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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31
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The Möbius recurrence: μP(x,x)=1 and both interval sums of μP vanish when x<y

Statement

For a locally finite poset P and x≤y,

μP(x,x)=1,

and, when x<y,

∑x≤z≤yμP(x,z)=0,∑x≤z≤yμP(z,y)=0.

Equivalently, off the diagonal,

μP(x,y)=−∑x≤z<yμP(x,z)=−∑x<z≤yμP(z,y).

Either recurrence together with the diagonal values uniquely determines μP interval by interval.

Facts & Assumptions

Given: A locally finite poset P and comparable elements x≤y.

[F1]

μP∗ζ=δ=ζ∗μP (The integer-valued Möbius function μP of a locally finite poset).

[F2]

Convolution is the finite interval sum, ζ is constantly 1, and δ is 1 on the diagonal and 0 off it (The incidence functions I(P,R) of a locally finite poset and their convolution, The delta and zeta incidence functions, A finite sum in a commutative monoid indexed by an arbitrary finite set).

Proof

technique · direct
1.1

Evaluating either inverse equation at (x,x) gives μP(x,x)=1.

F1F2
1.2

Evaluating μP∗ζ=δ at x<y gives ∑x≤z≤yμP(x,z)=0.

F1F2
1.3

Evaluating ζ∗μP=δ at x<y gives ∑x≤z≤yμP(z,y)=0.

F1F2
2.1

Isolating the term z=y in step 1.2 and the term z=x in step 1.3 yields the two displayed recursive formulas.

step 1.2step 1.3algebra
3.1

Each right-hand side uses only proper subintervals, so induction on the finite cardinality of [x,y] shows that either recurrence and the diagonal clause determine at most one function.

step 1.1step 2.1
4.1

Steps 1.1 through 3.1 prove both sums, both recurrences and uniqueness.

step 1.1step 1.2step 1.3step 2.1step 3.1∎

Depends on

Used by

Dependency tree · two levels

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Sources