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Is the following statement correct?

Let $\Sigma$ be an infinite set of propositions such that for every partition of $\Sigma$ into two subsets: $$\Sigma=\Sigma_{1}\cup\Sigma_{2}$$ At least one of the parts $\Sigma_1,\ \Sigma_2$ has a model.

Then $\Sigma$ has a model.

My attempt: I thought about some $\Sigma$ that has an infinite number of tautologies and a single contradiction C. so for every partition of $\Sigma$ into two subsets $\Sigma_1,\ \Sigma_2$ one subset will contain C and the other won't. so the other subset has a model, but there is no model for $\Sigma$.

Is my example correct? (I'm confused because i have notes here that this statement is correct)

Thanks!

dave
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1 Answers1

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I edited your statement to make it clearer (to me). I think you mean:

Let $\Sigma$ be any infinite set of propositions such that for every partition of $\Sigma$ into two subsets $\Sigma_1$ and $\Sigma_2$ (i.e., $\Sigma=\Sigma_{1}\cup\Sigma_{2}$), at least one of the parts $\Sigma_1,\ \Sigma_2$ has a model. Then $\Sigma$ has a model.

If that's what you mean, then I think you've successfully constructed a counterexample that shows the statement to be false.

EDIT: Just on the off chance that there might be confusion, the compactness theorem (referenced in the question header) is this statement:

Let $\Sigma$ be any set of propositions (finite or infinite) such that every finite subset $\Sigma_1$ has a model. Then $\Sigma$ has a model.

That statement is true.

StumpyLeg
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