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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?

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Maybe I'm missing something, but isn't this just the diagonal argument?


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 $Z$.

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  • $\begingroup$ Yes but $D$ need not be a proper class? $\endgroup$ Jan 28, 2020 at 20:34
  • $\begingroup$ @ZuhairAl-Johar The emphasis on proper classes isn't essential - we can always "merge" two enumerations, and since we have an enumeration of all sets on hand the issue remains. (I've edited to address this.) $\endgroup$ Jan 28, 2020 at 21:15
  • $\begingroup$ what does the "rows" of $Z$ mean? $\endgroup$ Jan 29, 2020 at 4:34
  • $\begingroup$ @ZuhairAl-Johar The classes indexed by $Z$ - the classes of the form $\{y: \langle a,y\rangle\in Z\}$ for some fixed $a$. $\endgroup$ Jan 29, 2020 at 5:22
  • $\begingroup$ Yes! That works! Although it is NOT "just" the diagonal argument. It is a version of it, but not the usual one! So as I expected MK can indeed prove having more proper classes than sets! Thanks! $\endgroup$ Jan 29, 2020 at 12:33

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