Let G consist of just a single ε-regular pair (A, B) of density d > 0, with |A| = |B| = l. Prove that for any d, we have that if ε is sufficiently small and ` is sufficiently large, then we can find a $K_{3,3}$ subgraph of $A \cup B$.
I am trying to use the lemma below to prove the above statement:
"For ε-regular pair (A, B) of density d > 0, and $Y \subset B$ have size $|Y| \ge \epsilon|B|$, all but fewer than $\epsilon|A|$ of the vertices in A have $\ge (d-\epsilon)*|Y|$ neighbors in Y."
So I let $Y \subset B$ with size $|Y| \ge \epsilon|B|$. Then I assume $ε=d/2$ and $l \ge 6/d$, which means that almost all vertices in A have greater or equal than $(d-\epsilon)*|Y|=d/2*6/d=3$ neighbors in Y, which is a subset of B.
However, in order to form a $K_{3,3}$, I need to show that more than 3 vertices in Y have at least 3 neighbors in A as well, which I am stuck on. Any help is appreciated.