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Here is the theorem I want to prove:

For $x,y \in P$ and $i \geq 0,$ let $c_i(x,y)$ be the number of chains $x=x_0 < x_1 < \dots < x_i = y$ of length $i$ from $x$ to $y.$ Let $$\phi (x,y) = c_0(x,y) - c_1(x,y) + c_2(x,y) - c_3(x,y) + \dots. $$ Prove that $\mu (x,y) = \phi (x,y).$

Where $\mu$s the Mobius Function and $P$ is a finite poset.

My thoughts are:

I found a proof for it in the following link here: Phillip Hall's theorem on chains but my formula has chains of length zero, is always the number of chains of length zero is zero? What is an example of a chain of length zero?

How also I can proof the statement by induction, could anyone show me the details please?

Intuition
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    I am totally unfamiliar with the subject matter that you are questioning, as well as the syntax used in your posting. However, it seems as if the theorem is based on Inclusion-Exclusion; I could easily be wrong. If it is based on Inclusion-Exclusion, then the following articles may be helpful to you: See this article for an introduction to Inclusion-Exclusion. ...see next comment – user2661923 Nov 13 '23 at 09:40
  • Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 Nov 13 '23 at 09:41
  • how will this help me in the proof by induction? @user2661923 – Intuition Nov 13 '23 at 09:47
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    I don't know whether an Inclusion-Exclusion approach will be of any help in an Induction proof of the theorem that you are trying to prove. – user2661923 Nov 13 '23 at 11:36

1 Answers1

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So $\phi (x,y) = \displaystyle \sum _{i = 0}^{\infty}(-1)^ic_i(x,y),$ where clearly the sum is finite, because the poset is finite.

  1. $\phi (x,x)=1$ :

    There is only one chain of size $0$ from $x$ to $x$, so $c_0(x,x)=1$ and $c_i(x,x)=0$ for $i>0$, hence $\phi (x,x)=1$.
  2. $\phi (x,y) = \displaystyle -\sum _{x\leq z<y}\phi (x,z)$ :

    Using the definition on the RHS, we have that $$\displaystyle -\sum _{x\leq z<y}\phi (x,z) = \displaystyle -\sum _{x\leq z<y}\sum _{i = 0}^{\infty}(-1)^ic_i(x,z),$$ exchanging the sums (they are finite and independent), we have that $$RHS = \sum _{i = 0}^{\infty}(-1)^{i+1}\sum _{x\leq z<y}c_i(x,z),$$ so if we show that $c_{i+1}(x,y)=\displaystyle \sum _{x\leq z<y}c_i(x,z)$, we are done, but this is obvious from the definition as if you consider $x=x_0<\cdots <x_i=z<y$ this is a chain of size $i+1$.

Because $\mu$, the Möbius function, also respects this recursion and these initial conditions, they are the same function.

Phicar
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