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Let $A\subset \mathcal{P}(\{1, \dots, n\}) $ and $ B \subset \{1, \dots, n \}$

We say $A$ shatters $B$ if $\forall y \subset B, \exists x \in A$ such that $x \cap B = y$.

I am asked to show that if $A$ does not shatter the sets: $\{1,2,3\},\{2,3,4\},\ \dots, \{n-2,n-1,n,\}, \{n-1,n,1\}, \{n,1,2\}$ and $n$ is a multiple of $3$ then $|A| \leq 7^{\frac{n}{3}}$

My current thinking is that, for each of these $3$-sets, we have to miss at least one of their subsets.

Specifically, for each $a \subset \{x,y,z\}$ there are $2^{n-3}$ subsets of $\{1,\dots,n\}$ that intersect with $\{x,y,z\}$ to give $a$. (Call the set of there $2^{n-3}$ subsets $C_{\{x,y,z\}}(a)$) Hence if $A$ does not shatter $\{1,2,3\}$ because we are missing $a$, then $A$ cannot contain these $2^{n-3}$ subsets.

I want to say that there is some subset $B \subset \mathcal{P}(\{1,\dots,n\}$ of size $8* 7^{\frac{n}{3}}$ such that we must have $A \subset B$, and we may only have at most $\frac{7}{8}$ of the elements of $B$. I suspect we have something like:

$B = \bigcup_{\{x,y,z\} \text{ mentioned earlier}}\bigcup_{a \subset \{x,y,z\}} C_{\{x,y,z\}}(a)$

However, at this point I am stuck and I'm not sure how to proceed. I can't think of a nice way to count the size of $B$ and show it is what I want because I can't think of an easy way to account for all of the overlaps occurring.

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  • $\begingroup$ Since $\mathcal{P}(\{1, \dots, n\})$ is a family of (non-empty) subsets of $\{1, \dots, n\}$ and $A$, $B$ are subsets of this family, they are also families of sets. Then $A$ does not shatter the sets ... means $A$ does not shatter the family consisting of these sets, right? $\endgroup$ – Alex Ravsky Dec 21 '18 at 3:43
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    $\begingroup$ Anyway, I don’t understand the definition of $A$ shatters $B$. If $A$ shatters $B$ then, $B\subset B$ so there exists $x\subset A$ such that $x\cap B=B$, that is $x\supset B$. So $A\supset B$. Conversely, if $A\supset B$ than for any $y\subset B$ we have $y\subset A$ so if we put $x=y$ then $x\cap B=y$. That is $A$ shatters $B$ iff $A\supset B$. $\endgroup$ – Alex Ravsky Dec 21 '18 at 3:44
  • $\begingroup$ @AlexRavsky my apologies, I wrote the definition wrong. Firstly, $A$ is a family of subsets and $B$ is a subset. Secondly, $A$ shatters $B$ if for all $y \subset B \; \exists x \in A$ such that $x \cap B = y$ $\endgroup$ – user366818 Dec 22 '18 at 15:11
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Consider the $n/3$ sets $$ \{1,2,3\},\{4,5,6\},\ldots,\{n-2,n-1,n\}. $$ Since $A$ doesn't shatter $\{1,2,3\}$, there exists some subset $T_{123}$ such that $S \cap \{1,2,3\} \neq T_{123}$ for all $S \in A$. Define $T_{456},\ldots$ similarly. Then $$ A \subseteq [\mathcal{P}(\{1,2,3\}) \setminus \{T_{123}\}] \times [\mathcal{P}(\{4,5,6\}) \setminus \{T_{456}\}] \times \cdots \times [\mathcal{P}(\{n-2,n-1,n\}) \setminus \{T_{n-2,n-1,n}\}]. $$ Each of the $n/3$ factors on the right-hand side contains $2^3-1=7$ elements, and so the right-hand side consists of $7^{n/3}$ sets.

When $n > 3$, this bound isn't tight, since the right-hand side does shatter all other adjacent triplets.

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