The set of all sets exists.

[godelian]

The word "all" indicates that the set S also contains itself.
But a set cannot contain itself.

Axiom 2 actually axiomatizes that possibility away. It explicitly says that S cannot contain itself. For all intents and purposes, it turns S into a proper class.

So, S does not contain itself. In general, there is no set that is allowed to contain S. Therefore, this axiom actually modifies the definition of every set. S is some kind of collection but not a set.

ZFC has this problem in general. The domain of sets is itself not a set. It is the universe of discourse for set theory, carefully designed as not being a set itself. That is an artificial hack.

In that sense, I do not consider the idea of the collection of all sets to be outlandish. For reasons of mere formalism, bureaucracy, and paradox avoidance, we strip it of its status of "set".

[Magnus Anderson]

Axiom 2 actually axiomatizes that possibility away. It explicitly says that S cannot contain itself. For all intents and purposes, it turns S into a proper class.

It actually does not. It merely introduces a different membership relation, one which returns "undefined" whenever you ask "Is the set of all sets a member of the set Y""?

S is internally consistent. There is no need for the Axiom of Regularity. And no other axiom can modify its internal consistency.

S is not expressed using a string of meaningless symbols that have yet to be imbued with meaning. Each word is already defined -- explicitly or implicitly. The word "all", for example, means that S contains all sets, including itself. If you later say that it does not, that's a contradiction.

Of course, one can change the meaning of the word "all" . . . but then, one would no longer be talking about the set of all sets in the standard sense. And it would be unnecessarily convoluted as well as deceptive.

[godelian]

Axiom 2 actually axiomatizes that possibility away. It explicitly says that S cannot contain itself. For all intents and purposes, it turns S into a proper class.

It actually does not. It merely introduces a different membership relation, one which returns "undefined" whenever you ask "Is the set of all sets a member of the set Y""?

S is internally consistent. There is no need for the Axiom of Regularity. And no other axiom can modify its internal consistency.

S is not expressed using a string of meaningless symbols that have yet to be imbued with meaning. Each word is already defined -- explicitly or implicitly. The word "all", for example, means that S contains all sets, including itself. If you later say that it does not, that's a contradiction.

Of course, one can change the meaning of the word "all" . . . but then, one would no longer be talking about the set of all sets in the standard sense. And it would be unnecessarily convoluted as well as deceptive.

Yes, the following:

Axiom 1: S is the set of all sets

Should probably be rephrased into something like:

Axiom 1: S is the proper class of all sets

Using axiom 2 to fix the problems in axiom 1 is not the best approach. It is indeed confusing. However, it is not necessarily wrong.

[Skepdick]

Yay! New toys!

https://claude.ai/public/artifacts/e99d ... 625e432a72

[Martin Peter Clarke]

Oooh! Barber paradox! Wrongly comes to mind I'm sure.

[Impenitent]

barber paradox?

A logician with some time to kill in a small town decided to get a haircut. The town had only two barbers, each with his own shop. The logician glanced into one shop and saw that it was extremely untidy. The barber needed a shave, his clothes were unkempt, and his hair was badly cut. The other shop was extremely neat. The barber was freshly shaved and spotlessly dressed, his hair neatly trimmed. Why did the logician return to the first shop for his haircut?

-Imp