A corollary from Sperner’s Theorem

combinatoricsmatching-theory

How can I conclude this fact:

If $$\mathcal{F}$$ is an antichain consisting of sets of size at most $$k \leq n/2$$, then $$|\mathcal{F}|\leq {n\choose k}$$

from the Sperner's Theorem:

Let $$\mathcal{F}$$ be a family of subsets of an $$n$$ element set. If $$\mathcal{F}$$ is an antichain, then $$|\mathcal{F}|\leq {n\choose \lfloor n/2 \rfloor }$$.

Let $$[n]=\{1,\dots,n\}$$, and let $$\binom{[n]}{k}$$ denote the set of subsets of $$[n]$$ with size $$k$$.
One way to prove Sperner's theorem is to find a sequence of order matchings $$f_k$$. When $$2k, this means that $$f_k$$ is a injective function $$\binom{[n]}{k}\to \binom{[n]}{k+1}$$ which satisfies $$A\subset f_k(A)$$ for all $$A\in \binom{[n]}{k}$$. When $$2k>n$$, the mappings go in reverse; $$f_k:\binom {[n]}{k}\to \binom{[n]}{k-1}$$, but we still require $$f_k$$ to be injective and satisfy $$A\subset f_k(A)$$. These order matchings can be found using Hall's marriage theorem, or using a linear algebra argument (see Stanley, Algebraic Combinatorics, Lemmas 4.5 - 4.7, https://klein.mit.edu/~rstan/algcomb/index.html).
The existence of these mappings for all $$k\in \{0,\dots,n\}\setminus \{n/2\}$$ implies Sperner's lemma. Namely, given an antichain $$\mathcal F$$, you can repeatedly apply these order matchings to the sets in $$\mathcal F$$ to transport them to the middle level of $$2^{[n]}$$, getting a new antichain $$\mathcal F'$$ which lives entirely in $$\binom{[n]}{\lfloor n/2\rfloor}$$. The properties of the order matchings imply $$|\mathcal F|=|\mathcal F'|$$, so we must have $$|\mathcal F|\le \binom{n}{n/2}$$.
This same idea easily applies to proving $$|\mathcal F|\le \binom{n}{k}$$ when all subsets in $$\mathcal F$$ have size at most $$k$$, where $$k. This time, you use the order matchings to transport each set in $$\mathcal F$$ to level $$k$$, and conclude using the same logic.