# Antichains and the LYM inequality

• May 7th 2012, 01:46 PM
hairymclairy
Antichains and the LYM inequality
Hello,

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

• May 7th 2012, 02:21 PM
Plato
Re: Antichains and the LYM inequality
Quote:

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.
• May 8th 2012, 01:01 AM
hairymclairy
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 $\mathscr{A}$ on [n], let $a_k=\left|\mathscr{A}\cap{[n]\choose k}\right|$ i.e the number of sets of size $k$ in $A.$ Then
$\sum_{k=0}^n\frac{a_k}{{n\choose k}}\leqslant 1$.

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

In the above notation, $[n]=\{1,2,3,...,n\}$. Each implies the other, but I'm struggling to show properly that the first inequality implies the second.
• May 8th 2012, 05:57 AM
hairymclairy
Re: Antichains and the LYM inequality
I've solved this now.

Thanks anyway.