Can All Things Be Counted?

Abstract

In this paper, I present and motivate a modal set theory consistent with the idea that there is only one size of infinity.

Appendix: Consistency of M

Let

$$M_{0} \vDash V = L + \mu \text{ is the least Mahlo cardinal}$$

be a ctm, and let $M = V_{\mu }^{M_{0}} = L_{\mu }^{M_{0}}$.

Let Inac(α) abbreviate the claim that α is inaccessible.

Fact 1

For each formula φ, we have

$$M \vDash \forall \alpha \exists \beta \geq \alpha [Inac(\beta) \wedge \forall x \in V_{\beta}[\varphi(x) \leftrightarrow \varphi^{V_{\beta}}(x)]]$$

Definition 1 (Generic Multiverse)

Let the generic multiverse $\mathbb {M}$ on M be the closure of {M} under N↦N[G], G generic from some $\mathbb {P}$ in N.¹

Remark 1

Since we chose M to be a model of V = L, we have that the generic multiverse on M consists of precisely those structures of the form M[G] for G generic over M by Usuba’s ‘Downward Directed Ground’ theorem [23].

Lemma 1

Every N in $\mathbb {M}$ satisfies Fact 1.

Proof

Any forcing in M satisfies the λ chain condition for some λ < μ, and so does not destroy μ s Mahlo-ness in M₀. Hence the lemma obtains in any model in $\mathbb {M}$. □

Remark 2

We use Mahlo cardinals because we want reflection to inaccessibles in all models of the generic multiverse. A weaker analogue of this result can be had for inaccessibles (by simultaneously substituting ‘Mahlo’ for ‘inaccessible’ and ‘inaccessible’ for ‘worldly’ everywhere). But since (as Hamkins has shown) worldly cardinals are not downwards absolute in the generic multiverse, we would run into difficulties below, and a Mahlo cardinal will therefore prove much more convenient. I do expect, however, that equi-consistency with ZFC should be attainable.

Definition 2 (The model $\mathbb {K}$)

Let $\mathbb {K}$ be a Kripke model with the following components.

• W, the set of possible worlds, is the set of all pairs 〈α,N〉, where $N \in \mathbb {M}$ and α is inaccessible in N. Write w₀, w₁ for the first and second component of w ∈ W respectively.

• D, the domain function, applies to a possible world and returns $\langle V_{w_{0}}^{w_{1}}, V_{w_{0} + 1}^{w_{1}} \rangle$ as the singular and plural domain of the world respectively.

• w is vertically accessible from u, written w ≤_vu, when w₁ = u₁ and u₀ ≤ w₀.

• w is horizontally accessible from u, written w ≤_hu, when w₀ = u₀ and $\exists \mathbb {P} \in V_{u_{0}}^{u_{1}} \exists G_{\mathbb {P}}[w_{1} = u_{1}[G_{\mathbb {P}}]]$.

• w is generally accessible from u, written w ≤_gu, when u₀ ≤ w₀ and $\exists \mathbb {P} \in V_{w_{0}}^{w_{1}} \exists G_{\mathbb {P}}[w_{1} = u_{1}[G_{\mathbb {P}}]]$.

Here, the notation $G_{\mathbb {P}}$ means that G is generic for $\mathbb {P}$ (relative to the obvious ground model).

Given a world w and (singular and plural) variable assignment v we define truth at w in the standard Kripke-Tarskian way.

I will use $w \vDash \varphi$ to mean that the world w satisfies φ in a Kripke model, and use the very same symbol $\vDash$ for things like $w_{1} \vDash \varphi$ to mean that a classical first order model satisfies φ. Context will always make it clear what is meant.

Note that since the plural domain of a world 〈α,N〉 is always the rank $V_{\alpha +1}^{N}$, we immediately get that $\mathbb {K}$ satisfies impredicative comprehension for formulas in non-modal (so plural and first order) fragment of the language. This in turn means first order truth is definable and that the forcing theorem holds for formulas in the plural language. See e.g. Krapf [11, p30].²

We need a few lemmas on the interactions between 〈v〉 and 〈h〉. The first says that w is generally accessible to v iff it is a forcing extension of a height extension of v.

Lemma 2

w ≤_gv iff there is a u with u ≤_vv and w ≤_hu.

Proof

The right to left is immediate. So suppose w ≤_gv. The world u = 〈w₀,v₁〉 is vertically accessible from v, and w is horizontally accessible from u. □

Remark 3

We used the downward absoluteness of inaccessibility in the previous proof; the argument would fail were ‘inaccessibility’ replaced by ‘worldly’ in the definition of $\mathbb {K}$.

Corollary 1

$\mathbb {K} \vDash \Diamond \varphi \leftrightarrow \langle v \rangle \langle h \rangle \varphi$.

Proof

Immediate from the previous lemma. □

The second says that if something can be forced to hold in all vertical extensions, then in all vertical extensions it can always be forced.

Lemma 3

$\mathbb {K} \vDash \langle h \rangle [ v ] \varphi \rightarrow [ v ] \langle h \rangle \varphi$.

Proof

If there’s a forcing extension such that all its height extensions φ, then this forcing will be available at all vertically accessible worlds. □

Lemma 4

For every expression ψ of ${\mathscr{L}}_{\Diamond }$ there is an expression t(ψ)(α) of the pure language of set theory (with free variable α) such that $w_{1} \vDash t(\psi )(w_{0})$ if and only if $w \vDash \psi$.

Proof

We define t(ψ) recursively. Here, ρ represents the standard rank function on sets.

• t(x ∈ y) = x ∈ y

• t(x ≺ xx) = x ≺ xx

• t(φ ∧ 𝜃) = t(φ) ∧ t(𝜃)

• t(¬φ) = ¬t(φ)

• $t(\forall x \varphi ) = (\forall x t(\varphi ))^{V_{\alpha }}$

• t(〈v〉φ) = ∃β > α[Inac(β) ∧ t(φ)(β)]

• $t(\langle h \rangle \varphi (x , xx)) = \exists \mathbb {P}, p \in \mathbb {P}[\rho (\mathbb {P}) < \alpha \wedge p \Vdash t(\varphi (\check {x}, \check {xx}))]$

• t(♢φ) = t(〈v〉〈h〉φ).

Here, plural check names are defined just as standard second-order check names are; see [11] here again for more details.

We now prove the claim by induction, assuming as inductive hypothesis that for all worlds u, $u \vDash \varphi$ iff $u_{1} \vDash t(\varphi )(u_{0})$ for φ of lower complexity than ψ.

Only the modal operators are significant. So suppose first that ψ := 〈v〉φ and $w \vDash \psi$. Then there is u ≤_vw with $V^{u_{1}}_{u_{0}} \vDash \varphi$. By induction, $u_{1} \vDash t(\varphi )(u_{0})$ and hence by definition $u_{1} = w_{1} \vDash \exists \beta \geq w_{0} [Inac(\beta ) \wedge t(\varphi )(\beta ) ]$, as required. In the other direction, suppose

$$w_{1} \vDash \exists \beta \geq w_{0} [Inac(\beta) \wedge t(\varphi)(\beta)].$$

Then there is a vertically accessible world from w that satisfies t(φ)(β), and the result follows by induction.

Turning to 〈h〉, suppose first that $w \vDash \langle h \rangle \varphi (x)$. Then there is u ≤_hw with $u \vDash \varphi$. By induction, $u_{1} \vDash t(\varphi )(u_{0})$ which is precisely to say $u_{1} \vDash t(\varphi )(w_{0})$. By the forcing theorem, there is a condition p for the relevant order $\mathbb {P} \in V_{w_{0}}^{w_{1}}$ such that $w_{1}\vDash p \Vdash t(\varphi (\check {x}))(\check {w_{0}})$. Thus

$$w_{1} \vDash \exists \mathbb{P}, p \in \mathbb{P} [\rho(\mathbb{P}) < w_{0} \wedge p \Vdash t(\varphi(\check{x}))(\check{w_{0}})]$$

as required. The converse is similar to the previous.

The ♢ modality follows immediately using Corollary 2. □

Lemma 5

The modal logic of ≤_v is S4.2.

Proof

S4 is immediate from the definitions. For .2, if two distinct worlds are v-accessible from a common one, whichever has the greater ordinal rank is accessible from the other. □

Lemma 6

If $N \in \mathbb {M}$ has $N \vDash p \Vdash \Box \varphi$ (some $\mathbb {P} \in N$), and N[G] is any forcing extension of N, then $N[G] \vDash p \Vdash \Box \varphi$.

Proof

Assume that $N \vDash p \Vdash \Box \varphi$ and $N[G] \vDash \neg (p \Vdash \Box \varphi )$. Then there is $q, \mathbb {Q} \in N$ such that $N \vDash q \Vdash \text {``}\exists r \leq p[r \Vdash \Diamond \neg \varphi ]\text {''}$. But now a product forcing in the order $\mathbb {P} \times \mathbb {Q}$ on a generic containing the condition 〈r,q〉 yields a contradiction. □

Lemma 7

The modal logic of ≤_h is S4.2.

Proof

S4 is immediate from the definitions. For .2, we cannot make use of convergence of ≤_h since it is not convergent.³ Hence instead we make use of the t-translation. Suppose that $w \vDash \langle h \rangle [ h ] \varphi$. Then

$$w_{1} \vDash \exists \mathbb{P}, p \in \mathbb{P} [\rho(\mathbb{P}) < w_{0} \wedge p \Vdash [ h ] t(\varphi)(\check{w_{0}})].$$

Consider any u ≤_hw with u₁ = w₁[G], some generic G. We must show $u \vDash \langle h \rangle \varphi$. Well, since $\mathbb {P}$ and p are in the domain of u, we can construct a u₁-generic filter H for $\mathbb {P}$ containing p. By Lemma 12, we still have $p \Vdash [ h ] t(\varphi )(\check {w_{0}})$ in u₁, and so the world 〈w₀,u₁[H]〉 is an accessible world from u satisfying φ as required. □

Lemma 8

≤_g satisfies S4.2.

Proof

Suppose $\Diamond \Box \varphi$ holds at w. Then 〈v〉〈h〉[v][h]φ holds at w, and by Lemma 9 〈v〉[v]〈h〉[h]φ holds at w. Whence [v]〈v〉[h]〈h〉φ holds at w, and so finally we conclude $\Box \Diamond \varphi$ using Lemma 9 again (contraposed). □

Lemma 9

For each formula φ(x,y,yy) in the language ${\mathscr{L}}_{\Diamond }$, any world w of $\mathbb {K}$ satsifies

$$\Box \forall y \Box \forall yy \exists xx [x \prec xx \leftrightarrow \varphi(x, y, yy)]$$

and similarly for any other number of parameters.

Proof

We already observed that every world satisfies impredicative comprehension for the pure second order fragment. So, let w be any world containing y, yy. Then the set of x with t(φ(x,y,yy)) is a member of the plural domain of any w (relative to any valuation v on w). The validity of the t-translation then implies the result. □

Lemma 10

$\mathbb {K} \vDash$ VE_v.

Proof

Any xx that don’t form a set at w form a set in the world given by the next inaccessible after w₀ in w₁. □

Lemma 11

$\mathbb {K} \vDash$HE.

Proof

Given a partial order and some dense sets in it in a world w, we know we can find a generic G that intersects all of them. The structure $V_{w_{0}}^{w_{1}[G]}$ is then a horizontally accessible world with the required features. □

Lemma 12

$\mathbb {K} \vDash$Comp($\subseteq$)_v

Proof

Given any x at a possible world, the set of all its subsets exists at the same world (since all of them are inaccessible ranks); and vertical expansion never adds further subsets. □

Lemma 13

Every instance of the 〈v〉 translation of replacement holds in $\mathbb {K}$.

Proof

For the purposes of this proof, let ♢ stand for 〈v〉. Then suppose:

$$w \vDash \forall x \in a \Diamond \exists y \varphi^{\Diamond}(x, y)$$

For each x ∈ a, we can therefore find a world w(x) with

$$w(x) \vDash \exists y \varphi^{\Diamond}(x,y)$$

by reflection to inaccessibles, we can find an inaccessible rank κ beyond all the w(x)₀ such that V_κ of w₁ reflects the situation for t(φ^♢(x,y)) and hence for φ^♢(x,y) in w₁. The world u = 〈κ,w₁〉 will then have

$$u \vDash \forall x \in a \exists y \varphi^{\Diamond}(x,y)$$

so, by replacement in u and Lemma 6

$$u \vDash \exists b \Box \forall x \in a \Diamond \exists y \in b \varphi^{\Diamond}(x,y)$$

as required. □

Lemma 14

Every instance of the 〈h〉-translation of the replacement scheme holds in $\mathbb {K}$.

Proof

For the purposes of this proof, let ♢ stand for 〈h〉; and let the clause for 〈h〉 in the t-translation temporarily be replaced by

• $t(\langle h \rangle \varphi (x , xx)) = \exists \mathbb {P}, p \in \mathbb {P}[ p \Vdash t(\varphi (\check {x}, \check {xx}))]$

so now there is no need for reference to an ordinal parameter and in fact we have

$$w \vDash t(\varphi^{\Diamond})\text{ iff } w\vDash \varphi^{\Diamond}$$

since the only relevant modality is ♢ = 〈h〉.

Suppose

$$w \vDash \forall x \in a \Diamond \exists y \varphi^{\Diamond}(x,y).$$

Then

$$w \vDash \forall x \in a \exists \mathbb{P}, p \in \mathbb{P}[p \Vdash t(\exists y\varphi^{\Diamond}(\check{x}, y))].$$

For each x ∈ a let $\mathbb {P}_{x}, p_{x}$ be a witnessing order and condition for which this holds. By replacement in w, we may form the products $\mathbb {Q} = \varPi _{x \in a} \mathbb {P}_{x}$ and q =π _x∈ap_x. Let G be generic for $\mathbb {Q}$ containing q. Then if u = 〈w₀,w₁[G]〉 it is easy to see that $u \vDash t(\forall x \in a\exists y \varphi (x,y))$. By replacement in u,

$$u \vDash \exists b \forall x \in a \exists y \in b[t(\varphi^{\Diamond}(x,y))].$$

Using Lemma 6, we get

$$u \vDash \exists b \Box \forall x \in a \Diamond \exists y \in b[\varphi^{\Diamond}(x,y)]$$

and we’re done.⁴□

Lemma 15

Every instance of the ♢-translation of the replacement scheme holds in $\mathbb {K}$.

Proof

Suppose

$$w \vDash \forall x \in a \Diamond \exists y \varphi^{\Diamond}(x,y).$$

Then for every x ∈ a there is w(x) with

$$w(x) \vDash \exists y \varphi^{\Diamond}(x,y).$$

By Corollary 2, w(x) is a forcing extension of a height extension of w; and so in particular we can find a corresponding u(x) ≤_vw with:

$$u(x) \vDash \exists p, \mathbb{P}[p \Vdash \exists y \varphi^{\Diamond}(\check{x},y)] $$

We then take κ an inaccessible rank above u(x)₀ which reflects the situation for $\exists p, \mathbb {P}[p \Vdash \exists y \varphi ^{\Diamond }(\check {x},y)]$. By product forcing, there is a horizontally accessible world v = 〈κ,w₁[G]〉 with

$$v \vDash \forall x \in a \exists y \varphi^{\Diamond}(x, y)$$

and the rest is straightforward. □

Theorem 1

M is consistent relative to the existence of a Mahlo cardinal.

Proof

The only difficult parts are Lemmas 11, 13–21. The rest is routine. □