# How to calculate the mobius function of a Poset using Hall's theorem

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.

In an interval $$[x,y]$$ in a poset, with $$x\neq y$$, there is only one chain of length one. It is the chain $$x. Since the chain must begin with the element $$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.
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 $$a, so $$C_1(a,d)=1$$. There are two chains of length two, $$a and $$a, so $$C_2(a,d)=2$$. Finally, there is one chain of length three, i.e. $$a, so $$C_3(a,d) = 1$$. Using Hall’s theorem we can now compute:
$$\mu(a,d) = C_0 -C_1 + C_2 -C_3= 0-1+2-1 =0$$