Let $n\geq 3$ be an integer. We call a family ${\cal C}$ of subsets of $\{1,\ldots,n\}$ *intersecting* if it has the following properties:


1. $A, B\in {\cal C}$ implies $A\cap B \neq \emptyset$, and
2. $A \in {\cal C}, B\subseteq A$ implies $B\notin {\cal C}$.


One way to construct an intersecting family with a large number of elements is the following. For any positive integer $k$ we say ${\cal A}\subseteq {\cal P}(\{1,\ldots,k\})$ is an [antichain](https://en.wikipedia.org/wiki/Antichain) if no member of ${\cal A}$ is contained in another member of ${\cal A}$. Let ${\cal M}$ be an antichain in ${\cal P}(\{1,\ldots,n-1\})$ with the maximum number of elements amongst all antichains in ${\cal P}(\{1,\ldots,n-1\})$. We denote this number by $m\_{n-1}$. Let $${\cal C} = \big\{X\cup\{n\}: X\in {\cal M}\big\}.$$ So ${\cal C}$ is an intersecting family on $
\{1,\ldots,n\}$ with $m\_{n-1}$ elements.


**Question.** Are there intersecting families on $
\{1,\ldots,n\}$ with more than $m\_{n-1}$ elements?



Maximum intersecting set families of $\{1,\ldots,n\}$
The intersecting family in your example has $\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}$ members by Sperner's theorem. An example that achieves a larger value would be to take all the subsets of $[n]$ that have $1+\lfloor \frac{n}{2}\rfloor$ elements. This is best possible.


Milner [proved](https://pdfs.semanticscholar.org/4f85/bedf628d70dc0d7b67f87c289966dfa46a90.pdf) that the largest size of a $k$-intersecting antichain in $\mathcal P([n])$ is $\binom{n}{\lfloor\frac{n+k+1}{2}\rfloor}$. Katona gave a simple proof of the case $k=1$ (the one relevant to the OP) in [this paper](https://www.sciencedirect.com/science/article/pii/S009731659792848X).


