# How does Russell's paradox affect Cantor's set theory?

I've been studying naive set theory and I have been told that Russell's paradox causes problems in Cantor's set theory when sets get "too big".

I don't understand why this causes a problem. I know how the paradox effected Frege's work in terms of logic but not Cantor.

Any help will be appreciated, Thanks

• Maybe they meant that due to Russell's paradox there cannot be a 'universal set': a set of 'everything'... that would certainly be a set that can be said to be 'too big' :) – Bram28 Apr 26 '18 at 15:12
• What do you call Cantor's set theory ? (I'm not that well-versed in history, I thought that Cantor's set theory was essentially a naive version of Frege's version - and that being given, any problem to Frege's theory is a problem to Cantor's theory) – Max Apr 26 '18 at 15:13

Maybe they meant that due to Russell's paradox there cannot be a 'universal set': a set of 'everything'.

Here's why: Assuming there would be a set of everything $U$, then we can consider the set $D$ defined as $\{ x \in U | x \not \in x \}$. We could thus say that $D$ contains all 'normal' sets, where a 'normal' set is a set that does not contain itself as an element.

Now, since $U$ is universal, we have $D \in U$. But is $D \in D$? Well, if $D \in D$, then by definition of $D$, we don't put $D$ in $D$, i.e. $D \not \in D$. But if $D \not\in D$, then by definition of $D$ we would put $D$ in $D$, i.e. $D \in D$. Hence, we obtain $D \in D$ if and only if $D \not \in D$, which is a contradiciton.

So, using the logic behind Russell's paradox, there cannot be a universal set. ... And yes, that would certainly be a set that can be said to be 'too big'.

• That's indeed what I thought, but how would i make this connection with the fact that Cantor proposed that there exists no largest cardinal number? Is it because the cardinal will get so large that it will start containing all elements, making it a universal set? – Maths Apr 26 '18 at 15:29
• @Desperados Yes, of course, thanks! – Bram28 Apr 26 '18 at 15:42
• @Maths Cantor showed that the cardinality of the power set of any set has a greater cardinality than the set itself .. very much using this same logic .. See the Proof in the Wikipedia page on Cantor's Theorem – Bram28 Apr 26 '18 at 15:49
• @Maths By Cantor's Theorem, there cannot be a 'greatest' set, since if there were, we immediately have a contradiction, since its powerset would be 'greater' yet. – Bram28 Apr 26 '18 at 15:55
• The introductory text by Suppes refers to the "Axiom of Abstraction": the (naive) assumption that for any stated property $P$ there exists $\{x: P\}.$ Russel showed an instance of this assumption that is paradoxical . If we want Comprehension, that is, for any set $S$ and any property $P$ there exists $T=\{x\in S:P\}$ then we can't have a unversal set $U.$ else there exists $T=\{x\in U: x\not \in x\}=\{x:x\not\in x\}$. – DanielWainfleet Apr 26 '18 at 19:17

These problems were all solved when axioms were stated for Set Theory. These axioms made it very clear what is and what is not a set. Objects that are not sets are called classes and they have this characteristic property: They cannot be elements of any set.

So Russel's paradox says something like this: Let $R:=\{x| x\not\in x\}$ and the "paradox" is that $R\in R$ iff $R\not\in R$. After the axiomatic approach of set theory this was not a problem: no one tells us that $R$ is a set and not a class. Actually, assuming $R$ was a set, Russel's paradox would be a contradiction, hence $R$ is a class.

Most of the sets that one deals with on Set Theory are bounded, in the sense that they are expressed in the form $\{x\in A: \phi(x)\}$, meaning, the elements of the set $A$ that satisfy the property $\phi(x)$. Russel's set is unbounded, it refers to all sets in the universe $V$. (Note: this is not enough to say that $R$ is not a set, I'm just bringing it up). Actually one of the axioms, the axiom of subsets, says the following:

Axiom (schema) : For each formula $\phi(x)$ of the Language $L=\{\in\}$ of set theory:

$(\forall a)(\exists b)(z\in b \leftrightarrow (z\in a \wedge \phi(z))$.

Translated to common language, this says that for any set $a$, there exists a set $b$ that consists of precisely all those elements of $a$ that satisfy the property $\phi(x)$.

If you want to have some fun with these stuff, try this exercise: is $C_1=\{x: |x|=1\}$ (The collection of all sets that have exactly $1$ element) a set or a class?

Edit in order to solve the puzzle, you will need the following axiom:

Axiom of union: If $x$ is a set, then there exists a set that consists exactly of all the elements of each element of $x$, or in formal language:

$(\forall x)(\exists y)(\forall z)(z\in y \leftrightarrow (\exists u)(u\in x \wedge z\in u))$