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This is an idea which I've had sitting on my desk for a while, but hadn't gotten to until now.

Suppose that there two types of set-theoretic objects, which we'll call "elements" and "antielements." For any element $a$, there is a unique antielement $b$ such that $a=b^*$ and $\{a\}\cup\{b\}=\emptyset$, and vice-versa ('$^*$' is an involutory bijection from the class of elements to the class of antielements).

The motivating principle behind this idea was that it is naturally possible to "remove" an element from the empty set if we think of it as sort of "loan." When we take an element out of the empty set we "owe" the universe that element back - similar to that old joke about how a mathematician might interpret two people walking into an empty building followed by three people walking out.

My question is this: are there any specific, non-obvious problems incurred by the addition of antielements to ZF (or NBG)?

There are of course some surface issues like "how should cardinality work with antielements?" and the definitions of "union" and "difference" have to be modified; but these aren't too daunting on their own. What I'm concerned about is any intolerable contradictions which necessarily arise from the existence of antielements - something like "if (antielements), then not (axiom of extensionality)" - which might not be immediately obvious.

I'm not the first person to suggest this idea, either. There are a few references to the concept of antielements/antisets scattered across the literature, but I haven't been able to find anything definitive (e.g. "Introduction to ZF with Antielements", etc.). If there is a fully developed set theory with antielements, I would greatly appreciate a reference.

Papers:

https://arxiv.org/pdf/0906.3120.pdf

https://arxiv.org/pdf/1701.02993.pdf

https://hal.archives-ouvertes.fr/hal-00853859/document

https://hal.archives-ouvertes.fr/hal-00946521/document

https://www.csz.com/cyber/html/negsets.pdf

Edit:

Following Noah Schweber's answer, I decided to look into the logic behind the concept of "antielements" to see where it would lead. I started by considering the declaration of a set as a generic case of comprehension for which the formula to be satisfied is "belongs to the set":

$$\forall X.X=\{x\mid x\in X\}$$

and proceeded to define the union accordingly. What I found was that if the empty set is unique, then there cannot be a non-empty set $A$ s.t. $A\cup B=\emptyset$, because it leads to the contradiction $a\in A\implies\neg(a\in A)$.

A possible way around this is to rewrite the axiom of the empty set as:

$$\exists X.\forall x.x\in X\iff x^*\in X$$

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    $\begingroup$ Since your "antielements" are defined by how they act under unions, can you state how the axiom of union would look like in your proposal? And what would the elements of $a^*$ be? If you need to modify the axiom of extensionality, that could be a pretty deep change. $\endgroup$ – hmakholm left over Monica Aug 16 '19 at 17:05
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    $\begingroup$ By the time you finesse it enough to actually work, it'll basically be group theory with different notation. That's only a problem if you're looking for a completely new area of math, of course. $\endgroup$ – Kevin Aug 16 '19 at 22:19
  • $\begingroup$ @HenningMakholm The axiom of union can be stated as it normally would. After working a bit on this, I found that the axiom of the empty set presents a bigger problem. $\endgroup$ – R. Burton Aug 19 '19 at 1:28
  • $\begingroup$ It seems to me this question asks the same, using the name "negative members". $\endgroup$ – Jozef Mikušinec Nov 13 '20 at 0:58
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Consider the following. On the one hand we have $$(\{a\}\cup\{a\})\cup\{a^*\}=\{a\}\cup\{a^*\}=\emptyset$$ but on the other hand we have $$\{a\}\cup(\{a\}\cup\{a^*\})=\{a\}\cup\emptyset=\{a\}.$$ Associativity of "$\cup$" is built into the underlying framework ($\cup$ isn't a primitive operation, it's defined in terms of $\in$, and the logical associativity of $\vee$ directly leads to the algebraic associativity of $\cup$). So this is a genuine problem.

One hope is to fix this using multisets, so that $\{a\}\cup\{a\}=\{a,a\}\not=\{a\}$. However, even this doesn't really solve things: what about infinite unions? E.g. for $z\in\mathbb{Z}$ let $x_z=\{a\}$ if $z$ is even and let $x_z=\{a^*\}$ if $z$ is odd; what is $$\bigcup_{z\in\mathbb{Z}}x_z?$$ A case can be made for it being $\{\}$, or $\{a\}$, or $\{a^*, a^*, a^*\}$, or ... by looking at the union in different ways. And while there might be a natural way to handle this specific case, how are you going to treat the general situation where the index set doesn't have any obvious ordering (or similar disambiguating mechanism)?

The only way I see to make antielements work properly is to not only pass to multisets, but also drop the axiom of union. But this is a huge loss.

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  • $\begingroup$ Note that in the axiom of union, you cannot have two instances of $\{a\}$, as in that axiom you are taking the union over the elements of a set. However, the problem persists by simply replacing the first instance by $\{a,b\}$. $\endgroup$ – celtschk Aug 16 '19 at 14:42
  • $\begingroup$ As it happens, one of the applications of antielements was simplifying the construction of the integers. If we look at $(\omega,\cup,^*)$ as a representation of $(\Bbb{N},+,-)$ with Peano arithmetic, then these are the exact same problems you encounter going from the naturals to the integers ($(a+b)+(-c)\ne a+(b+(-c))$ loss of associativity, and $\sum_{z\in\Bbb{Z}}1^z$ undefined). I wonder if this says something fundamental about the concept of negative numbers? $\endgroup$ – R. Burton Aug 16 '19 at 15:08
  • $\begingroup$ Sorry, I mean PA suitably modified to accommodate negative numbers. $\endgroup$ – R. Burton Aug 16 '19 at 15:13
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    $\begingroup$ @R.Burton Um, $((a+b)+(-c)=a+(b+(-c))$ in the integers. You don't lose associativity there. And $(\omega, \cup)$ should not be thought of as a representation of $(\mathbb{N}, +)$. $\endgroup$ – Noah Schweber Aug 16 '19 at 15:23
  • $\begingroup$ You are absolutely right. I think I might have mixed up additive inverse and subtraction, and union and disjoint union. $\endgroup$ – R. Burton Aug 16 '19 at 15:30

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