A problem about an intersecting family of subsets of set that is an antichain Suppose $\Delta$ is an intersecting family of subsets of a set of $A$, where $A$ is a set with $n$ elements.Also suppose that $\Delta$ is an antichain, i.e.: $$A_1,A_2\in\Delta \thinspace  (A_1\neq A_2)\Rightarrow \thinspace (A_1\not\subset A_2\thinspace \thinspace and\thinspace \thinspace A_2\not\subset A_1 ) $$and each element of $\Delta$ has size at most $k$ where $k\leq n/2$ is a positive integer. Now suppose $\Gamma$ is the set of all subsets of $A$ with $k$ elements. Is there a $1-1$ function $f$ from $\Delta$ to $\Gamma$ such that:$$x\in \Delta \thinspace \Rightarrow x\subseteq f(x)?$$
 A: There is an injective function from $\Delta$ to $\Gamma$ such that $f(\Gamma)$ such that for all $x\in \Delta$ we have $A\subseteq f(x)$.
Proof: Define $P_i$ as the set of $i$-subsets of $A$. For each $i$ such that $i+1\leq n/2$ consider the bipartite graph with parts $P_i$ and $P_{i+1}$, and where the edges are given by inclusion. Notice all the vertices in a given side have the same degree, so there is a matching that saturates $P_i$ (provable by hall easily).
Now consider the graph with vertex set $P_1\cup P_2\dots P_{\lfloor n/2 \rfloor}$ and select one matching for each layer, notice if we only take those edges the graph is broken up into a bunch of disjoint paths, and for each subset of $x\in A$ of size $k$ and for each $j>k$ there is exactly one vertex of size $j$ connected to $k$. If we are given $\Delta$ and $k$ we can define $f(x)$ in this way, and we will only have $f(x) = f(y)$ if $x$ contains $y$ or visceversa.
Notice we do not need to use that $\Delta$ is intersecting, it is sufficient to use it is an antichain. You can see Dilworth's theorem for a lot of related stuff.
