7a. Take any $s$ sets in $\mathcal{F}_{s,t,\ell}$. By definition, each misses at most $\ell$ elements from $[t + s\ell]$. Thus their intersection misses at most $s \ell$ elements from $[t + s\ell]$ and thus has size at least $t$.

7b. Extend the definition of $S_{ij}$ to sets in the natural way. It's obvious that if $A$ is in $\mathcal{F}_{s,t,\ell}$ then $S_{ij}(A)$ is too. Hence shifting has no effect on $\mathcal{F}_{s,t,\ell}$.

7c. This one definitely takes much more explanation, but I don't think it's actually tricky -- you just have to follow your nose all the way through. However, you do have to avoid the following two extremely common mistakes:

"$S_{ij}\mathcal{F}$ is given by taking $S_{ij}(A)$ for all $A \in \mathcal{F}$." -- Not true.

"For any $A' \in S_{ij}\mathcal{F}$, there is an $A \in \mathcal{F}$ such that $S_{ij}(A) = A'$." -- False false false! Consider $i = 1$, $j = 2$, $n = 2$, $\mathcal{F} = \{ \{1\}, \{2\} \}$. Then $S_{12} \mathcal{F} = \mathcal{F}$, and yet for $\{2\} \in S_{ij} \mathcal{F}$, there is no set $A \in \mathcal{F}$ such that $S_{12}(A) = \{2\}$!

Here is my two-part proof:

Part 1.

Suppose by way of contradiction that there are $s$ sets $B_1, ..., B_s$ in $S_{ij} \mathcal{F}$ whose intersection is of size less than $t$. Let $A_1, ..., A_s \in \mathcal{F}$ be the sets they "came from". So for each $k$, either $B_k = S_{ij}(A_k)$ or $B_k = A_k$, the latter occurring when $S_{ij}(A_k)$ is also in $\mathcal{F}$. Clearly, the $A_k$'s are all be distinct, and so the intersection of the $A_k$'s has size at least $t$.

How could the $A_k$'s have intersection at least $t$ and yet the $B_k$'s have intersection less than $t$? Certainly nothing changes on coordinates other than $i$ or $j$. It's also clear $S_{ij}$ can't cause coordinate $i$ to leave the intersection. This means that it must be the case that coordinate $j$ left the intersection. In particular, $j$ is in the intersection of the $A_k$'s and not in the intersection of the $B_k$'s.

Further, since we know that $j$ is in the intersection of the $A_k$'s, it follows that $i$ is not in the intersection of the $A_k$'s. Otherwise, we have both $i$ and $j$ in each $A_k$, which surely means that $i$ and $j$ are both in each $B_k$. But we've already established that $j$ is not in the intersection of the $B_k$'s.

Finally, since $i$ is not in the intersection of the $A_k$'s, and since shifting caused the intersection size to go down, $i$ had better not be in the intersection of the $B_k$'s.

To summarize, $j$ but not $i$ is in the intersection of the $A_k$'s, and neither $i$ nor $j$ is in the intersection of the $B_k$'s. Further, it must be the case that the $A_k$'s have intersection size exactly $t$.

Part 2.

Focus on the coordinates $i$ and $j$ in the $A_k$'s now. For each $A_k$ with a $0$ in the $i$ coordinate and a $1$ in the $j$ coordinate (and there is at least one such $A_k$), how come all those $1$'s didn't shift over into the $i$th coordinate, causing all the $B_k$'s to have $1$'s in their $i$th coordinate?

It must be because there is at least one $A_k$ -- say $A_1$ WOLOG -- such that $S_{ij}(A_1)$ was already in $\mathcal{F}$. So let's look at the intersection of the following $s$ distinct sets in $\mathcal{F}$: $S_{ij}(A_1), A_2, ..., A_s$. These $s$ sets must also have intersection size at least $t$. Now the intersection of these is very similar to the intersection of $A_1, ..., A_s$; however, $S_{ij}(A_j)$ does not have a $0$ in the $j$th coordinate. This means $j$ is not in the intersection of $S_{ij}(A_1), A_2, ..., A_s$.

But this is a risky situation, because we concluded in Part 1 that $A_1, ..., A_s$ had intersection size only $t$, and this included the fact that $j$ was in the intersection. The only way $S_{ij}(A_1), A_2, ..., A_s$ could still have intersection size $t$ is if miraculously, putting in $S_{ij}(A_1)$ causes $i$ to be in the intersection, where it wasn't before.

But the only way for this to happen is if $A_1$ was the only set among $A_1, ..., A_s$ with the pattern $0, 1$ in coordinates $i, j$. In this case, every other $A_k$ has the pattern $1, 1$, which certainly implies $B_k = A_k$. Further, we have $B_1 = A_1$ by assumption (since $S_{ij}(A_1)$ was already in $\mathcal{F}$). So in fact all $B_k$'s equal their $A_k$'s. This is a contradiction because in Part 1 we concluded that $j$ was in the intersection of the $A_k$'s but not in the intersection of the $B_k$'s.

## Thursday, February 28, 2008

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