Isolated type, Stone Space, topology, open set

Question

A $p(x)$ type is isolated by $\phi(x)$ iff $\forall \psi(x) \in p(x):\ \phi(x)\to \psi(x)$ . How can I see that a type is isolated if and only if $\{p\}$ is open set in the Stone space? Open sets are of the form $\{p \in S_n(\emptyset)\ |\ \phi\in p \}$.

Answer 1

(Isolated $\Rightarrow$ $\{p\}$ is open.) Assume $p$ is isolated by $\varphi$. We know that the set of complete types containing $\varphi$ is an open set. However, any complete type containing $\varphi$ must contain every consequence of $\varphi$ and therefore every formula in $p$. Therefore $p$ is the only complete type containing $\varphi$. Therefore—also using the fact that $p$ contains $\varphi$—$\{p\}$ is the set of all complete types containing $\varphi$, which is open.

($\{p\}$ is open $\Rightarrow$ isolated.) Assume $\{p\}$ is open. It is therefore a union of basic open sets (i.e., those of the form $\{q \mid \varphi \in q\}$ for some $\varphi$). One of those basic open sets must contain $p$, and that basic open set must therefore be $\{p\}$. Therefore $\{p\}$ is a basic open set, i.e. we can find $\varphi$ such that $\{p\} = \{q \mid \varphi \in q\}$. That is, $p$ is the unique type containing $\varphi$.

I claim that $\varphi$ isolates $p$. We must only show that $p$ contains $\varphi$ (which is already known) and that every formula in $p$ is a consequence of $\varphi$. Assume we have $\psi \in p$ so that $\psi$ is not a consequence of $\varphi$. That is, $\varphi \wedge \neg\psi$ is consistent. We can then extend $\varphi \wedge \neg\psi$ to a complete type that will contain $\varphi$ and is not $p$, contradicting that $p$ is the unique type containing $\varphi$.