6
$\begingroup$

As I understand the ZFC solution to Russell's paradox, since $\{x\mid x\notin x\}$ must be $\{x\mid x\notin x\}\cap S$ for some set $S$, the paradox goes away, but in Morse-Kelley, although again there must be some $M$ such that $\{x\mid x\notin x\}\cap M$, this $M$ may be a proper class, which no longer is as limiting as the ZFC version, and hence no longer gives the same solution. So MK must handle Russell's Paradox in a different way. How? I would be grateful for enlightenment. Thanks.

$\endgroup$
  • $\begingroup$ In MK, the proof of nonexistence of $\{x:x\notin x\}$ (in ZFC) states that the class $\{x:x\notin x\}$ is a proper class. $\endgroup$ – Hanul Jeon Dec 29 '13 at 4:52
7
$\begingroup$

Morse-Kelley set theory works like this: You have unrestricted comprehension, so for any formula $\phi$, you are allowed to construct $$\{x\mid \phi(x)\}$$ (“the set of all $x$ that have property $\phi$”) whereas in ZF set theory you may only construct $$\{x\in A\mid\phi(x)\}$$ where $A$ is some previously-constructed set (“the set of all $x$ in $A$ that have property $\phi$”).

But having done that, what are the members of the thing that you constructed?

In ZF one has that, for each $y$:

$$y\in \{x\in A \mid \phi(x)\}\\\text{ if and only if }\\y\in A\text{ and }\phi(y)$$

But in Morse-Kelley set theory membership is restricted in a different way: one has, for each $y$: $$y\in \{x\mid \phi(x)\}\\\text{ if and only if }\\\text{$y$ is a set and }\phi(y)$$

So the $\{x\mid \phi(x)\}$ notation in Morse-Kelley set theory is a bit misleading: It looks like you're forming the class of everything with property $\phi$, but actually you are only forming the class of all sets with property $\phi$.

Whereas in ZF you can't use specification to construct a set without having another set first, in Morse-Kelley you can construct it, but then you can't prove that any $y$ is a member of it unless you first prove that $y$ is a set. In the memorable words of Irving Kaplansky, good sets are allowed to be members of other sets, but bad sets are blackballed from membership. (Kaplansky was talking about Von Neumann set theory, not Morse-Kelley, but they are very similar.) To prove that $y\in S$, you have to first show that $y$ is a "good" set, something that is allowed to be a member of things.

Now let's see what happens when we try to construct the Russell set $R$ in Morse-Kelley theory. We want $$R \stackrel{\mathrm{def}}= \{x\mid x\notin x\}$$

so we will have, for each $y$, that $$y\in R\\ \text{if and only if}\\ \text{$y$ is a set and $y\notin y$}.$$ Then we take $y=R$ and see what happens. We will have $$R\in R\\ \text{if and only if}\\ \text{$R$ is a set and $R\notin R$}.$$ There is no contradiction here; we have merely proved that $R$ is not a set. It is therefore "blackballed", and $R\notin C$ holds for every class $C$. In particular $R\notin R$.

There is no contradiction because Morse-Kelley theory only requires that $R$ contain the sets with the Russell property, and $R$, not being a set, does not fit the bill.

Similarly Morse-Kelley theory has a class of all sets, called $\mathscr U$, but it does not include itself, since it is not itself a set.

If you want to read about this in more detail, there is a wonderfully brief explanation of it in the Appendix of Kelley's General Topology (1955) pp. 250–258.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.