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Two set families $\mathcal{A},\mathcal{B}$ are cross-Sperner when no member of either is a subset of a member of the other, including equality. Thus $\mathcal{A}\cap \mathcal{B}=\varnothing$. Neither family need itself be an antichain. On the Boolean lattice of subsets of $[n]$, the Cross-Sperner inequality bounds $\sqrt{|\mathcal{A}|}+\sqrt{|\mathcal{B}|}$ by $2^{n/2}$.

Choose disjoint subsets $K_1,K_2\subseteq [n]$ of size $k$ and let $R=[n]\setminus (K_1\cup K_2)$. The two-block extremisers for the cross-Sperner inequality areFor $A\in \mathcal{A}$ and $B\in \mathcal{B}$, some point of $K_1$ belongs to $A$ but not $B$, so $A⊈B$. Some point of $K_2$ belongs to $B$ but not $A$, so $B⊈A$. Thus these form a cross-Sperner family pair.

An element of $\mathcal{A}$ fixes its intersections with both blocks and freely chooses its subset of $R$. An element of $\mathcal{B}$ chooses any proper subset of $K_1$, any nonempty subset of $K_2$, and any subset of $R$. HenceThe hypotheses ensure that the blocks exist and both set families are nonempty, including when $n=2k$. Moreover,so this construction attains equality in the Cross-Sperner inequality.

The printed inclusion symbol here means non-strict inclusion: the PDF explicitly excludes a common member of the two set families. We use $\subseteq$ throughout. Thus a cross-Sperner family pair need not consist of internal antichains; the restriction is between the two families.

Give the Boolean lattice the uniform probability measure, and write $N=2^n$. Form the upward closure of a set family and downward closure of a set family of $\mathcal{A}$:These are respectively an increasing event and a decreasing event. We have $\mathcal{A}\subseteq U\cap D$, whereas the cross-Sperner family condition gives $\mathcal{B}\subseteq U^c\cap D^c$. Set $u=\mathbb{P}(U)$ and $d=\mathbb{P}(D)$. By negative correlation of increasing and decreasing events, first for $U,D$ and then for $U^c,D^c$,Apply the Cauchy-Schwarz inequality to the unit vectors $(\sqrt{u},\sqrt{1-u})$ and $(\sqrt{d},\sqrt{1-d})$. It yieldsTherefore the Cross-Sperner inequality isIf the printed inclusion symbol were read as strict inclusion while the explicit exclusion of common members were discarded, this conclusion would fail, for example with both families equal to the middle uniform set family when $n=2$. The PDF's parenthetical clause fixes the intended convention.

Use the four families from the hint. Put $N=2^n$, $u=|{\mathcal{U}}_1|/N$, and $v=|{\mathcal{V}}_1|/N$. The families ${\mathcal{U}}_1,{\mathcal{V}}_1$ are up-sets, while their complements ${\mathcal{U}}_2,{\mathcal{V}}_2$ are down-sets. Applying the Harris-Kleitman inequality to the up-sets, and equivalently to the down-sets after taking complements of the ground set, givesThe cross-incomparability assumption givesConsequentlyThe Cauchy-Schwarz inequality now yieldsThis is the sharp two-family Cross-Sperner inequality.

Choose disjoint subsets $K_1,K_2\subseteq [n]$ of size $k\ge 1$ with $2k\le n$. Let $\mathcal{A}$ consist of subsets containing all of $K_1$ and none of $K_2$; let $\mathcal{B}$ consist of subsets omitting at least one point of $K_1$ and containing at least one point of $K_2$. These form a cross-Sperner family pair, with sizes $2^{n-2k}$ and $(2^k-1{)}^2{2}^{n-2k}$. Their square roots sum to $2^{n/2}$, proving sharpness of the Cross-Sperner inequality. The free coordinates outside the two blocks account for the factor $2^{n-2k}$.