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Math Help - Antichains and the LYM inequality

  1. #1
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    Antichains and the LYM inequality

    Hello,

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

    Thanks in advance.
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  2. #2
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    Re: Antichains and the LYM inequality

    Quote Originally Posted by hairymclairy View Post
    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.
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    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.
    Last edited by hairymclairy; May 8th 2012 at 05:57 AM.
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  4. #4
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    Re: Antichains and the LYM inequality

    I've solved this now.

    Thanks anyway.
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