Show that gamma has a model if and only if each set in gamma has a model

Question

Let $\Gamma$ be a theory that is closed under provability. That is, if there are sentences $\varphi_{1},...,\varphi_{n}$ in $\Gamma$ such that $\varphi_{1},...,\varphi_{n} \vdash \phi$ it applies that $\phi \in \Gamma$. Prove that $\Gamma$ has a model if and only if each set in $\Gamma$ has a model.

I want to prove the above statement and the only theorem I can think of to prove this is the compactness theorem, that a set of sentences $\Gamma$ where every finite subset of $\Gamma$ has a model, then there is a model for the whole $\Gamma$.

But this statement deals not only with finite sets in $\Gamma$ but all sentences, infinite and finite. But my question is, can I still use the compactness theorem to prove this statement? I reason that if all sets have a model, that means that all finite sets also has a model, which results in $\Gamma$ having a model?

Also, do I need to in any way take into account that the theory is closed under provability when I do this?

Answer 1

I assume that you meant to write the following:

Let $\Gamma$ be a theory that is closed under provability. Prove that $\Gamma$ has a model if and only if each sentence in $\Gamma$ has a model.

Suppose $\Gamma$ has a model $M$. Then by definition of $M\models \Gamma$, we have $M\models \varphi$ for each sentence $\varphi\in \Gamma$, so each sentence in $\Gamma$ has a mdoel.

Conversely, suppose that every sentence in $\Gamma$ has a model. By the compactness theorem, to show that $\Gamma$ has a model, it suffices to show that every finite subset of $\Gamma$ has a model. Let $\Delta\subseteq \Gamma$ be finite, and write $\Delta = \{\psi_1,\dots,\psi_n\}$. Then $T\vdash \bigwedge_{i=1}^n \psi_i$, so $\bigwedge_{i=1}^n \psi_i$ is in $T$, since $T$ is closed under provability. Thus $\bigwedge_{i=1}^n \psi_i$ has a model $M$. Now since $M\models \bigwedge_{i=1}^n \psi_i$, also $M\models \psi_i$ for each $i$, so $M\models \Delta$, as desired.

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If instead you mean to write:

Let $\Gamma$ be a theory that is closed under provability. Prove that $\Gamma$ has a model if and only if each subset of $\Gamma$ has a model.

... then the proof is trivial.

Suppose $\Gamma$ has a model $M$. Let $\Gamma'\subseteq \Gamma$ be a subset. For all $\varphi\in \Gamma'$, $\varphi\in \Gamma$, so $M\models \varphi$, and hence $M\models \Gamma'$. So every subset of $\Gamma$ has a model.

Conversely, suppose every subset of $\Gamma$ has a model. Since $\Gamma\subseteq \Gamma$, $\Gamma$ has a model.