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Is it possible for a multiset to have a "negative" number of one or more elements? If so, how are such multisets defined, and what terminology exists for them?

J.P.
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    That's not a multiset, that's $\mathbb{Z}^S$, where $S$ represents the index set. – vadim123 May 05 '13 at 18:40
  • @vadim123 Sorry, could you explain your comment a bit? Why is it not a multiset? – J.P. May 05 '13 at 19:05
  • JP, because a multiset is $\mathbb{N}^S$. See @MJD's answer for more details. – vadim123 May 05 '13 at 20:37
  • I imagine that it could be useful to represent members of $\mathbb Q$ as a multiset with possibly negative multiplicity indicating the prime factorization of both the numerator and denominator, where a negative multiplicity indicates that the factor is in the denominator. – dfan May 05 '13 at 23:39
  • No, you're not allowed. – Jacob Wakem Dec 01 '16 at 04:54

3 Answers3

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As I pointed out a while back, there isn't much standard terminology or notation even for ordinary multisets. Mathematicians usually resort to one of two workarounds when they need to deal with multisets: they either replace the multiset of elements of $S$ with a (not strictly) monotonic sequence of elements of $S$, or they replace it with a function $c:S\to\Bbb N$ that counts how many of each element of $S$ there is in the multiset. For example, $c(\bullet) = 3$ means that the multiset $c$ contains three instances of the element $\bullet$.

As vadim123 pointed out in a comment, it's easy to adjust the latter workaround from $c:S\to\Bbb N$ to $c:S\to\Bbb Z$, but at that point the object you are dealing with is a lot more like a function than it is like a set, and it's not clear what benefit you would get from trying to think of it like some weird kind of set.

Community
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MJD
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Yes, such multisets with negative multiplicities are considered and used. For example, see this link .

user64494
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One way to describe such multisets, at least those which only consist of a finite number of elements, is as elements of the free abelian group on a set $S$ (this is not the same as the set of functions $S \to \mathbb{Z}$ when $S$ is infinite, and in any case it is a covariant functor rather than a contravariant one). This is a common construction in algebraic topology and algebraic geometry.

Qiaochu Yuan
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