# Thread: Antichains and the LYM inequality

1. ## Antichains and the LYM inequality

Hello,

Can anyone help me understand how to derive the local LYM inequality from the LYM inequality?

2. ## Re: Antichains and the LYM inequality

Originally Posted by hairymclairy
Can anyone help me understand how to derive the local LYM inequality from the LYM inequality?
I think that you need to supply several definitions.

3. ## Re: Antichains and the LYM inequality

They're fairly standard theorems, I thought they'd be well-known as they trivially imply Sperner's theorem/lemma about the maximal size of an antichain.

The LYM inequality states that given an antichain $\displaystyle \mathscr{A}$ on [n], let $\displaystyle a_k=\left|\mathscr{A}\cap{[n]\choose k}\right|$ i.e the number of sets of size $\displaystyle k$ in $\displaystyle A.$ Then
$\displaystyle \sum_{k=0}^n\frac{a_k}{{n\choose k}}\leqslant 1$.

Now the local LYM inequality states that if $\displaystyle A\subseteq{[n] \choose r}$ then $\displaystyle \frac{|\partial A|}{{n \choose r-1}}\geqslant\frac{|A|}{{n\choose r}}$, where $\displaystyle \partial A$ is the lower shadow of the antichain $\displaystyle A$, in symbols $\displaystyle \partial A=\{B\in{[n]\choose r-1}:B\subseteq A'\in A\}$.

In the above notation, $\displaystyle [n]=\{1,2,3,...,n\}$. Each implies the other, but I'm struggling to show properly that the first inequality implies the second.

4. ## Re: Antichains and the LYM inequality

I've solved this now.

Thanks anyway.