Is the following provable in MK?


$\not \exists S: \\ \text{ } \\1. \ \ \forall s \in S \exists a,b (s=\langle a,b \rangle) \\ \text{ } \\2. \ \ \forall x (set(x) \to \exists! X (\neg set(X) \land \{x\} \times X \subset S \land \forall k(\langle x,k \rangle \in S \rightarrow k \in X))) \\ \text{ } \\3. \ \ \forall X (\neg set(X) \to \exists! x ( \{x\} \times X \subset S \land \forall k(\langle x,k \rangle \in S \rightarrow k \in X))) $


In English: there exists a class $S$ of ordered pairs, such that for every set $x$ there exist a unique proper class $X$ such that $\{x\} \times X$ is a maximal subclass of $S$ that is a cartesian product of $\{x\}$ by a class. And the opposite direction also holds, i.e. for every proper class $X$ there exists a unique set $x$ such that $\{x\} \times X$ is a maximal sublass of $S$ that is a cartesian product of $\{x\}$ by a class. 


Now it is known that there can exist a class $S^\*$ that fulfills only the first two conditions, simply take the union of all cartesian product classes of singletons by their complements. 


The idea here is to express that there are more proper classes than sets, internally within MK. Since existence of $S^\*$ can be taken to mean that there are at least as many proper classes as sets are there. But failure of condition 3, tell's us that we don't have as many sets as proper classes are there, so from both situations we infer that there are more proper classes than sets. However is non existence of $S$ provable in MK? What's the proof?



Does MK prove internally that there are more proper classes than sets?
Maybe I'm missing something, but isn't this just the diagonal argument?




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Suppose $S$ were as in your question. Let $E=\{\langle x,y\rangle: set(x)\wedge y\in x\}$, and consider the set $$Z=\{\langle \langle x,0\rangle, y\rangle: \langle x,y\rangle\in S\}\cup\{\langle\langle x,1\rangle, y\rangle: \langle x,y\rangle\in E\}.$$ $Z$ has the property that for each class (proper or non) $C$ there is some $a$ with $$C=\{y: \langle a,y\rangle\in Z\}.$$


We now apply diagonalization to this new $Z$: the class $$D=\{x: \langle x,x\rangle\not\in Z\}$$ cannot appear as any of the "rows" of $Z$.


