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The idea of this question is if we can go in the direction opposite to that of Cantor's argments for sets being strictly smaller than their powers, through building two disjoint hierarchies on top of two distinct empty objects, then we add an axiom saying that for every object in one hierarchy there is an object in the other hierarchy with an isomorphic set picture (i.e. the $\in$ graph on its inclusive transitive closure), the question is:

Can we add a bijective function $J$ over some infinite stage of the first hierarchy to the power set of its $\in$-copy in the second hierarchy, and axiomatize that $J$ can be used in replacement and separation axioms of both hierarchies from parameters coming from the the union of both hierarchies with all quantifiers relativized to that union?

This daunting question supplies the superficial appearance of in some sense circumventing Cantor's diagonal argument, albeit under different set constructions! $J$ basically is a bijection between a set and the power set of its $\in$-copy. To cut a possible paradox, it is hoped that the restriction of parameters and quantification to the union of the two hierarchies will prevent using crossover relations and functions that link elements of one hierarchy to the other. The only permitted crossover function is $J$. Without another crossover relation or function, it appears difficult to prove a paradox. However, the question is about if this can work?

Formal workup

Language: $\mathsf {FOL} (=, \in, \emptyset, \varnothing, J)$

$\emptyset, \varnothing$ are constants; $J$ is a partial function symbol.

Axioms: those of ID theory +

  1. Weak Extensionality: $\forall x (x \in a \iff x \in b) \land \exists x \, (x \in a) \to a=b$

Define: $set(y) \iff \exists s (y \in s)$

  1. Comprehension: $\exists x \forall y (y \in x \iff \phi(y) \land set(y))$, if $\phi$ doesn't have $x$ free.

  2. Emptyness: $\forall x: \not \exists y \, (y \in x) \iff x=\emptyset \lor x= \varnothing$

  3. Sets existence: $set (\emptyset), set(\varnothing), \emptyset \neq \varnothing$

Define: $x \text{ is } \ _\emptyset set \iff set(x) \land \forall y \in TC(x) [\emptyset \in TC(\{y\})] \land \varnothing \not \in TC(\{x\})$

Define: $x \text{ is} \ _\varnothing set \iff set(x) \land \forall y \in TC(x) [\varnothing \in TC(\{y\})] \land \emptyset \not \in TC(\{x\})$

Where $TC$ stands for 'transitive closure' defined in the usual manner.

Define: $\ _\emptyset V = \{x: x \text{ is } \ _\emptyset set\}$

Define: $\ _\varnothing V = \{x: x \text{ is } \ _\varnothing set\}$

  1. Pairing: $ a,b \text { are } \ _\emptyset sets \to set ( \{a,b\} ) \\ a,b \text { are } \ _\varnothing sets \to set ( \{a,b\} )$

  2. Union: $set(x) \to set(\bigcup x)$

  3. Power: $ \ _\emptyset set(x) \to set(\ _\emptyset \mathcal P(x)) \\ \ _\varnothing set(x) \to set(\ _\varnothing \mathcal P(x)) $

Where: $\ _\emptyset \mathcal P (x) = \{set(y): \forall z \in y (z \in x) \land y \neq \varnothing \} ; \\\ _\varnothing \mathcal P (x) = \{set(y): \forall z \in y (z \in x) \land y \neq \emptyset \} $

  1. Infinity: $set (_\emptyset \omega);set (_\varnothing \omega) $

  2. Replacement: if $f$ is a definable function by a formula with all quantifiers relativized to $ \ _\emptyset V \ \cup \ _\varnothing V$ , with parameters from $\vec{p},a$, then: $$\forall \vec{p},a \in \ _\emptyset V \ \cup \ _\varnothing V :\\ \forall x \in a (f(x) \in \ _\emptyset V) \lor \forall x \in a (f(x) \in \ _\varnothing V) \\\to set(\{f(x): x \in a\})$$

  3. Copying: $\forall x \in \ _\emptyset V \exists y \in \ _\varnothing V \exists f \ ( f: picture(x) \approx picture(y))$

Where the picture of a set $x$ is the $\in$-graph on the transitive closure of $\{x\}$; $`` \approx "$ stands for "isomorphism" on membership.

The copying relation shall be denoted using $``^*"$, so for every $x \in \ _\emptyset V$, we define $x^*$ as the element of $\ _\varnothing V$ whereby an isomorphism $f$ exists from the set picture of $x$ to the set picture of $x^*$. We call $x^*$ as the $\in$-copy of $x$.

Define recursively: $$ \ _\emptyset V_\emptyset = \emptyset \\ \ _\emptyset V_{\ _\emptyset \alpha + 1} = \ _\emptyset\mathcal P (V _{_\emptyset \alpha}) \\ \ _\emptyset V_{\ _\emptyset \lambda} = \bigcup_{\alpha \in \ _\emptyset \lambda} \ _\emptyset V_\alpha $$, for limit $_\emptyset \lambda$

Similarly define $\ _\varnothing V_{\ _\varnothing \alpha}$ by replacing $\emptyset$ by $\varnothing$ in the above definitions.

  1. Foundation: $$\ _\emptyset V = \bigcup_{\alpha \in \ _\emptyset \mathsf {ORD}} \ _\emptyset V_\alpha \\ \ _\varnothing V = \bigcup_{\alpha \in \, _\varnothing \mathsf {ORD}} \ _\varnothing V_\alpha $$

Where $\alpha \in \ _k \mathsf{ORD}$ iff $\alpha$ is a transitive $ _k set$ of transitive sets, having every nonempty subclass of it having a least element with respect to $\in$; where $k \in \{\emptyset, \varnothing\}$

  1. Anti-Cantorian: $$\exists \text{ limit } \ _\emptyset \alpha \, ( J: \ _\emptyset V_{\ _\emptyset \alpha} \rightarrow \ _\varnothing \mathcal P ((\ _\emptyset V_{\ _\emptyset \alpha} )^*),\\ J \text{ is a bijection })$$

So the question is whether this theory is consistent relative to some extension of $\sf ZF$?

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  • $\begingroup$ I have deleted my "answer" as it was incorrect because quantifiers were not restricted as required in the replacement scheme. $\endgroup$ Sep 6 at 18:18
  • $\begingroup$ In the 2 lines after 3., do you want 𝑇𝐶(𝑦) replaced bye 𝑇𝐶({𝑦})? $\endgroup$ Sep 6 at 19:01
  • $\begingroup$ In 6., does ∀t(tεy-->tεx) imply y is in the power set of x? $\endgroup$ Sep 6 at 19:30
  • $\begingroup$ the definitions after 3 seems OK, no need to change them. About the powers those are defined below so the $\varnothing$-power is the usual power with taking $\emptyset$ out, and the converese for the other power $\endgroup$ Sep 7 at 15:41
  • $\begingroup$ With regard to 6., I want to know if the power set of x includes all y with the property that ∀t(tεy-->tεx). An answer of yes or no would be helpful. $\endgroup$ Sep 7 at 17:47
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This theory is inconsistent.

Let A(x) hold if x is the first empty set or the first empty set is the only empty set in the transitive closure of {x}. Let B(x) hold if x is the second empty set or the second empty set is the only empty set in the transitive closure of {x}.

Let Aβ be the be the set of those elements x such that A(x) and the rank of x is less than β.

(a)Suppose s is transitive, t is transitive, A(s), and A(t). Suppose u is transitive, v is transitive, B(u), and B(v). If f is an isomorphism from s to u and g is an isomorphism from t to v,

then fx=gx for all xεs∩t.

Proof: Suppose not and let yεs∩t for which such f and g exist with fy not equal to gy. By 2., there is a Y such that xεY<-->(xεTC({y}) and such f and g exist for y). By 6.,Since Y is contained in s,

   set(Y). Let x be an element of Y of least rank. Then fx=f"x=g"x=gx.

(b)Suppose s is a transitive set and A(s). Then there is a unique transitive set y such that (B(y) and there is an isomorphism from sU{s} to yU{y}), and any 2 such isomorphisms are the same.

Proof: By (9.), there is a y such that B(y) and there is an isomorphism f from su{s} to the transitive closure of {y}. Then fs=y and y is transitive. The uniqueness follows from (a).

Let 𝛼 be the ordinal whose existence is guaranteed by 11.

Suppose that t is the unique transitive set such that there is an isomorphism f from A𝛼U{A𝛼} to tU{t}. Suppose that T is the unique transitive set such that there is an isomorphism g from

A(𝛼+1)U{A(𝛼+1)} to TU{T}. Then T={x|x⊆t and B(x)}.

Proof:Suppose x⊆t and B(x) By 2., there is a z be such cεz iff (the unique isomorphism from A𝛼U{A𝛼} to tU{t}, sends c to an element of x). set(z), since z is contined in A(𝛼)U{A(𝛼)}. Then gz=x and

thus xεT. Now suppose xεT. Then x=gz for some zεA(𝛼+1). Then gz=g"z=f"z⊆t.

By 2., there is an I consisting of those (x,y)ε(A𝛼XA(𝛼+1)) such that there is a z with Jx=z and z is the image of y under the isomorphism from A(𝛼+1)U{A(𝛼+1)} to {x|x⊆t and B(x)}U{{x|x⊆t and B(x)}}.

Since I is contained in A𝛼XA(𝛼+1), set(I). But I is a function from A𝛼 onto A(𝛼+1).

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  • $\begingroup$ I understand now your question about powers. I see that the definition of powers must stipulate its members to be sets beforehand and collect them, so I'll amend the definition of powers. Thanks $\endgroup$ Sep 10 at 9:05
  • $\begingroup$ I just wanted to say that in the amended definition (which is actually the intended one, anyhow) of powers, you cannot prove that I is a set even though its contained in a set $\endgroup$ Sep 10 at 9:15
  • $\begingroup$ One the other hand if the you find the amended version consistent and found a model of it, then this would be a very great result. $\endgroup$ Sep 10 at 13:16
  • $\begingroup$ So I just wanted to emphasize, that the question as it stands now, is still unsolved, the inconsistency featured in this answer is against the older improper way of presenting powers on my side, so it pertains to that older version, and not to the one standing as of now. $\endgroup$ Sep 11 at 18:35
  • $\begingroup$ I've just wanted to point out that the above answer is excessive, indeed if every subclass of a set is a set (which is not the case here), then of course we'll easily reach into an inconsistency, simply define by 2. the class of all ordered pairs $(x,y)$ where $x \in A_\alpha$ and $y \in A_{\alpha+1}$ where there exists $z$ such that $z=y^*$ for $z=J(x)$, this is contained in $A_\alpha \times A_{\alpha+1}$, its a bijection from $A_\alpha$ to $A_{\alpha+1}$, BUT it is not a set! $\endgroup$ Sep 24 at 10:37

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