False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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FALSE: the positive-size multiset product always encodes a valid multiset class

Statement

False claim: once one knows the positive-size counts $(a_{n})_{n \geq 1}$ , the formal product

$\prod_{n \geq 1} (1 - x^{n})^{- a_{n}}$

automatically is the ordinary generating function of the multiset construction, with no further local-finiteness hypothesis on the underlying class.

The formal product itself is coefficientwise well defined. What is false is its unconditional interpretation as a multiset generating function: omitted size- $0$ behaviour can destroy local finiteness completely while leaving the displayed positive-size sequence unchanged.

Facts & Assumptions

Given: The multiset product theorem assumes that the underlying class has no size-zero objects (If $A$ has no size-zero objects then $MSET (A)$ has generating function $\prod_{n \geq 1} (1 - x^{n})^{- a_{n}}$ ).

[L1]

Well-defined locally finite products are the ones licensed by the summability machinery (Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products).

Refutation

technique · direct

1.1

Let $A$ have one object $z$ of size $0$ and one object $u$ of size $1$ . Its positive-size counting sequence is $a_{1} = 1$ and $a_{n} = 0$ for $n > 1$ , so the displayed product is $(1 - x)^{- 1}$ .

given

2.1

But $MSET (A)$ has infinitely many size- $0$ objects: the multiplicity functions with $m (z) = 0, 1, 2, \dots$ and $m (u) = 0$ are all distinct and all have total size $0$ . So the would-be multiset class is not locally finite in degree $0$ , and [L1] does not license a generating function for it.

step 1.1L1given

3.1

The displayed product therefore does not automatically encode a valid multiset construction from the bare positive-size sequence alone. The omitted no-size-zero hypothesis matters, so the claim is false.

step 2.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics (standard reference, not scraped)