Axiom Of Choice

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

Back

Notes