Calculating Moebius inversion of a poset with a minimum element.

Question

If P is a poset that has a minimum element. We let x be an element of P that covers 1 single element y.Assume that y is not the minimum element, how do I prove that μ(minimum element,x) = 0?

So μ(min, min) + μ(min , closest element that covers min) = 0, and μ(x,z) = $-∑_{x<y<z}μ(x,y)$ but how do i prove that the value of μ(minimum element, y)is fixed?

I've spent some time looking at this and can't figure it out. Any hints?

Answer 1

Let $0$ be the minimum element of $P$.

Since $x$ only covers $y$, $$ \mu(0,x) \;=\; \sum_{0\leqslant z\leqslant x}\mu(0,z) \:-\: \sum_{0\leqslant z\leqslant y}\mu(0,z) \;=\; 0 \:-\: 0 \;=\; 0 . $$