Sets and supersets

You’re gonna need a bigger boat.

(Jaws)

Abstract

It is a commonplace of set theory to say that there is no set of all well-orderings nor a set of all sets. We are implored to accept this due to the threat of paradox and the ensuing descent into unintelligibility. In the absence of promising alternatives, we tend to take up a conservative stance and tow the line: there is no universe (Halmos, in: Naive set theory, 1960). In this paper, I am going to challenge this claim by taking seriously the idea that we can talk about the collection of all the sets and many more collections beyond that. A method of articulating this idea is offered through an indefinitely extending hierarchy of set theories. It is argued that this approach provides a natural extension to ordinary set theory and leaves ordinary mathematical practice untouched.

Appendix: axiomatising superset theories

In this appendix, we state axiomatisations and some basic results for SETS $+$ SETS and $SETS\,\times \, SETS$.

1.1 An axiomatisation of SETS $+$ SETS

Recall that we want to build one notion of membership $\eta$ on top of another $\in$. As we’ve discussed earlier, it is probably easier to visualise what we are doing by thinking of $\in$ as being about a lower universe. Moreover, we could have introduced a constant symbol U, in the manner of Feferman and Kreisel (1969) or Bourbaki (1972) and produced an axiomatisation from there. This would have worked and, indeed, we’ll exploit this relationship below; however, we opted against this in our official axiomatisation in order to emphasise that the ambiguity of supersets is located in notion of membership itself.

Let ${\mathscr {L}}_{\in ,\eta }=\{\in ,\eta \}$.

Definition 2

Let us say that x is an $\in$ -set if

$$\begin{aligned} \exists y\ x\in y. \end{aligned}$$

Otherwise, we shall say that x is a proper $\eta$ -class.

Remark 3

The idea here is that x is an $\in$-set if it can be, so to speak, covered using the $\in$ relation. We’ll place further constraints on $\in$-sets in the axioms below.

Write $\forall ^{\in }x\ \varphi (x)$ in place of $\forall x(x \text{ is } \text{ an } \in \text{-set }\rightarrow \varphi (x))$.

Let $ZFC(\in )$ be the result of writing out the axioms of ZFC and using $\forall ^{\in }$ instead of $\forall$.

Let $ZFC(\eta )$ be the result of writing out the axiom of ZFC with $\eta$ replacing $\in$. If t is any (defined) term of ordinary set theory, let $t^{\eta }$ be the result of defining it with $\eta$ instead of $\in$.

Let our theory $\Gamma _{SETS\,+\,SETS}$ consist of the following axioms:

• (1) $ZFC(\in )$.

• (2) $ZFC(\eta )$.

• (3) $\forall x\forall y(x\in y\rightarrow x\eta y)$ (Cumulativity).

• (4) $\forall x\forall y(x\in y\ \wedge \ x \text{ is } \text{ an } \in \text{-set }\rightarrow y \text{ is } \text{ an } \in \text{-set })$ (End-extension). ⁵¹

• (5) $\forall x\forall y(rank^{\eta }(x)\le ^{\eta }rank{}^{\eta }(y)\wedge y \text{ is } \text{ an } \in \text{-set }\ \rightarrow \ x \text{ is } \text{ an } \in \text{-set })$ (Top extension).⁵²

• (6) $\exists x(\forall ^{\in }y\ y\eta x)$ (Closure).

Lemma 4

(i) There is an r such that for all $\in$-sets y

$$\begin{aligned} y\eta r\ \leftrightarrow \ y\notin y. \end{aligned}$$

Proof

By (6), let u, be such that $\forall ^{\in }y\ y\eta u$. Using $\eta$-separation, we see that there is some r such that⁵³

$$\begin{aligned} \forall y(y\eta r\ \leftrightarrow \ y\eta u\wedge \lnot y\eta y). \end{aligned}$$

$\square$

Theorem 5

Suppose there are two inaccessible cardinals $\kappa _{1}<\kappa _{2}$. Let ${\mathscr {M}}=\langle V_{\kappa _{2}},\in \upharpoonright (V_{\kappa _{1}}\times V_{\kappa _{1}}),\in \upharpoonright (V_{\kappa _{2}}\times V_{\kappa _{2}})\rangle$. Then ${\mathscr {M}}$ is a natural model and

$$\begin{aligned} {\mathscr {M}}\models \Gamma _{SETS\,+\,SETS.} \end{aligned}$$

In fact, if we forgo the restriction to natural models, consistency may be established more easily via an adaptation of a trick from Feferman and Cohen (1966, 1969).

Theorem 6

$Con(ZFC)\rightarrow Con(\Gamma _{SETS\,+\,SETS})$.

Proof

To prove this we first observe that $\Gamma _{SETS\,+\,SETS}$ is mutually interpretable with the theory, $\Delta$, articulated in the a language ${\mathscr {L}}=\{\in ,U\}$ which admits a constant for universes, and consisting of the following axioms:

• (1) ZFC;

• (2) ZFC(U) - where we restrict all of the quantifiers in axioms to U; and

• (3) $\exists \alpha \ U=V_{\alpha }$.

Thus, it suffices to show that $Con(ZFC)\rightarrow Con(\Delta )$. Suppose not, then Con(ZFC) and for some finite $\Lambda \subseteq \Delta$, $\Lambda$ is unsatisfiable. Let $\Lambda =\Lambda _{0}\cup \Lambda _{1}\cup \{\exists \alpha \ U=V_{\alpha }\}$ where $\Lambda _{0}\subseteq ZFC$ and $\Lambda _{1}\subseteq ZFC(U)$. Since Con(ZFC) we may fix some ${\mathscr {M}}\models ZFC$. Clearly, ${\mathscr {M}}\not \models \Lambda$.

Let $\Lambda _{1}^{\dagger }$ be the result of substituting ordinary quantifiers for the restricted quantifiers in $\Lambda _{1}$. Then, using reflection we know that $ZFC\vdash ``\exists \alpha \ V_{\alpha }\models \Lambda _{1}^{\dagger }$”. Thus, it can be easily seen that ${\mathscr {M}}\models \Lambda$: contradiction. $\square$

1.2 An axiomatisation of $SETS\,\times \, SETS$

Recall that we are trying to stack indefinitely many membership relations on top of each other and that we are indexing our membership relation to do this.

Let ${\mathscr {L}}=\{\in \}$ where $\in$ is now a three-place predicate, where the third argument is written as a subscript on $\in$.

Definition 7

• (i) For all x, y

  $$\begin{aligned} x\ll y\ \leftrightarrow \ \exists z\ x\in _{z}y. \end{aligned}$$

  This gives us a notion of universal membership: membership according to any relativisation.

• (ii) x is a super ordinal, abbreviated SupOn(y), if there is some y such that x is $\in _{y}$-transitive and an $\in _{y}$-linear order. We are going to use the super ordinals to index the membership relation.

• (iii) If x is a super ordinal, we say that y is an $\in _{x}$ -set if

  $$\begin{aligned} \exists z\ y\in _{x}z. \end{aligned}$$

The idea here is that y is an $\in _{x}$-set if it can, so to speak, be covered using the $\in _{x}$ relation.

Let us write $\forall ^{y}z\,\varphi (z)$ for

$$\begin{aligned} \forall z(z \text{ is } \text{ a } \in _{y}\text{-set } \rightarrow \varphi (z)). \end{aligned}$$

Let $\forall y(SupOn(y)\rightarrow ZFC(\in _{y}))$ be the result taking each axiom of ZFC:

• adding the subscript $_{y}$ to each $\in$;

• changing quantifiers to $\forall ^{y}$; and

• substituting that into the space in $\forall y(SupOn(y)\rightarrow$...).

Remark

We should note that $\forall y(SupOn(y)\rightarrow ZFC(\in _{y}))$ is not a single sentence but an infinite collection of sentences.

Let $ZFC(\ll )$ be the result of replacing $\in$ by $\ll$ in every axiom of ZFC. Let $\Gamma _{SETS\,\times \, SETS}$ comprise of the following axioms:

• (1) If $x\in _{z}y$, then z is a super-ordinal.

• (2) There is some x such that for all y and for all z, $x\notin _{y}z$, which we denote $\emptyset$.

• (3) $\emptyset$ is a super-ordinal.

• (4) $\forall y(SupOn(y)\rightarrow ZFC(\in _{y})$ for all y.

• (5) $ZFC(\ll )$.

• (6) If x and y are super-ordinals, and $x\ll y$, then $\forall u\forall w(u\in _{x}w\rightarrow u\in _{y}w)$ (Cumulativity).

• (7) If $rank^{\ll }(x)\,\le ^{\ll }\, rank^{\ll }(y)$ and y is a $\in _{z}$-set, then x is a $\in _{z}$-set (Top extension).

• (8) If x is a super-ordinal, then there is some super-ordinal $y\gg x$ and some u such that $\forall _{x}z\ z\in _{y}u$; moreover, for every $y\gg x$ there is such a u (Closure).

Proposition 8

• (i) Any ordinary ordinal (i.e., $\in _{\emptyset }$-ordinal) $\alpha$ is a super ordinal.

• (ii) For all super-ordinals x, there is a Russell’s set.

Theorem 9

Let $\kappa$ be an inaccessible limit of inaccessible cardinals and let $\langle \lambda _{\alpha }|\alpha <\kappa \rangle$ enumerate all the inaccessible cardinals below $\kappa$. Then

$$\begin{aligned} \langle V_{\kappa },E\rangle \models \Gamma _{SETS\,\times \, SETS} \end{aligned}$$

where

$$\begin{aligned} E=\{(x,y,\alpha )\in V_{\kappa }\times V_{\kappa }\times On^{V_{\kappa }}\ |\ \alpha \text{ is } \text{ inaccessible } \wedge x\in y\wedge rank(y)<\lambda _{\alpha }\}. \end{aligned}$$

Moreover, this is the smallest natural model of $\Gamma _{SETS\,\times \, SETS}$.

Again, if we forgo the natural models requirement, the theory is no stronger than ZFC.

Theorem 10

$Con(ZFC)\rightarrow Con(\Gamma _{SETS\,\times \, SETS})$.

Proof

As with $\Gamma _{SETS\,+\,SETS}$, we first note that $\Gamma _{SETS\,\times \, SETS}$ is mutually interpretable with a theory that it a little easier to work with. Let $\Delta$ be a theory articulated in ${\mathscr {L}}=\{\in ,u\}$ where u is a one-place function symbol. Let $\Delta$ consist of the following sentences:

• (1) ZFC;

• (2) $\forall \alpha \ \varphi ^{u(\alpha )}$ where $\varphi$ is an axiom of ZFC;

• (3) $\forall \alpha \exists \kappa \ u(\alpha )=V_{\kappa }$.

• (4) $\forall \alpha <\beta \ u(\alpha )\subsetneq u(\beta )$.

It then suffices to show that $Con(ZFC)\rightarrow Con(\Delta )$. Suppose not. Then Con(ZFC) and there is some finite $\Lambda \subseteq \Delta$ such that $\Lambda$ is unsatisfiable. Let $\Lambda =\Lambda _{0}\cup \Lambda _{1}\cup \{\forall \alpha \exists \kappa \ u(\alpha )=V_{\kappa },\forall \alpha <\beta \ u(\alpha )\subsetneq u(\beta )\}$ where $\Lambda _{0}\subseteq ZFC$ and $\Lambda \subseteq \{\forall \alpha \varphi ^{u(\alpha )}\ |\alpha \in On\wedge \varphi \in ZFC\}$. Using our assumption, fix some ${\mathscr {M}}$ such that ${\mathscr {M}}\models ZFC$. Then clearly ${\mathscr {M}}\not \models \Lambda$.

Let $\Lambda _{1}^{\dagger }$ be the result removing the initial $\forall \alpha$ and replacing the restricted quantifiers by ordinary quantifiers for sentences in $\Lambda _{1}$. By reflection, we know that $ZFC\vdash ``\forall \alpha \exists \beta >\alpha \ V_{\alpha }\models \Lambda _{1}^{\dagger }$”. Use this to fix the interpretation of u function. Then from here it can be shown that ${\mathscr {M}}\models \Lambda$: contradiction. $\square$