In ZFC set theory, we specify a collection of sentences within a first-order language to count as our axioms. The models of this collection of sentences are the set-theoretic universes (and many of these models are “unintended” – pesky perversions in which the set of naturals ω is uncountably large, as one example – but we’ll put this aside for this post). Most of the axioms act as constraints that all sets must follow. For instance, the axiom of pairing says that “For any sets x and y, there must exist another set containing as elements just x and y and nothing else”. This is an axiom that begins with universal quantification over the sets of the universe, and then states some requirement that must hold of all these sets.
The anti-set program is what you get when you take each of these restriction axioms, and negate the restriction it imposes on all sets. So, for instance, the axiom of anti-pairing says that for any sets x and y, there must NOT exist any set {x, y}. Contrast this with the simple negation of pairing, which would tell us only that there exist two sets x and y such that their pair doesn’t exist. The anti-axiom is much stronger than the negated axiom, in that it requires NO pairs to exist.
Not all axioms begin with universal quantifiers, in particular the axiom of infinity, which simply asserts that a set exists that satisfies a certain property. To form the axiom of anti-infinity, we simply negate the original axiom (so that no sets with that property exist).
As it turns out, the anti-set program, if applied to ALL the axioms of ZFC, ends in disaster and paradox. In particular, a contradiction can be derived from anti-comprehension, from anti-replacement, and from anti-extensionality. We don’t handle these cases the same way. Anti-comprehension and anti-replacement are simply discarded, being too difficult to patch. By contrast, anti-extensionality is replaced by ordinary extensionality. What’s up with that? The philosophical justification is simply that extensionality, being the most a priori of the bunch, is needed to justify us calling the objects in our universe sets at all.
There’s one last consideration we must address, which regards the axiom of anti-choice. I am currently uncertain as to whether adding this axiom makes the theory inconsistent. One thing that is currently known about the axiom of anti-choice is that with its addition, out go all finite models (there’s a really pretty proof of this that I won’t include here). In the rest of this post, I will be excluding anti-choice from the axioms and only exploring models of anti-ZF.
With that background behind us, let me list the axioms of anti-ZF.
Anti-Pairing
∀x∀y∀z∃w (w ∈ z ∧ w ≠ x ∧ w ≠ y)
No set is the pair of two others.
Anti-Union
∀x∀y∃z (z ∈ y ∧ ¬∃w (z ∈ w ∧ w ∈ x))
No set is the union of another.
Anti-Powerset
∀x∀y∃z (z ∈ y ∧ ∃w (w ∈ z ∧ w ∉ x))
No set contains all existing subsets of another.
Anti-Foundation
∀x∀y (y ∈ x → ∃z (z ∈ x ∧ z ∈ y))
Every set’s members have at least one element in common with it.
Anti-Infinity
∀x∃y ((y is empty ∧ y ∉ x) ∨ (y ∈ x ∧ ∃z (z = S(y) ∧ z ∉ x)))
Every set either doesn’t contain all empty sets, or has an element whose successor is outside the set.
If the axioms of ZF are considered to be a maximally nice setting for mathematics, then perhaps the axioms of anti-ZF can be considered to be maximally bad for mathematics.
We need to now address some issues involving the axiom of anti-infinity. First, the abbreviations used: “y is empty” is shorthand for “∀z (z ∉ y)” and “z = S(y)” is shorthand for “∀w (w ∈ z ↔ (w ∈ y ∨ w = y))” (i.e. z = y ⋃ {y}).
Second, it turns out that from the other axioms we can prove that there are no empty sets. So the first part of the disjunction is always false, meaning that the second part must always be true. Thus we can simplify the axiom of anti-infinity to the following statement, which is logically equivalent in the context of the other axioms:
Anti-Infinity
∀x∃y (y ∈ x ∧ ∃z (z = S(y) ∧ z ∉ x))
Every set has an element whose successor is outside the set.
Let’s now prove some elementary consequences of these axioms.
Theorems
>> There are no empty sets.
Anti-Union ⊢ ¬∃x∀y (y ∉ x)
Suppose ∃x∀y (y ∉ x). Call this set ∅. So ∀y (y ∉ ∅) Then ⋃∅ = ∅. But this implies that the union of ∅ exists. This contradicts anti-union.
>> There are no one-element sets.
Anti-Pairing ⊢ ¬∃x∃y∀z (z ∈ x ↔ z = y)
Suppose that ∃x∃y∀z (z ∈ x ↔ z = y). Then ∃x∃y∀z (z ∈ x ↔ (z = y ∨ z = y)). But this is a violation of anti-pairing, as then x would be the pair of y and y. Contradiction.
>> There are no two-element sets.
Anti-Pairing ⊢ ¬∃x∃y∃z∀w (w ∈ x ↔ (w = y ∨ w = z))
Suppose that ∃x∃y∃z∀w (w ∈ x ↔ (w = y ∨ w = z)). Then w is the pair of y and z, so anti-pairing is violated. Contradiction.
>> No models of anti-ZF have just N-element sets (for any finite N).
Suppose that a model of anti-ZF had only N-element sets. Take any of these sets and call it X. By anti-infinity, X must contain an element with a successor that is outside the set. Call this element Y and its successor S(Y). Y cannot be its own successor, as then its successor would be inside the set. This means that Y ∉ Y. Also, Y is an N-element set by assumption. But since Y ≠ S(Y), S(Y) must contain all the elements of Y in addition to Y itself. So S(Y) contains N+1 elements. Contradiction.
>> There is no set of all sets.
Suppose that there is such a set, and call it X. By Anti-Infinity, X must contain an element Y with a successor that’s outside of X. But no set is outside of X, by assumption! Contradiction.
>> No N-element model of anti-ZF can have N-1 sets that each contain N-1 elements.
Suppose that this were true. Consider any of these sets that contain N-1 elements, and call it X. By the same argument made two above, this set must contain an element whose successor contains N elements. But there are only N sets in the model, so this is a set of all sets! But we already know that no such set can exist. Contradiction.
>> Every finite set of disjoint sets must be its own choice function.
By a choice function for a set X, I mean a set that contains exactly one element in common with each element of X. Suppose that X is a finite set of disjoint sets. Let’s give the elements of X names: A1, A2, A3, …, AN. By anti-foundation, each An must contain at least one element of X. Since the Ans are disjoint, these elements cannot be the same. Also, since there are only N elements of X, no An can contain more than one element of X. So each element of X contains exactly one element of X. Thus X is a choice function for X!
The next results come from a program I wrote that finds models of anti-ZF of a given size.
>> There are no models of size one, two, three or four.
>> There are exactly two non-isomorphic models of size five.
Here are pictures of the two:
Now, some conjectures! I’m pretty sure of each of these, especially Conjecture 2, but haven’t been able to prove them.
Conjecture 1: There’s always a set that contains itself.
Conjecture 2: There can be no sets of disjoint sets.
Conjecture 3: In an N-element model, there are never less than three sets with fewer than N-1 elements.
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