If a formula has a Morley Rank then it is less then $|T|^+$ 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$.
