I am trying to understand the proof of Theorem 8.30. in the book "The theory of ultrafilters" by Comfort and Negrepontis.
Theorem. If $p$ is a normal ultrafilter on a measurable cardinal $\alpha$, then $\{\beta < \alpha : \beta \rightarrow (\beta )^{<\omega}_2\} \in p$.
I will try to summarize the proof leading up to the detail that I do not understand.
The proof goes by contradiction, so it is assumed (after showing that the set of all cardinals smaller than $\alpha$ is in $p$) that $$I=\{\beta < \alpha : \beta \not\rightarrow (\beta )^{<\omega}_2 \} \in p.$$ For every $\beta\in I$, we take a partition witnessing the relation $\beta \not\rightarrow (\beta )^{<\omega}_2 $, and construct from this a partition of $\alpha$. As $\alpha $ is Ramsey, there is some set $A\subset \alpha$, $\lvert A\rvert=\alpha$, that is homogeneous with respect to this partition.
We consider the assumption that $\{\beta \in I : \lvert \beta \cap A \rvert < \beta\}\in p$, and get by normality of $p$ that there is some $\gamma < \alpha$ such that $\{\beta \in I : \lvert \beta \cap A \rvert = \gamma\}\in p$.
Now the book says that this is a contradiction to the fact that $p$ is uniform. This is the part that I do not understand. Why is it immediately clear that $\lvert \{\beta \in I : \lvert \beta \cap A \rvert = \gamma\}\rvert <\alpha$?