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It is clear that König's lemma, [10 F] implies [10 K] and using the usual proof that the Completeness Theorem implies the Compactness Theorem, it follows that [10 K] implies [10 L]. To show that [10 L] implies Form 10, \(C(\aleph_{0},<\aleph_{0})\), let \(X=\{x_i: i\in\omega\}\) be a denumerable set of finite sets, where each \(x_i = \{x_{i,0}, x_{i,1},\cdots x_{i,n_i}\}\). Let \(\cal L\) be a first order language with a single function symbol \(f\) and constant symbols for each \(x_i\in X\) and each element \(x_{i,j}\) of \(\bigcup X\).  Let \(S\) be the following set of sentence in \(\cal L\):
For each \(i\in\omega\), the conjunction of

  • \(\land\left(\{ x_{i,j } \ne x_{i,k} : j,k \le n_i\hbox{ and } j\ne k \}\right)\) and
  • \(\vee\left(\{f(x_i) = x_{i,0},\ldots,f(x_i) = x_{i,n_i}\}\right)\).
Each finite subset of \(S\) has a model, so [10 L] implies that the denumerable set of sentences has a model. In that model, \(f\) is a choice function on \(X\).

Howard-Rubin number: 19

Type: Statement of proofs

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