Deducing this better version of the regularity lemma from the usual one

Question

Szemeredi's regularity lemma states that for every $\epsilon >0$ there is $M\in \Bbb N$ such that all graphs have an equipartition $\{V_1,...,V_k\}$ for some $k$ satisfying $\frac 1\epsilon \leq k \leq M$ such that at most $\epsilon k^2$ of the pairs $(V_i,V_j)$ are not $\epsilon$-regular.

By an equipartition we mean that the sets are as equal as possible, i.e. $||V_i|-|V_j||\leq 1$, and by an $\epsilon$-regular pair $(V_i, V_j)$ we mean that for every $A\subseteq V_i, B\subseteq V_j$ with $|A| \geq \epsilon |V_i|, |B| \geq \epsilon |V_j|$ we have that $|d(V_i,V_j)-d(A,B)|<\epsilon$, where $d(X,Y)$ denotes the density of the bipartite graph on $X,Y$.

Deduce from the regularity lemma that we can even require that every $V_i$ will have at most $\epsilon k$ irregular pairs $(V_i,V_j)$, which strengthens the traditional formulation above (in which we only claim that there are at most $\epsilon k^2$ irregular pairs).

This fact is referred to as a well-known (and easy) observation right before theorem 2 in this paper. It is claimed explicitly this holds when we take $M^3$ instead of $M$ in the statement.

This is also given as problem 2 here.

I'd like to know why this is true.

Answer 1

Given a graph $G$. By Szemeredi's regularity lemma, there is $M(\epsilon^3)$ such that $G$ has an equipartition $\{V_1, \dots, V_k\}$ such that $1/\epsilon^3 \le k \le M(\epsilon^3)$ such that at most $\epsilon^3 k^2$ of the pairs $(V_i, V_j)$ are not $\epsilon^3$-regular.

Consider graph $H$ with vertex set $[k]$ such that $i\sim j$ if $(V_i, V_j)$ is not $\epsilon^3$-regular. We repeatedly remove vertices from $H$ in the following way:

1. Suppose $h_1, \dots, h_{i-1}$ have been removed from $H$.
2. Choose $h_i$ be the vertex in $H$ with degree $> \epsilon k$ if any.
3. Remove vertex $h_i$ from $H$ and all adjacent edges.

Notice that every time we remove a vertex $h_i$ from $H$, more than $\epsilon k$ many adjacent edges. Because $H$ has $\le \epsilon^3 k^2$ edges, we have removed $< \epsilon^2 k$ vertices.

Without loss of generality, we may assume that vertices $1, \dots, l$ are left in $H$. It is clear that $l > (1-\epsilon^2)k$ and that for each $i \in [l]$, for all, but $\epsilon k$, $j\in [l]$, $(V_i, V_j)$ is $\epsilon^3$-regular in $G$.

Now we distribute the vertices in $\cup_{i > l} V_{i}$ to $V_1, \dots, V_l$ and get $U_1, \dots, U_l$ so that $\{U_1, \dots, U_l\}$ is an equipartition of $G$. It suffices to show that for all $i, j\in[l]$ if $(V_i, V_j)$ is $\epsilon^3$-regular, then $(U_i, U_j)$ is $2c\epsilon$-regular, where $c > 1$ is some absolute constant.

Suppose $(V_i, V_j)$ is $\epsilon^3$-regular. Take $A \subset U_i$ and $B\subset U_j$ with $|A| \ge \epsilon |U_i|$ and $|B| \ge \epsilon |U_j|$. Note that $V_i \subset U_i$, $|V_i| \approx |V(G)|/k$, $|U_i| \approx |V(G)|/l$, and so $$|V_i|/|U_i| \approx l/k > 1-\epsilon^2.$$ Therefore $$|A\cap V_i|/|V_i| > (|A|-|U_i\setminus V_i|)/|U_i|> \epsilon - \epsilon^2 > \epsilon^3.$$ Similarly, $|B\cap V_j|/|V_j| > \epsilon^3$. Since $(V_i, V_j)$ is $\epsilon^3$-regular, $$|d(V_i, V_j)-d(A\cap V_i, B\cap V_j)| < \epsilon^3.$$ One can also check that $$|d(U_i, U_j)-d(V_i, V_j)|\lesssim c\epsilon^2 \text{ and }|d(A\cap V_i, B\cap V_j)-d(A, B)|\lesssim c\epsilon$$ for some absolute constant $c>1$. Hence $$|d(U_i,U_j)-d(A,B)|\le 2c\epsilon.$$

Answer 2

If I read the paper right, there are still $\epsilon k^2$, not $\epsilon k$ irregular pairs allowed becuse they allow $\epsilon k$ pairs $(V_iV_j)$ for each $V_i$. So the only difference to the original lemma is that the $\epsilon k^2$ pairs are additionally somewhat equally distributed among the possible pairs ...