An upset, or upward closed set, is a subset $\mathcal{A}$ of $\mathcal{P}(n)$ such that, if the set $S$ is contained in $\mathcal{A}$, then all of the elements of $\mathcal{P}(n)$ which contain $S$ are contained in $\mathcal{A}$.
A downset, or downward closed set, is a subset $\mathcal{B}$ of $\mathcal{P}(n)$ such that, if the set $S$ is contained in $\mathcal{B}$, then all of the subsets of $S$ are contained in $\mathcal{B}$.
Let $\mathcal{A} \subset \mathcal{P}(n)$ be an upset and $\mathcal{B} \subset \mathcal{P}(n)$ be a downset. Prove that $|\mathcal{A} \cap \mathcal{B}| \leq \frac{|\mathcal{A}|\cdot|\mathcal{B}|}{2^n}$.
The only proof I have seen of this is using that the complement family $\mathcal{A}^c$ is a downset and then applying Harris' lemma for two downsets.
However, I am interested in seeing a direct proof of this (i.e. by induction on $n$, as the proof for downsets in Kleitman's Theorem on downsets). Unfortunately, in my tries I always get an inequality in the bad direction (or perhaps something else is wrong).
Any help appreciated!
The following inductive proof is taken from D.J. Kleitman Families of non-disjoint subsets J. Combin. Theory, 1 (1966), pp. 153-155
Suppose that the result holds true for $n=k$, and let $\mathcal A$ be an upset and $\mathcal B$ be a downset in $\mathcal{P}({1,2,...,k+1})$. Let $x=k+1$, $\mathcal A_x$ be the set of sets within $\mathcal A$ which contain $x$, and $\mathcal A_{\overline x}$ be the set of sets within $\mathcal A$ which do not contain $x$.
$\mathcal A=\mathcal A_x \cup \mathcal A_{\overline x}$
$\mathcal B=\mathcal B_x \cup \mathcal B_{\overline x}$
It is easy to see that $|\mathcal A_x| \ge |\mathcal A_{\overline x}|$ and $|\mathcal B_x| \le |\mathcal B_{\overline x}|$; any member of $\mathcal A_{\overline x}$ can be turned into a member of $\mathcal A_x$ by appending $x$ to each of its element subsets and so on.
By the induction hypothesis, $2^k|\mathcal A_{\overline x}\cap\mathcal B_{\overline x}| \le |\mathcal A_{\overline x}| |\mathcal B_{\overline x}|$ and $2^k|\mathcal A_{x}\cap\mathcal B_{x}| \le |\mathcal A_{x}| |\mathcal B_{x}|$.
So, putting it all together, $2^{k+1}|\mathcal A\cap\mathcal B|=2^{k+1}|\mathcal A_x\cap\mathcal B_x|+2^{k+1}|\mathcal A_{\overline x}\cap\mathcal B_{\overline x}|\le 2|\mathcal A_{\overline x}| |\mathcal B_{\overline x}|+2|\mathcal{A}_x||\mathcal{B}_x|\le|\mathcal{A}||\mathcal{B}|$