\) and \(\phi\in\cal G\) which fixes \(F-\{k\}\) pointwise and \(\phi(t)\not=t\). Choose \(\eta\in\) fix\((E)\) such that \(\eta(K)\) is disjoint from \(K\cup\{a\in A: \psi(a)\not= a\}\). We have that \(\eta(t)\not= t\) because \()\psi(t)\not=t\) but \(\psi(\eta(t))=t\). Assume \(K=\{k_1,k_2,\cdots,k_n\}\) and for \(i=1,2,\cdots, n\), let \(f_i\) be the permutation of \(A\) that interchanges \(k_i\) and \(\eta(k_i)\). Then \(\prod_{i=1}^{n}f_i\) agrees with \(\eta\) on a support of \(t\), so \(\gamma(t)=\eta(t)\not=t\). Therefore, for at least one of the \(f_i\)'s, \(f_i(t)\not= t\).
Now, for \(\gamma, \sigma\in\hbox{ fix}(F-\{k\})\), \(\gamma(t)=\sigma(t)\) iff \(\gamma(k)=\sigma(k)\). It follows that \(y=\{\gamma(t): \gamma\in\) fix\()(F-\{k\})\}\) is an infinite subset of \(x\) that is in \(\cal N1\). Moreover, \(y\) is an antichain, for suppose \(\gamma(t)<\sigma(t)\), where \(\gamma(t),\sigma(t)\in y\). Let \(\rho\in\) fix\( (F-\{k\})\) be a permutation that interchanges \(\gamma(t)\) and \(\sigma(t)\). Then \(\sigma(t)<\gamma(t)\) which is a contradiction.
Howard-Rubin number: 105
Type: result
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