# The set of all sets which are elements of themselves

Russell's paradox concerns itself with the set $$S=\{x|x\notin x\}\;\;\;S\in S ?$$ but it is supposedly solved in ZFC theory.

Now, what about the set $$U=\{y|y\in y\}\;\;\;U\in U ?$$

Is U an element of itself?

Russell's paradox concerns itself with the set $$S=\{x|x\notin x\}\;\;\;S\in S ?$$ but it is supposedly solved in ZFC theory.

Now, what about the set $$U=\{y|y\in y\}\;\;\;U\in U ?$$

Is U an element of itself?

The paradox is "resolved" by calling U a class instead of a set. A class doesn't contain itself.

CompuChip
Homework Helper
The paradox arises in the first place, because $S \in S$ if and only if $S \not\n S$. It goes to show that defining a set as any expression of the form
$$\{ x \mid x \text{ satisfies } P(x) \}$$
is not a good idea.

If you repeat that trick for U, you get $U \in U$ if and only if $U \in U$, and that is not a problem.

The paradox arises in the first place, because $S \in S$ if and only if $S \not\n S$. It goes to show that defining a set as any expression of the form
$$\{ x \mid x \text{ satisfies } P(x) \}$$
is not a good idea.

If you repeat that trick for U, you get $U \in U$ if and only if $U \in U$, and that is not a problem.

Yes I know. It is an element of itself iff it is an element of itself. It is not an element of itself iff it is not an element of itself. So, you cannot make any definite conclusions about U.

micromass
Staff Emeritus
Homework Helper
Russell's paradox concerns itself with the set $$S=\{x|x\notin x\}\;\;\;S\in S ?$$ but it is supposedly solved in ZFC theory.

Now, what about the set $$U=\{y|y\in y\}\;\;\;U\in U ?$$

Is U an element of itself?

If we are working in ZFC set theory, then your U doesn't exist. It is not a set.

The rule for forming sets goes as follows: If A is a set and if P is a formula, then

$$\{x\in A~\vert~P(x)\}$$

is a set.

So for example,

$$\{x\in A~\vert~x\notin x\}~\text{and}~\{x\in A~\vert~x\in x\}$$

are both sets.

However, things like $$A=\{x~\vert~x\notin x\}$$ and $$B=\{x~\vert~x\in x\}$$ are not sets since we don't have a suitable set A. So those two things do not exist in ZFC set theory.

Those two things are something called "proper classes" and can be worked with. But if C is a proper class, then it is not allowed that C is an element of something.

Deveno
If we are working in ZFC set theory, then your U doesn't exist. It is not a set.

The rule for forming sets goes as follows: If A is a set and if P is a formula, then

$$\{x\in A~\vert~P(x)\}$$

is a set.

So for example,

$$\{x\in A~\vert~x\notin x\}~\text{and}~\{x\in A~\vert~x\in x\}$$

are both sets.

However, things like $$A=\{x~\vert~x\notin x\}$$ and $$B=\{x~\vert~x\in x\}$$ are not sets since we don't have a suitable set A. So those two things do not exist in ZFC set theory.

Those two things are something called "proper classes" and can be worked with. But if C is a proper class, then it is not allowed that C is an element of something.

of course, that just "kicks the problem upstairs", now we need a new name for the collection of all classes (that is, there is no class of all classes, which is a problem that actually comes up; for example, if you want to talk about all possible algebraic structures, many of which form proper classes (the class of all sets, the class of all groups, the class of all vector spaces, etc. -most of these arise because there is often a "generic" way to turn a set into an algebraic structure of some sort)).

i note in passing that the question of the thread-starter is settled by the axiom of regularity (rather than, let's say, the axiom of limited comprehension needed to deal with the Russell set) which states that every set A contains an element B such that B and A are disjoint.

if we apply this axiom to {A}, which just has the single element A, we get that A∩{A} = Ø, which means that A is not an element of A.

micromass
Staff Emeritus
Homework Helper
of course, that just "kicks the problem upstairs", now we need a new name for the collection of all classes (that is, there is no class of all classes, which is a problem that actually comes up; for example, if you want to talk about all possible algebraic structures, many of which form proper classes (the class of all sets, the class of all groups, the class of all vector spaces, etc. -most of these arise because there is often a "generic" way to turn a set into an algebraic structure of some sort)).[/tex]

Note that in ZFC set theory, there is no notion of a class. A class is just a shorthand for a formula. So

$$y~\{x~\vert~\psi(x)\}~\Leftrightarrow~\psi(y)$$

is the same thing. Classes are no real entities. So there is not really a problem anymore. In formalized theory, there should be no notion of classes.

If you do want to work with classes and with collections of classes (which is a conglomerate, other names are possible though), then you are working with a strong kind of set theory. That is: you are working in a set theory with an inaccessible cardinal. The existence of this cannot be proven consistent with ZFC. This does not prevent mathematicians to work with them anyway.
i note in passing that the question of the thread-starter is settled by the axiom of regularity (rather than, let's say, the axiom of limited comprehension needed to deal with the Russell set) which states that every set A contains an element B such that B and A are disjoint.

if we apply this axiom to {A}, which just has the single element A, we get that A∩{A} = Ø, which means that A is not an element of A.[/QUOTE]

The axiom of regularity is quite a useless axiom, if you ask me. It is not really needed as 99.999% of mathematics can be done without the axiom of regularity. It might me interesting sometimes, but I choose not to accept the axiom. Almost everything in ZFC- can also be done in ZFC+.

Deveno