Description: We shall give a proof that [1 BU] (In every vector space over \(\Bbb Q\), every generating set includes a basis.) implies Form 1, AC.
Content:
We shall give a proof that [1 BU] (In every vector space over \(\Bbb Q\), every generating set includes a basis.) implies Form 1, AC. Bleicher [1964] has shown that [1 BU] implies \(C(\infty,<\aleph_0)\) (Form 62), so it is sufficient to prove that [1 BU] implies MC (Form 67). (The proof is due to K.~Keremedis.)
Let \(X = \{x_i: i\in I\}\) be a set of pairwise disjoint infinite sets. For each \(i\in I\), let \(V_i\) be the set of all functions from \(x_i\) to \(\Bbb Q\) such that for all \(f\in V_i\), \(\exists r\in\Bbb Q\) such that \(\{u\in x_i : f(u)\ne r\}\) is finite. Then \(V_i\) is a vector space. Let \(V\) be the weak direct product of the \(V_i\)'s. For each \(i\in I\), define \[ G_i = \{F\in V:(\forall j)[(\forall u\in x_j)(j\ne i \to F_j(u) = 0)\wedge \] \[ (j=i\to (\exists u\in x_j)(F_j(u)=1\wedge (\forall v\in x_j)(v\ne u \to F_j(v)=0)\vee \] \[ (F_j(u)=0\wedge (\forall v\in x_j)(v\ne u\to F_j(v)=1)))]\} \] Let \(G= \bigcup\{G_i: i\in I\}\). Then \(G\) is a generating set for \(V\). Let \(B\subseteq G\) be a basis for \(V\). For each \(i\in I\), let \(H^i\) be the function in \(V\) such that \(\forall u\in x_j\), \[ H^i(u) = \begin{cases} 0, & j\ne i \\ 1, & j = i \end{cases} \] For each \(i\in I\), \(H^i\) can be uniquely represented as a linear combination of a finite number of basis elements. It follows from the definition of \(G_i\) that each basis element uniquely specifies an element in \(x_i\). Thus, each \(H^i\) uniquely specifies a finite subset of \(x_i\).Howard-Rubin number: 137
Type: Proof
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