Poset and Mobius function

Question

Suppose that $P = (X,\leq)$ is a poset such that for $x,y \in X, x \neq y$, we have that either $|[x,y]| = 0$ or $|[x,y]| = 2$ or $|[x,y]| = 6$. I asked to compute the Möbius function for this poset. I originally thought it to be $\mu(x,y) = 1, x = y$, $\mu(x,y) = -1, y = x + 1$, and $\mu(x,y) = 0$ otherwise. I have been debating with a classmates who believes it is $\mu(x,y) = 1, x = y$, $\mu(x,y) = -1, y = x + 1$, $\mu(x,y) = 3, y = x + 5$ and $\mu(x,y) = 0$ otherwise. Looking for some help on truly understanding this problem.

Answer 1

I'm afraid your classmate is (mostly) right.

You either forgot to mention that the elements of the poset are numbers, or you abused notation in adding integers to them. I'll solve the problem as stated without assuming that the elements of the poset are numbers.

If $|[x,y]|=0$, then $x\nleq y$ and $\mu(x,y)=0$.

If $|[x,y]|=1$, then $x=y$ and $\mu(x,y)=1$.

If $|[x,y]|=2$, then the interval consists only of $x$ and $y$, and $\mu(x,y)=-1$.

If $|[x,y]|=6$, then there are $4$ elements between $x$ and $y$. Let one of them be $a$. Then $|[x,a]|=2$, since the cases $0$ and $1$ can be excluded and $|[x,a]|=6$ would imply that there are further elements in $[x,y]$. Then by Hall's theorem (see How to calculate the mobius function of a Poset using Hall's theorem), since $[x,y]$ contains $C_0=0$ chains of length $0$, $C_1=1$ chain of length $1$ and $C_2=4$ chains of length $2$, in this case we have $\mu(x,y)=0-1+4=3$.