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Hall's Theorem states that:

$u(x,y) = C_0-C_1+C_2-C_3+\cdots$

where $C_k$ is number of chains of length $k$

If $x\neq y$ then $C_0=0$ and $C_1=1$

But my question is why does $C_1$ have to equal one? I've drawn a few different Posets where $C_1$ appears to be zero.

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In an interval $[x,y]$ in a poset, with $x\neq y$, there is only one chain of length one. It is the chain $x<y$. Since the chain must begin with the element $x$ and end with the element $y$, there are no other options.

As an example let’s compute the Mobius value $\mu(a,d)$ in the chain $a<b<c<d$.

First, there is no chain of length zero beginning at $a$ and ending at $d$, so $C_0(a,d)=0$. There is one chain of length one, namely $a<d$, so $C_1(a,d)=1$. There are two chains of length two, $a<b<d$ and $a<c<d$, so $C_2(a,d)=2$. Finally, there is one chain of length three, i.e. $a<b<c<d$, so $C_3(a,d) = 1$. Using Hall’s theorem we can now compute:

$$\mu(a,d) = C_0 -C_1 + C_2 -C_3= 0-1+2-1 =0$$

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  • $\begingroup$ So if i had a Poset that looked like a diamond, for instance, then going from the bottom to the top is a chain of length one regardless of whether there is a line connecting them (im coming from graph theory so I think im confusing "paths" and "chains") Thank you very much. $\endgroup$ – Wombles Apr 15 '19 at 5:32
  • $\begingroup$ Yes exactly, chains are different from paths. You may want to compare the definitions. $\endgroup$ – M47145 Apr 15 '19 at 6:04

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