Möbius function of a poset

Remarks

Clause 2 is a recursion over the finite interval $[x, y]$, so $\mu$ is well defined. Equivalently, $\mu$ is the inverse of the zeta function $\zeta$ in the incidence algebra of $P$, where $\zeta(x, y) = 1$ for $x \le y$ and $0$ otherwise. From this one also gets $\mu(x, y) = -\sum_{x < z \le y} \mu(z, y)$ for $x < y$.

Example: in the lattice of subsets of a finite set, $\mu(S, T) = (-1)^{\abs{T \setminus S}}$ for $S \subseteq T$. Check: for $\abs{T \setminus S} = k \ge 1$, the sum $\sum_{S \subseteq U \subseteq T} (-1)^{\abs{U \setminus S}} = \sum_{j=0}^{k} \binom{k}{j} (-1)^j = (1 - 1)^k = 0$, as clauses 1 and 2 require.

Non-example: the function $\nu(x, y) = (-1)^{y - x}$ for $x \le y$ on the chain $\N$ is not the Möbius function. It satisfies clause 1, but clause 2 fails at $(0, 2)$: there $\mu(0, 2) = -(1 - 1) = 0$, while $\nu(0, 2) = 1$.

For the positive integers ordered by divisibility, $\mu(d, n)$ is the number-theoretic Möbius function of $n / d$.