I learned that the average size in any ideal of subsets of $[n]$ is at most $n/2$, but I think the downward closed family of the subsets of $[n]$ also satisfied. I want to know how to proof it or it is wrong. A downward closed family $\mathcal{F}$ means for any $A \in \mathcal{F}, B \subseteq A$, we have $B \in \mathcal{F}$, and the ideal's definition is here <https://en.wikipedia.org/wiki/Ideal_(set_theory)>.



The average size of downward closed family of the subsets of $[n]$ is at most $n/2$?
Here is a pedestrian answer. If $\def\cF{\mathcal F}\cF$ is a downward closed subset of $\mathcal P([n])$, we have


$$\frac1{|\cF|}\sum\_{A\in\cF}|A|=\sum\_{i\in[n]}\Pr\_{A\in\cF}[i\in A].$$


Now, for any $i\in[n]$,


$$\Pr\_{A\in\cF}[i\in A]\le\frac12,$$


because the mapping $A\mapsto A\smallsetminus\{i\}$ provides an injection


$$\{A\in\cF:i\in A\}\to\{A\in\cF:i\notin A\}.$$


