Skip to main content
2 of 3
added 97 characters in body
Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47

Is $\sf MK$ bi-interpretable with this modification of it posing an Ur-proper class for every set?

Let's take $\sf MK$ set theory.

Adopt the notation of upper case ranging over all objects, lower case only range over sets, and $\frak A,B,C,..$ to range only over proper classes.

Define: $X \approx Y \iff \forall m \, (m \in X \leftrightarrow m \in Y)$

Weaken Extensionality to:

Class Extensionality: $\forall X \exists! {\frak A} : X \approx \frak A $

Set Extensionality: $\forall x \exists! y : x \approx y$

Modify limitation of size to:

Limitation of size: $|X| \neq |V| \to \exists x: x \approx X$

Where $V$ is the class of all sets, and $||$ stands for cardinality function.

Add all other axioms of $\sf MK$.

Call this theory $\sf MK'$.

So, $\sf MK'$ have an Ur-proper class for every set.

Is $\sf MK$ bi-interpretable with $\sf MK'$?

I think they are equi-interpretable! The proof is:

Define a new membership relation $\in^*$ by: $$ y \in^* x \iff \exists z: z \in x \land y \approx z$$

I think that $\sf MK'$ would prove all axioms of $\sf MK$ with $\in$ replaced by $\in^*$ and $=$ replaced by $ \approx $.

For the other direction, we need to define a new membership relation $\in'$ along the following lines:

Define: $\operatorname {binarytuple}(X) \iff \\ X: V \to \{0,1\} \land \\ \exists x \forall y (y \in x \leftrightarrow \langle y, 1 \rangle \in X)$

Define: $y \in' X \iff \\ \operatorname {binarytuple}(X) \land \langle y,1 \rangle \in X \lor \\ \neg \operatorname {binarytuple}(X) \land y \in X$

Accordingly, $\sf MK$ would prove that for every set there is a unique proper class that is $\in'$-coextensional with it. And so, $\sf MK$ would prove all axioms of $\sf MK'$ with $\in$ replaced by $\in'$.

Of course, this doesn't prove them bi-interpretable. Hence, the question raised above.

Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47