Description:
We shall prove that Form 10(\(C(\aleph_0,<\aleph_0)\)) + Form 163 (Every non-well orderable set has an infinite Dedekind finite subset.) implies Form 231 (\(UT(WO,WO,WO)\)) and also that Form 133 (Every set is either well orderable or has an infinite amorphous set.) implies Form 231.
Content:
We shall prove that
Form 10(\(C(\aleph_0,<\aleph_0)\)) + Form 163 (Every non-well orderable set has an infinite Dedekind finite subset) implies Form 231 (\(UT(WO,WO,WO)\)) and also that Form 133 (Every set is either well orderable or has an infinite amorphous set) implies Form 231.
(It is shown in Brunner [1982a] that Form 10 + Form 163 implies Form 165 (\(C(WO,WO)\)).)
Suppose \(X\) is a well ordered set of pairwise disjoint well orderable sets. Assume \(Y=\bigcup X\) cannot be well ordered. Form 163 implies that \(Y\) has an infinite Dedekind finite set \(Z\). Let \( T = \{w\cap z: w\in X \wedge w\cap z\ne\emptyset\}\). For each \(w\in Z\), \(w\cap Z\) is a well ordered Dedekind finite set so it must be finite. If \(T\) is finite, then \(Z\) is a finite union of finite sets and is therefore finite. This contradicts the definition of \(Z\). If \(T\) is infinite, then \(T\) is well ordered because \(X\) is well ordered. Therefore \(T\) must have a denumerable subset, \(U\). \(U\) is a denumerable set of finite sets, so by Form 10, \(U\) has a choice set. This choice set is a denumerable subset of \(Z\). This contradicts the definition of \(Z\). Thus, the union of \(X\) can be well ordered.
Since Form 133 (every set is either well orderable or has an infinite amorphous subset) implies Form 10 Brunner [1984b[) and Form 133 implies Form 163 (clear), it follows that Form 133 implies Form 231.
Howard-Rubin number:
123
Type:
Proof
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