Can you have negative sets?

I figure that since you can, of course, have members in a set, have only a single member in a set, and then have no members in a set, it seems not then a big step forward (or backwards depending how you think of it) to think of a set with negative members.

I shall elucidate. Since set theory deals with membership, and it deals not with the quantity, but the quality of those members, perhaps it be possible to have a set with negative members which subtract members from another set whose positive counterparts is contained therein.

For example, the union of the sets $A$ and $B$, where set $A = \{1,2,3\}$ and set $B =\{-3\}$ would result in the set $A ∪ B = {1,2}$.

Two notes: First, you can arbitrarily construct any set one desires, but when applied to the real world, perhaps this may be of use?; Second, the empty set seems frivolous but turned out to be quite useful, maybe the same may be said for negative sets?

As someone pointed out, and they are of course correct, the set would actually be $\{1,2,3,-3\}$. However, in sticking with the principle, is what am describing denotable?

• Why don't you use {!3} (in parallel to the negation operator in C) instead of {-3}; the latter already has an established meaning. – Peter A. Schneider Nov 29 '16 at 18:03
• @JimJam That is a new notation suggestion superior to the one OP used. Right? I don't think it has been used before. – kaine Nov 29 '16 at 18:32
• your introduction to the problem seems ill-motivated to me. You could have many members of a set, but each of them is a member "just once", which would correspond to the number $1$. No member would correspond to the number $0$ (and then you could identify a set with its characteristic function, on a suitable domain). Multisets is a different story (when a member may be counted more than once). – Mirko Nov 29 '16 at 19:03
• @JimJam What's that called? Gee, of course it hasn't been seen before; you just came up with these negative sets. Why don't you give it a name. I would call that operator the "anti-element" operator. It creates an anti-element which annihilates a matching element. – Kaz Nov 29 '16 at 22:13
• Not possible without modifying the properties of $\cup$, since $A\cup B = (A \cup A) \cup B = A \cup (A \cup B) = A \cup \{1, 2\} = A \neq \{1, 2\}$ – fabian Nov 30 '16 at 16:20

This is possible. Essentially you want to extend the definition of a multiset to also include negative numbers as multiplicities. But I don't know if there is any useful application of this.

• I could think of some application in computing. E.g. you have a data structure with items $A=\{a, b, c\}$ and want to update it to a new version with the "update set" $B=\{-a, d\}$, i.e. removing $a$ and adding $d$, thus $A \cup B = \{b,c,d\}$. – Adrian Nov 29 '16 at 15:47
• Is it really an extension of a multiset though? It seems like what he's describing is more like an extension of a set in which a given element can be negatively included. That is, whereas for a standard set, any given element is either in the set or not in the set, this adds a third option of being negatively in the set. With a multiset, there are more options (corresponding to various multiplicities). We can think of this new type of set (call it, say, a "mixed set") as an ordered pair of sets $C=(A,-B)$ where the second set is what we might call a "negative set" <cont> – Gabriel Burns Nov 29 '16 at 15:52
• in which union $A \cup -B$ is irreducible for disjoint $A$ and $B$ (it's the mixed set $(A,-B)$.) and otherwise reduces to $(A \setminus B )\cup -(B\setminus A)$. – Gabriel Burns Nov 29 '16 at 15:57
• See en.wikipedia.org/wiki/Multiset#Generalizations for a sampling of generalizations of the notion of multiset. See also projecteuclid.org/download/pdf_1/euclid.ndjfl/1093635499 – Barry Cipra Nov 29 '16 at 16:11
• If you don't allow arbitrary (integer) multiplicities, then (edit: as noted in Aminopterin's answer) unions are no longer associative since $(-X \cup X) \cup X = 0 \cup X = X$ but $-X \cup (X \cup X) = -X \cup X = 0$. That's a pretty big problem. – Robin Saunders Nov 29 '16 at 23:27

What you describe is isomorphic (using this word loosely) an operation (in the normal sense) of the class of all functions $f:\; S \mapsto \{-1,0,1\}$. For example, if $f(a) =m$, then the set $f$ has $m$ copies of element $a$.

Call $$\mathcal{L} := \big\{ f: S \mapsto \{-1,0,1\} \big\}$$

Then union becomes "addition and truncate to 1". I.e.,

$$\cup : \mathcal{L} \times \mathcal{L} \mapsto \mathcal{L} \\\\ (f \cup g)(a) := \begin{cases} 1,\; f(a)+g(a) \geq 1 \\\\ -1,\; f(a)+g(a) \leq -1 \\\\ 0,\; \textrm{otherwise} \end{cases},\; a \in S.$$ However, this doesn't preserve the property $f \cup (g \cup h) = (f \cup g) \cup h$.

So I have another suggestion. Allow set elements be repetitive, even negative copies. Then, $$\mathcal{L} := \big\{ f: S \mapsto \mathbb{Z} \big\}$$

Then simply define union to be addition:

$$\cup : \mathcal{L} \times \mathcal{L} \mapsto \mathcal{L} \\\\ (f \cup g)(a) := f(a)+g(a),\; a \in S.$$ I don't know how to define intersection, though. Maybe an intersection of sets is $$\cap : \mathcal{L} \times \mathcal{L} \mapsto \mathcal{L} \\\\ (f \cap g)(a) := f(a) g(a),\; a \in S.$$ So I have preserved $$f \cap g = g \cap f \\\\ f \cup g = g \cup f \\\\ (f \cup g) \cap h =(f \cap h) \cup (g \cap h)$$ Unfortunately, I did not preserve $$(f \cap g) \cup h =(f \cup h) \cap (g \cup h)$$ The traditional count of element is the number of element truncated by one. If $f*$ be traditional count of $f$, then $$f*(a) =\begin{cases} f(a),\; f(a)=-1,0,1 \\\\ 1,\; f(a) >1 \\\\ -1,\; f(a) <-1 \end{cases}$$

• +1 I was just writing up a comment to the same effect (though less thoroughly covering the idea). – Paul Sinclair Nov 29 '16 at 18:14
• Try $\min$ and $\max$ as your intersection and union operators ... it will fulfill the distributive laws, though not the union property of the question. – Paŭlo Ebermann Nov 29 '16 at 19:51
• Is it reasonable to summarize the math you did as "you can define anything you want, but it is difficult to define normal set operators such as union and intersection with 'negative elements' without breaking the existing properties of unions and intersections?" – Cort Ammon Nov 29 '16 at 23:15
• +1 Great answer shame it was later and will have trouble getting upvoted enough. We really should use a different method of answer ordering. – DRF Nov 30 '16 at 9:02
• @Aminopterin I understand completely. There has even been a debate about this on meta at some point. The issue is essentially that early (and reasonably good) answers get early points from people and later answers even if they are more detailed and better will have a hard time ever catching up since answers are ordered by points and many people don't necessarily scroll per the first or if they do give the first points too. – DRF Nov 30 '16 at 9:46

This is an interesting question.

Clearly you have no problem with a set that contains some negative numbers, like the $B$ in your question. You're asking instead about something like a set with a negative number of elements.

That's not possible with the usual meaning of "set" so to get such "sets" you would have to invent an extension of that meaning. That kind of invention in mathematics has happened before. The numbers $1, 2, \ldots$ are good for counting. That system was usefully extended to include $0$ and the negative integers, even though it's a stretch to think of those as good for counting.

As for sets: yes the empty set turns out to be very useful. You'd want to invent "negative sets" if you stumbled across a problem where the idea would help you to a solution. The problem you started with doesn't seem to need such an invention, since ordinary sets containing negative numbers do the job when combined with a new operation on sets that creates a new set by forming a union and "cancelling" when a number and its negative match.

• wait, negative numbers and set union don't work the way he described. as has been pointed out $\{1,2,3\}\cup \{-3\}=\{1,2,3,-3\}\neq\{1,2\}$ – Gabriel Burns Nov 29 '16 at 16:01
• @GabrielBurns Right. Edited thanks. – Ethan Bolker Nov 29 '16 at 16:13

There is a generalization of multisets that may be what you want, which I shall call countsets but is actually nothing more than a set of pairs such that the first member of the pairs is unique and the second member of each pair is a non-zero integer. The reason I describe it this way instead of as a function from the universe (whole domain of discourse) to the integers is that in ZFC there is no function because every function has domain being a set, and there is no universal set. $\def\none{\varnothing} \def\wi{\subseteq} \def\zz{\mathbb{Z}}$

More precisely, a set $S$ is a countset iff ( $S \wi U \times \zz_{\ne 0}$ for some set $U$ and ( for each $x \in U$ there is a unique $y \in \zz_{\ne 0}$ such that $(x,y) \in S$ ) ).

For convenience let $base(S) = \{ x : (x,y) \in S \}$ and $count(S,x) = \cases{ 0 & if$x \notin base(S)$\\ y & if$(x,y) \in S$for some$y$}$.

We can then define operations on countsets $S,T$ as follows:

• $S \cap T = \{ (x,m) : x \in base(S) \cup base(T) \land m = \min(count(S,x),count(T,x)) \land m \ne 0 \}$.

• $S \cup T = \{ (x,m) : x \in base(S) \cup base(T) \land m = \max(count(S,x),count(T,x)) \land m \ne 0 \}$.

• $S + T = \{ (x,m) : x \in base(S) \cup base(T) \land m = count(S,x)+count(T,x) \land m \ne 0 \}$.

• $S - T = \{ (x,m) : x \in base(S) \cup base(T) \land m = count(S,x)-count(T,x) \land m \ne 0 \}$.

• $-S = \none - S$.

These satisfy the properties:

• Normal sets embed into countsets via $S \mapsto \{ (x,1) : x \in S \}$. [So countsets extend normal sets.]

• $\cap,\cup,+$ are commutative and associative on countsets.

• $\cap,\cup$ are idempotent on countsets (when applied to identical countsets gives the same result).

• $\none$ is the identity for $+$ on countsets; namely $S + \none = S$ for any countset $S$.

• $-$ is the inverse operation for $+$; namely $S + (-S) = \none$ for any countset $S$.

• Countsets obey extensionality; namely $S = T$ iff $count(S,x) = count(T,x)$ for every $x \in base(S) \cup base(T)$. [The restriction of non-zero count is precisely to achieve this property, because it ensures that there is a unique representation of each countset.]

In particular we get what you want because:

$\{(1,1),(2,1),(3,1)\} + \{(3,-1)\} = \{(1,1),(2,1)\}$.

In short, we can use non-zero integer tags for each item, instead of thinking of them as negative items, and all this can be defined and manipulated in any reasonable set theory.

• Note that $S \cup T \ne S + T$ in general. One cannot mix the notions of maximum and counting without severely breaking all nice properties. – user21820 Nov 30 '16 at 3:19

I think the best way to think of this is to first define the concept of a negative set as follows: For any set $B$ we define the negative set $-B$, which supports the operations of union with any set $A\supseteq B$ (defined as $A\cup -B=A\setminus B$) or any other negative set (defined $-A\cup -B=-(A\cup B)$), and intersection with negative sets (defined as $-A\cap -B=-(A\cap B)$).

Once we have this definition, we can extend it to include the idea of "mixed sets" which we define as $[A\cup -B]$ for two disjoint sets $A$ and $B$. Then union of two mixed sets can be defined as $[A\cup -B] \cup [C\cup -D]= [((A\cup C)\setminus (B\cup D))\cup -((B\cup D)\setminus(A \cup C))]$ with intersection and complementation defined similarly.

Sets with negative elements thus correspond to ordered pairs of disjoint sets. Extending the notion of sets in this way is to me reminiscent of extending real numbers to complex numbers, as these new sets are a combination of a standard set and a new sort of object derived from a second standard set, just as a complex number is a real number combined with an imaginary number (which is derived from another real number through multiplication by $i$).

• It may be worth noting that "union" will not be associative. – quid Nov 29 '16 at 16:17
• Wouldn't you rather have $A \cup -B = (A\setminus B)\cup(-(B \setminus A))$ ? Oh sorry, you said $B$ is a subset of $A$. – Joce Nov 29 '16 at 16:22
• @Joce, yes, as explained in the next paragraph. However, in the first paragraph, we are only defining negative sets, not mixed sets, so $A\cup -B$ is undefined if $B\nsubseteq A$. The next paragraph defines it (just as you describe) creating the new class of "mixed sets" – Gabriel Burns Nov 29 '16 at 16:27
• $A=\{1,2\}\cup-\{3\}$, $B=\{3\}$, $C=\{3\}$. $A\cup B=\{1,2\}$, thus $(A\cup B)\cup C=\{1,2,3\}$. But, since $B \cup C=B$, $A\cup (B \cup C)=\{1,2\}$. – Joce Nov 29 '16 at 16:36
• @quid True. It seems like for any element $x$ and any family of mixed sets $\mathcal{C}$ with $\Gamma=[A_{\Gamma}\cup -B_{\Gamma}]=\bigcup \mathcal{C}$, we would want $x\in A_{\Gamma}$ iff $|{C\in \mathcal{C}: x\in A_C}|\gt|{C\in \mathcal{C}: x\in A_C}|$ and $x\in B_{\Gamma}$ iff $|{C\in \mathcal{C}: x\in B_C}|\gt|{C\in \mathcal{C}: x\in A_C}|$, but unfortunately there is no way to define binary union to ensure this. – Gabriel Burns Nov 29 '16 at 16:43

The '$-$' sign you define introduces a 'count' for any element, which can be present positively, absent, or present negatively. However, the operations that you would then use are the arithmetic operations plus and minus, which need the count to be over the set of integers, not just $\{-1,0,1\}$ : this is the multiplicity idea of the multisets.

In order to have such a concept, you first need to have a set of operations operating on $\{-1,0,1\}$ as the AND/OR operators are on $\{0,1\}$, or as plus and minus are on integers.

If you seek a generalization of multisets where elements can have negative membership count (signed multisets) then this has been investigated many times in the past. One place to start learning about the history is Wayne D. Blizard's paper Negative Membership, which cites prior work by T. Hailperin, M. Kline, E. Fischbein, H, Whitney, R. Rado, M. P. Schutzenberger, S. Eilenberg, R Feynman and others.

See also Blizard's The Development of Multiset Theory and also this prior question.

protected by Community♦Jan 11 '17 at 16:52

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?