# Logic question proving something about compactness

Let $\Sigma$ be a set of formulas. There's a finite set $\Lambda \subseteq \Sigma$.

I'm asked to prove or disprove that $\Sigma$ has a model if and only if $\Lambda$ has a model.

It seems to me it's just true using compactness, but that sounds too easy.

• It is certainly true that (in first-order logic) $\Sigma$ has a model iff every finite subset of $\Sigma$ has a model. And naturally it is perfectly possible for some finite subset of $\Sigma$ to have a model while $\Sigma$ doesn't. One needs a more precise statement of the problem. – André Nicolas Jun 23 '12 at 16:13

It is not enough that there is a finite $\Lambda$ that has a model. In order for compactness to be applicable (or for the claim to be true at all), you need to assume that for every possible finite $\Lambda$ it has a model.
Assume $\Lambda$ has a model. Let $\varphi$ be a sentence in $\Lambda$ and let $\Sigma = \Lambda \cup \left\{ \neg\varphi \right\}$. Clearly $\Sigma$ is inconsistent and thus is unsatisfiable (by the soundness theorem).