Let us think about how we would construct a family \(\mathcal{F}\subseteq \binom{\mathbb{N}}{3}\) with \(|\mathcal{F}|=m\) for small values of \(t\) which minimizes the cardinality of \(\partial \mathcal{F}\text{.}\) For \(m=1\text{,}\) the family \(\mathcal{F}\) consists of a single set of size three, and so the cardinality of the shadow is always equal to \(\binom{3}{2}\text{.}\)

Things start getting more interesting for \(m=2\text{.}\) Without loss of generality, assume that \(\mathcal{F}\) contains the set \(\{1,2,3\}\text{.}\) If the other element of \(\mathcal{F}\) is \(\{a,b,c\}\text{,}\) then

\begin{equation*}
\partial \mathcal{F} = \{\{1,2\},\{1,3\},\{2,3\}\} \cup\{\{a,b\},\{a,c\},\{b,c\}\}\text{.}
\end{equation*}

Therefore, if we are trying to make \(\partial\mathcal{F}\) as small as possible, we want the “overlap” between the collections \(\{\{1,2\},\{1,3\},\{2,3\}\}\) and \(\{\{a,b\},\{a,c\},\{b,c\}\}\) to be as large as possible. Clearly, this occurs when \(\{1,2,3\}\) and \(\{a,b,c\}\) have as many elements in common as possible and so it is optimal to take \(\{a,b,c\}=\{1,2,4\}\text{.}\) Note that, of course, this choice is not unique; e.g., \(\{a,b,c\}=\{2,3,7\}\) is also optimal. Note that this matches our earlier intuition that it is smart to avoid introducing new elements of \(\mathbb{N}\) into the union of sets in \(\mathcal{F}\) as much as possible.

Now, if we apply similar intuition to the case \(m=3\text{,}\) it seems to be a good idea to take three sets which have large pairwise intersections. For the sake of discussion, suppose that \(\mathcal{F}\) contains the sets \(\{1,2,3\},\{1,2,4\}\) and a third set \(\{a,b,c\}\text{.}\) Then

\begin{equation*}
\partial \mathcal{F} = \{\{1,2\},\{1,3\},\{2,3\},\{1,4\},\{2,4\}\}\cup\{\{a,b\},\{a,c\},\{b,c\}\}\text{.}
\end{equation*}

In particular, \(\partial \mathcal{F}\) already contains all elements of \(\binom{[4]}{2}\text{,}\) with the exception of \(\{3,4\}\text{.}\) So, if we take \(\{a,b,c\}\) to be a subset of \(\{1,2,3,4\}\text{,}\) then, at worst, the shadow will have cardinality \(6\text{.}\) On the other hand, if \(\{a,b,c\}\) contains an element of \(\{5,6,7\}\text{,}\) then at least two of the sets \(\{a,b\},\{a,c\}\) and \(\{b,c\}\) will not be contained in \(\binom{[4]}{2}\text{,}\) and so \(\partial \mathcal{F}\) will contain at least \(7\) elements. With a bit more thinking, these arguments lead us to the conclusion that an optimal choice for \(\mathcal{F}\) is \(\{\{1,2,3\},\{1,2,4\},\{1,3,4\}\}\text{.}\) But also, taking any three sets from \(\binom{[4]}{3}\) would be equally good.

Given the answer for \(m=3\text{,}\) the case \(m=4\) actually becomes trivial. If we let \(\mathcal{F}=\binom{[4]}{3}\text{,}\) then the shadow again has cardinality \(6\text{,}\) which must be optimal, as it cannot be better than the best example in the case \(m=3\text{.}\)

Now, when \(m=5\text{,}\) the best thing to do is to take \(\mathcal{F}=\binom{[4]}{3}\cup\{\{1,2,5\}\}\text{,}\) which has a shadow of cardinality \(8\text{.}\) For \(m=6\text{,}\) the best is \(\mathcal{F}=\binom{[4]}{3}\cup\{\{1,2,5\},\{1,3,5\}\}\) and for \(m=7\) it is \(\mathcal{F}=\binom{[4]}{3}\cup\{\{1,2,5\},\{1,3,5\},\{2,3,5\}\}\text{.}\)

But why, you may ask? It is a bit tedious to continue to rule out all other possibilities as we have been doing, but the vague intuition is as follows. When

\(\binom{4}{3}\lt t\leq \binom{5}{3}\text{,}\) we have no choice but to introduce a fifth element of

\([n]\text{,}\) say, the number

\(5\text{,}\) into the union of sets in

\(\mathcal{F}\text{.}\) Assuming that

\(\mathcal{F}\) contains all of the sets in

\(\binom{[4]}{3}\) (which we haven’t justified, but let’s go with it), then the “new” sets that we get in the shadow are of the form

\(\{a,5\}\) where

\(\mathcal{F}\) contains a set of the form

\(\{a,b,5\}\text{.}\) So, if we let

\(\mathcal{F}'=\{\{a,b\}: \{a,b,5\}\in\mathcal{F}\}\text{,}\) then it seems reasonable that, to make the shadow of

\(\mathcal{F}\) small, we would want to minimize the shadow of

\(\mathcal{F}'\text{;}\) however, we already learned how to do that in

Example 2.3.1.