To build a tree in a countable theory with uncountable many types in finite variables

Question

I'm trying to prove that if $T$ is a countable consistent theory that has no binary trees, then $T$ is small. i.e $|S_n (T)|=\aleph _0$

In order to do this i assume toward contradiction that $S_n (T)$ is uncountable. Then i try to find a formula $\varphi(x_1,..,x_n)$ which it and its negation appear in uncountably many types. Or in other wards $[\varphi ]$ and $[\neg\varphi ]$ are uncountable. (base sets of the topology on $S_n (T)$) and by that inductively i can build a tree.

Since $T$ is countable it is easy to see that there is a formula such that $[\varphi ]$ is uncountable, since if not we get a contradiction to compactness of the topological space $S_n (T)$. But i can't see how to find such formula that also $[\neg\varphi ]$ is uncountable.

I would appreciate your help!

Answer 1

Here is a direct argument: Assume that $|S_n(T)|>\aleph_0$ and assume that every formula $\varphi$ has a small side, that is either $[\varphi]$ or $[\neg \varphi]$ is countable. Now take all the small formulas $\varphi_0, \varphi_1, \ldots $, so $[\varphi_i]$ is countable and for every $\psi$ there is $k$ such that $\psi=\varphi_k$ or $\neg \psi=\varphi_k$. Then $\bigcup [\varphi_i]$ is countable. Let $p$ be a type not in this union. Then $\neg \varphi_i \in p$ for all $i$. However by the above remark for every $\psi$ either $\psi$ or $\neg \psi$ occurs in this list and therefor there is only one possibility for $p$. This contradicts the assumption that the number of types in uncountable.

Answer 2

Thinking of types as paths through a tree, use the fact that every closed subset of $2^\omega$ is either countable or has a perfect subset. Then since $S_n(T)$ is closed, if it's uncountable it has a perfect subset - that is, there's a perfect tree $X\subseteq 2^{<\omega}$ such that $[X]\subseteq S_n(T)$. Then any splitting node of $X$ gives you such a formula.

I use "$[X]$" for the set of paths through $X$, for $X$ a tree.

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By the way, the construction of $X$ is very nice: just take every $\sigma\in 2^{<\omega}$ such that $[\sigma]\cap S_n$ is uncountable. Here I'm fixing some ordering $\varphi_i$ of all formulas in $n$ variables, so each element of $2^{<\omega}$ corresponds to a finite set of $\varphi_i$s and negated $\varphi_i$s in an obvious way.

It's a nice exercise to show that, if $S_n$ is uncountable, then this $X$ is in fact a perfect tree; the closure of $S_n$ then implies that $[X]\subseteq S_n$.