Can ZFC be interpreted in a set theory having finitely many ranks?

One of the most notable features of $ZFC$ is that it builds up sets in recursively defined $V_i$ stages (where $i$ is an ordinal), however the usual formulation of $ZFC$ has infinitely many stages. The rank of a set $X$ is the ordinal index $i$ of the stage $V_i$ where $X$ first appears as a subset of. So in $ZFC$ we can have a set having an infinite rank.

The question is: can we define a theory $T$ in the language of set theory and choice (i.e. choice can be added as a primitive function symbol to the language of set theory) such that $T$ speaks of sets that are built up in stages in generally a similar manner to the buildup of sets in $ZFC$ such that $T$ can interpret $ZFC$ but provided that $T$ has only finitely many stages?

I think that the answer is to the positive, the following is a definition of a theory having only 5 stages and in which all of the $\beth_i$ numbers definable in ZFC are definable in it.

We add to the language of set theory (first order logic with equality and membership) a primitive unary choice function $c$, and stipulate the following axioms:

Identity axioms +

1. Axiom of Choice: $(\exists y \in X) \implies c(X) \in X$.

1. Weak Extensionality: non empty sets having the same elements are equal

2. Base: There exists a set $B$ (for base) of all empty sets

3. Power: There exists $P(B)$, $PP(B)$, $PPP(B)$; ($P$ signifies power set)

4. Axiom schema of Separation (as in Zermelo)

5. Infinity: $N < B$

6. Accessibility by power: $X \subset B \wedge \exists x \in X \wedge X< B \implies P(X)\backslash B < B$

7. Accessibility by union:$X\subset P(B) \wedge X\not\subset B \wedge X < B \wedge \forall Y \in X (Y < B) \implies U(X) < B$

Where $N$ is the set of all natural numbers; $A < B$ is defined as "strictly subnumerous to" in the usual manner using Kuratowski's ordered pairs; $"\backslash"$ signifies: except, and $"U"$ signifies: set union.

A natural number is defined as the choice set of an equivalence class of finite sets of empty sets under equivalence relation bijection, where bijection is defined in the usual manner using Kuratowski ordered pairs; and finite is defined after Dedekind's also using Kuratowski ordered pairs.

Now sets belonging to the first rank are the empty sets, those belonging to the second rank are subsets of $B$ that do not belong to lower ranks, those belonging to the third rank are subsets of $P(B)$ that do not belong to lower ranks, and so on sets belonging to the $i^{th}$ rank are subsets of $P_{i-2}(B)$ that do not belong to lower ranks.

/Theory definition finished.

Now to understand what's going on here, the trick is to code each third rank set that is strictly subnumerous to $B$ down to a second rank set that is equinumreous to it (i.e. to that third rank set), this is done by defining the set of all unordered injections (denoted as injection* defined below) from the rank3 set to $B$, and then take the choice set of those injections and take the range of the selected injection which would be a rank2 set that is equinumerous to the rank3 set, and it stands as its code, we'll denote it by putting ' after the symbol for the rank3 set.

$Define: F \text{ is injective* from A to B } \iff \\ \forall X \in F \exists a \in A,\exists b \in B (X=\{a,b\}) \wedge \\ \forall X \in A \exists! Y \in B (\{X,Y\} \in F) \wedge \\ \forall a,b \in A \forall x \in B (\{a,x\} \in F \wedge \{b,x\} \in F ⟹ a=b)$

$Define: X \text{ is a cardinal } \iff \exists Y (Y \text{ is an equivalence class of "sets of empty sets" under bijection } \wedge X= c(Y))$

$Define: X=|Y| \iff \exists K (\forall Z (Z \in K \iff Z\subset B \wedge ( ( Y \not\subset B \wedge Z \text{ equinumerous to } Y') \lor (Y \subset B \wedge Z \text{ equinumerous to Y }))) \wedge X=c(K))$

$Define: X \text{ is an ordinal } \iff \forall Y \in X (Y \text { is a cardinal }) \wedge (X \text{ is a finite non empty set of natural numbers closed under relation < } \lor (X \text { is infinite } \wedge \forall \text{ natural number } Y (Y \in X) \wedge \forall Y(Y \in X \wedge Y \text { is not a natural number } \implies Y=N \lor \exists Z \in X (Y= |P(Z)\backslash B |))))$

$Define: \text{ For every ordinal }X:\\ X+1=\{x| x \in X \lor x=|X|\} \text{ if X is finite }\\ X+1= \{x| x \in X \lor x=|P(U(X))\backslash B|\} \text{ if X is infinite }$

Now for any ordinal $i$ we can define an injection* $S_i$ from $i$, such that for every element $\{j,x\}$ of $S_i$ where $j \in i$ there exists an element $\{j+1, |P(x)\backslash B|\}$ of $S_i$, and such that $\{0,N'\}$ is in $S_i$.

Now we can easily interpret the $\beth_i$ numbers as the cardinality of the set union of the range of $S_i$ (where the range of $S_i$ is the set of all non-ordinal elements of the set union of $S_i$)

Now as long as $i < B$ then $\beth_i < B$, according to the last two axioms.

To show that we can get to $\beth$ fixed points, first we need to define the least ordinal that is equinumerous to $\beth_i$, we call that $Ord(\beth_i)$, then we define a sequence of unordered pairs whose domain (the set of all ordinal elements of its union) is $N$, starting with $\{0,\beth_0\}$ and such that for every $\{j, \beth_i\}$ in it ,we have $\{j+1, \beth_{Ord(\beth_i)}\}$ in it. Now the union of the range of that sequence is the first $\beth$ fixed point $k=\beth_k$. All other $\beth$ fixed points of $ZFC$ are definable here.

Actually $B$ itself is inaccessible, and so it proves the consistency of $ZFC$.

• Apart from idea in the post, can't we interpret a model of ZFC inside any model of arithmetic having Con(ZFC), just by building the Henkin model there? This wouldn't need any higher ranks at all. – Joel David Hamkins May 3 '17 at 18:41
• @StellaBiderman, I don't think that $\exists x (N < x \wedge \forall y < x (P(y)<x) \wedge \forall y (y < x \wedge \forall z \in y (z < x) \implies U(y)<x)$, where "<" signifies "strict subnumreousity" is a theorem of ZFC. – Zuhair Al-Johar May 3 '17 at 19:03
• @Zuhair you're right. – Stella Biderman May 3 '17 at 19:09
• @JoelDavidHamkins, I don't think I can answer to your question, but wouldn't that be an interpretation using another language than that of set theory and choice? How we are to think of ranks under such different milieus? – Zuhair Al-Johar May 3 '17 at 19:22
• My point was that even finite set theory $(HF,\in)$, which is bi-interpretable with arithmetic, can interpret ZFC set theory, provided it satisfies Con(ZFC), since it can build a Henkin model of ZFC. So we don't need any infinite ranks at all. – Joel David Hamkins May 4 '17 at 12:19

In relation to the above question, it can be proven that for any theory formulated in first order logic, one can easily have a set theory that interpret it having only three ranks, and even one rank if primitive type level ordered pairs are used. The general workup is to weaken Extensionality as to allow existence of multiple empty sets, then define ordered pairs between those empty sets, without stipulating their existence, if we don't use primitive ordered pairs, then the least rank of a set theoretic pair of empty sets is three, for example define the Kuratowski ordered pairs of them, then define the primitive predicates of the theory in terms of those pairs and lay down all axioms about them in terms of those pairs over empty sets. Example if we want to interpret $ZFC$, then we define Kurastowski pairs, denoted by $(,)^k$ as:

$Define: p=(a,b)^k \iff \exists x \forall z (z \in x \iff z=a) \wedge \exists y \forall z (z \in y \iff z=a \lor z=b) \wedge \forall z (z \in p \iff z=x \lor z = y)$

$Define: y \in^k x \iff \exists p (p=(y,x)^k)$

We axiomatize weak extensionality and existence of an empty set and then add all axioms of $ZFC$ in terms of the membership relation $\in^k$ with the variables ranging over empty sets only, so for example Axiom of pairing would be written as:

$Pairing: \forall \text{ empty } a,b \space (\exists \text{ empty } x \space (\forall \text{ empty } y \space (y \in^k x \iff y=a \lor y=b)))$

The resulting theory clearly interpret $ZFC$. Now if instead of using Kuratowski pairs, we use a primitive type level ordered pair, then we'll have a theory that interpret $ZFC$ and yet having only ONE rank! Of course the drawback of the above theory is that it has theorems that are not true should $ZFC$ be true. Since it falsifies pairing over relation $\in$, i.e. there can exist a set $\{\{a\},\{a,b\}\}$ but not exist a set $\{\{b\},\{a,b\}\}$ (where $\{,\}$ is defined in terms of relation $\in$).

The answer above suppose 5 ranks, however it can be done in 4 ranks only, here is a proof: First we define a relation $<^*$ which denotes "strict subnumerousity" but after $injections^*$

$Define: X <^* Y \iff \exists F (F \text{ is injection}^* \text{ from } X\ to\ Y ) \wedge \not \exists G (G \text { is injection}^* \text{ from } Y\ to\ X)$

$Define: X =^* Y \iff \exists F (F \text{ is injection}^* \text{ from } X\ to\ Y ) \wedge \exists G (G \text { is injection}^* \text{ from } Y\ to\ X)$

Now when comparing sizes of rank3 sets with $B$ we use $<^*$ , while when comparing sizes rank2 sets with $B$ we use the ordinary $<$ that uses injections implemented in the usual manner as sets of kuratowski ordered pairs. So axiom 6,7,8 must be changed to

Infinity: $N <^* B$

Accessibility by power: $X \subset B \wedge \exists Y \in X \wedge X < B \implies P(X)\backslash B <^* B$

Accessibility by union: $X\subset P(B) \wedge X \not\subset B \wedge X\backslash B <^* B \wedge \forall Y \in X (Y < B) \implies U(X) < B$

Now in the above proof rank5 sets were used to define the $X'$ set code of $X$, and in defining the $S_i$ sequences. The former can be dispensed with by taking the 'cardinality' of $X$ to stand as its rank2 code. And the cardinality of $X$ can be defined without the need to use $X'$ in the following manner:

$Define: Y=|X| \iff \exists K (\forall Z (Z \in K \iff Z \subset B \wedge ( (X \not \subset B \wedge Z =^* X) \lor (X \subset B \wedge Z \text { equinumerous to } X))) \wedge Y=c(K))$

As regards the later, instead of $S_i$ being a set of {rank3,rank2} pairs we can define it as a set of {rank1, rank2} pairs, this mean that we need to define "ordinals" in a different manner.

First it can be proven that the set $Card$ of all cardinals is injective* to $B$.

$Define: Y=X^* \iff \forall Z (Z \in Y \iff Z \in S(X) \wedge \forall K \leqslant X (Z \not\in K))$

So for every cardinal $X$ the,$X^*$ is the set of all elements of the successor cardinal of $X$ (here denoted as $S(X)$) that are not elements of cardinals preceding $S(X)$.

Now since each successor cardinal must contain some elements that do not exist in its predecessor cardinals then for each $X, X^*$ is non empty.

Now we define an unordered map $G$ from $Card$ to $B$ that sends each cardinal $X$ to $c(X^*)$. Clearly $G$ is $injective^*$ from $Card$ to $B$.

Now for the ordinals defined in the original posting, we can define images of them under $G$, i.e. for each ordinal $d$, define $d^G$ (termed as the $G$ code of $d$) as the set of all images of elements of $d$ under $G$.

Now in a similar manner to how $Card$ is proven to be $injectable^*$ to $B$, we can prove the set of all $G$ codes of ordinals to be $injectable^*$ to $B$ by a specific $injection^*$. Let's denote that $injection^*$ by $H$, then for each ordinal $d$ we have $H(d^G)$ that codes it, and this is a rank1 set. Now for each rank1 code $H(d^G)$ we define $H(d^G) + 1 = H((d+1)^G)$. So each $\beth_i$ can be defined as above but with the rank1 ordinal codes instead of the ordinals in the $S_i$sequence and we'll have $\beth_i =|UU(S_i)|$. Now $Ord(\beth_i)$ would be re-defined as the rank1 code of the least ordinal $d$ such that $d =^* \beth_i$. In this way we obviate the need for a fifth rank and the whole proof proceed using 4 ranks only.