We saw in class that given a complete theory $T$, then if $MR\varphi\ge (2^{|T|})^+$ then $MR\varphi=\infty$ And we ware told that Lachlan improved this result to $|T|^+$.
To prove it, I assume towards contradiction that we have $\infty>MR\varphi\ge|T|^+$ and then to construct a binary tree, starting with $\varphi$, and from that we conclude that $MR\varphi=\infty$. But I don't see the right way to build the tree... help will be appreciated!
A theorem that is definitely under represented in the literature. Anyway here is as extract from by notes, its terse, but the basic idea is there:
Definition: $\alpha_T$ is the least ordinal $\alpha$ with the property that if $\alpha \leq \mbox{MR}(\varphi)$ then $\mbox{MR}(\varphi)= \infty$.
Theorem [Lachlan] If $\mathcal{L}$ is countable, then $$\alpha_T \leq \omega_1$$
Proof: Let $\psi$ be a formula such that $\omega_1 \leq \mbox{MR}(\psi)$. Then for every $\alpha < \omega_1$, $\alpha +1 \leq \mbox{MR}(\psi)$. Thus there are $\varphi_{\alpha}(\overline{x},\overline{b}_{\alpha})$ such that $$\alpha \leq \mbox{MR}(\psi(\overline{x}) \wedge \varphi_{\alpha}(\overline{x},\overline{b}_{\alpha}) ) $$ and $$\alpha \leq \mbox{MR}(\psi(\overline{x}) \wedge \neg \varphi_{\alpha}(\overline{x},\overline{b}_{\alpha}) ). $$ Since there are at most $\aleph_0$ many formulas there must be one formula that occurs uncountably often. Call this formula $\varphi_{\emptyset}(\overline{x},\overline{y})$. Thus for every $\alpha < \omega_1$ there is $\overline{b}$ such that
$$\alpha +1\leq \mbox{MR}(\psi(\overline{x}) \wedge \varphi_{\emptyset}(\overline{x},\overline{b}) ) $$ and $$\alpha +1 \leq \mbox{MR}(\psi(\overline{x}) \wedge \neg \varphi_{\emptyset}(\overline{x},\overline{b}) ). $$ By a repetition of the same argument we have $\varphi_{s}(\overline{x},\overline{y}_s)$ for $s \in 2^{<\omega}$ such that all paths through the tree are consistent. Thus by compactness there is an infinite tree and $ \mbox{MR}(\psi) =\infty$.