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

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?

• @RossMillikan Yes they should, thanks for noticing this! I have updated this now. Apr 15, 2021 at 13:56
• Each sentence is always finite. Each proof of $\phi$ only uses finitely many of the sentences in $\Gamma$, so you can use compactness ignoring the infinite subsets of $\Gamma$. You need deductively closed because otherwise there could be a $\phi$ that you can prove that is not part of $\Gamma$ Apr 15, 2021 at 13:57
• What do you mean by a "set in $\Gamma$"? Do you mean "sentence in $\Gamma$" instead? Apr 15, 2021 at 14:12
• @EricWofsey Yes you are right, I made a mistake when translating the text. Thanks! Apr 15, 2021 at 17:30
• @RossMillikan Thank you for your answer! I will do deductively closed. Apr 15, 2021 at 19:57

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.

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.