Description:
A proof that Form 314 implies Form 119
Content:
A proof that Form 314 (For every set \(X\) and every permutation \(\pi\) on \(X\) there are two reflections \(\rho\) and \(\sigma\) on \(X\) such that \(\pi =\rho\circ\sigma\) and for every \(Y\subseteq X\) if \(\pi[Y]=Y\) then \(\rho[Y]=Y\) and \(\sigma[Y]=Y\).(A reflection is a permutation \(\phi\) such that \(\phi^2\) is the identity.) implies Form 119 (\(C(\aleph_{0}\),uniformly orderable with order type of the integers)).
Assume that \(B=\{A_i: i\in\omega\}\) is a denumerable set of denumerable sets and \(f\) is a function defined on \(\omega\) such that \(\forall i\in\omega\),
\(f(i)\) is an ordering of \(A_i\) with order type that of the integers. We want to construct a choice function \(g\) for \(B\) so we may assume that the
elements of \(B\) are pairwise disjoint.
Let \(\eta\) be the permutation of \(\bigcup B\) defined by \(\eta(a) = \) the immediate successor of \(a\) under the
ordering \(f(A_i)\) where \(A_i\) is the unique element of \(B\) such that \(a\in A_i\). By Form 314, \(\eta\) is
the product of two reflections: \(\eta = \phi\circ\psi\).
Fix an \(i\in\omega\) and for each permutation \(\beta\) of \(\bigcup B\), let\(\beta' = \beta\mid A_i\). Then \(\eta' = \phi'\circ\psi'\) and \(\phi'\)
and \(\psi'\) are reflections. We complete the proof by showing that exactly one of\(\phi'\) or \(\psi'\) has exactly one fixed point. (This fixed point will
be \(g(A_i)\).)
Since \(\phi'\) and \(\psi'\) are both reflections, \(\phi'\) and \(\psi'\) can both be written as a product of disjoint cycles,\(\phi' = \prod_{i\in
I}(s_i,t_i)\) and \(\psi' = \prod_{j\in J}(c_j,d_j)\). Assume that \(A_i = \{\ldots,a_{-1},a_0,a_1,a_2,\ldots\}\) where the elements are listed in the order
determined by the ordering \(f(A_i)\). Then \(\eta'\) is the infinite cycle\(\eta' = (\ldots,a_{-1},a_0,a_1,a_2,\ldots)\). Let \(r\) be the smallest positive
integer such that for some integer \(k\) the transposition\((a_k,a_{k+r}) \) is one of the \((s_i,t_i)\) or one of the \((c_j,d_j)\)
We first show that \(r=1\). Assume that \(r>1\) and that \((a_k,a_{k+r})\) is one of the \((c_j,b_j)\). (The argument is similar if \((a_k,a_{k+r})\) is one
of the \((s_i,t_i)\).) This means that \(\psi'(a_k) = a_{k+r}\).Since \(\eta'(a_k) = a_{k+1}\) (and \(r\ne 1\)), \(\phi'(a_{k+r})\) must be\(a_{k+1}\), hence
the transposition \((a_{k+1},a_{k+r})\) must be one of the\((s_i,t_i)\). This contradicts the definition of \(r\).
Assume that \((a_k,a_{k+1})\) is one of the \((c_j,d_j)\). (The argument is similar if \((a_k,a_{k+1})\) is one of the \((s_i,t_i)\).) This means
that \(\psi'(a_k)=a_{k+1}\). Since \(\eta'(a_k) = a_{k+1}\) we conclude that\(a_{k+1}\) must be a fixed point of \(\phi'\). Further, since
\(\psi'(a_{k+1})=a_k\) and \(\eta'(a_{k+1}) = a_{k+2}\) we conclude that \(\phi'(a_k) = a_{k+2}\),that is, \((a_k,a_{k+2})\) is one of the \((s_i,t_i)\). By
this last conclusion,\(\psi'(a_{k+2})=a_k\). It follows that \(\psi'(a_{k-1})=a_{k+2}\), that is \((a_{k-1},a_{k+2})\) is one of the \((c_j,d_j)\). Continuing
in this way, we see that for every natural number \(j\), \(\psi'\) interchanges \(a_{k-j}\) and\(a_{k+j+1}\) and \(\phi'\) interchanges \(a_{k-j}\) and
\(a_{k+j+2}\). Therefore,\(\psi'\) has no fixed points and \(\phi'\) has exactly one fixed point.
Howard-Rubin number: 124
Type: Proof
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