Description:
Form 126 (\(MC(\aleph_0,\infty)\)) implies Form 82 and Form 185
Content:
In this note we prove
Form 126 (\(MC(\aleph_0,\infty)\)) implies Form 82 (If a set is infinite then its power set is Dedekind infinite.) and Form 185 (Every Dedekind finite, linearly ordered set is finite.)
First we prove:
\(MC(\aleph_0,\infty)\) implies that for every infinite set \(X\) there is a sequence \((Y_n)_{n\in \omega}\) of finite subsets of \(X\) such that \(\forall n\in \omega\), \(Y_n \subsetneq Y_{n+1}\).
Let \(X\) be an infinite set and for each \(n\in\omega\), let \(X_n = \{Y\subseteq X: |Y| = n\}\). By \(MC(\aleph_0,\infty)\)there is a function \(f\) such that \(\forall n\in \omega\), \(f(X_n)\) isa finite non-empty subset of \(X_n\). We define the sequence \((Y_n)_{n\in\omega}\) by induction as follows: \(Y_0 = \emptyset\). Assumethat \(Y_n\) is defined and let \(k\) be the least natural number such that \(|Y_n| < k\). We let \(Y_{n+1} = (\bigcup f(X_k))\cup Y_n\).Since f(\(X_k\)) is a set of \(k\) element sets \(|Y_n| < k \le|\bigcup f(X_k)| \le |Y_{n+1}|\).
It is clear from the lemma that \(MC(\aleph_0,\infty)\) implies that the power set of an infinite set is Dedekind infinite. To show that \(MC(\aleph_0,\infty)\) implies Form 185, let \(X\) be a Dedekind finite set which is linearly ordered by \(\le\). The assumption that \(X\) is infinite leads to a contradiction as follows: By the lemma we get a sequence \((Y_n)_{n\in \omega}\) of finite subsets of \(X\) such that \(Y_n \subsetneq Y_{n+1}\) for all \(n\in \omega\). If we define \(t_n\) to be the \(\le\)-least element of \(Y_{n+1} - Y_n\) then \((t_n)_{n\in\omega}\) is an infinite sequence of distinct elements of \(X\) and therefore \(X\) is Dedekind infinite.
Howard-Rubin number:
76
Type:
Proof
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