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Let $A_0=I,A_1,\dots,A_k$ be the assocation matrices of a $k-$ class association scheme $R_0,\dots,R_k$ on a set $X$. Let $K \subset \{0,1,\dots,k\}$. We say a subset $Y \subset X$ is a $K-$coclique if for any $(u,v) \in Y \times Y$, $(u,v) \not \in R_i$ for any $i \in K$. Similarly, $Y$ is a $K-$clique if for any $(u,v) \in Y \times Y$, there is some $i \in K$ with $(u,v)\in R_i$. Our goal is to prove that if $|A|$ is a coclique and $B$ is a clique, then $$ |A||B| \leq N $$ holds, where $N := |X|$.

In Van Lint and Wilson Theorem 30.4, the proof comes from the following fact.

Claim: Let $a$ and $b$ be the distribution vectors defined as $$ a_i := \frac{1}{|A|} |\{(A\times A) \cap R_i\}|, b_i := \frac{1}{|B|} |\{(B\times B) \cap R_i\}| $$ for $i = 0,1,\dots,k$. Then, $a_ib_i = 0$ for all $i > 0$.

I can see that if $i \in K$, then, since $A$ is a $K-$coclique, no $(u,v) \in A \times A$ satisfies $(u,v) \in R_i$, hence $a_ib_i = 0$ for each $i \in K$.

Question But, why for an index $i \not \in K$ do we have $a_ib_i = 0$? This looks like the only place to use the fact that $B$ is a clique, but I do not see how.

Mutasim Mim
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Say you consider relations $K \subseteq \{1,\dots,k\}$, you let $A$ be a coclique for $K$, and $B$ be a clique for $K$. This means that any two distinct elements in $A$ are in some relation $R_i$ with $i \notin K$, and any two distinct element of $B$ are in some relation $R_i$ with $i \in K$.

Therefore, for any $i > 0$, either $i \in K$ and $a_i = 0$ or $i \notin K$ and $b_i = 0$. In both cases $a_i b_i = 0$.

Note that the claim you are making only holds for $i >0$ since $a_0 = b_0 = 1$.

SamA3aensen
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