# Does the universal set exist without allowing the concept of a set belonging to itself?

I have studied some elementary set theory and encountered a proof that a universal set containing everything cannot exist, as follows:

Suppose, on the contrary, that there exists a set $$\mathbb U$$ containing everything. Then, by the axiom of specification, there exists a set $$\mathbb M= \{ x\in \mathbb U |\ x\ \text{is not a member of itself}\ \}$$ however, we can easily demonstrate that both $$\mathbb M \in \mathbb M$$ and $$\mathbb M \notin \mathbb M$$ lead to a contradiction.

I was wondering if such a contradiction still arises in a different way when we assume that it is impossible for any set to contain itself and consider $$\mathbb U$$ as the universal set that contains everything except itself.

• Cantor's theorem shows that the set $P(\Bbb U)$ has cardinality strictly larger than $\Bbb U$, so it would contain many sets that are not in $\Bbb U$ Sep 30 at 10:44
• @MW, Thank you; I understand now. To summarize, if we assume that $\mathbb U$ doesn't contain itself we can easily make a bigger set like $\mathbb V$, and the main difficulty is to find a contradiction when $\mathbb U$ does contain itself where we can use Russel's Paradox or Cantor's theorem to refute it, right?
– Aria
Oct 1 at 13:24
• There is an alternate approach to avoiding Russell's paradox and its many variations, called "Type Theory". This is the approach proposed by Russell himself in collaboration with Whitehead. In it, the formula "$X \in X$" cannot be constructed. Instead of just "sets", you get a hierarchy of different types of collections, each of which can only have members of the lower types in the hierarchy as its elements. For more information, consult Russell & Whitehead's "Principia Mathematica". Oct 1 at 13:32
• New Foundations is a set theory with a universal set that also avoids Russell's paradox. The set $\mathbb{M}$ cannot be formed because separation is only valid for stratified formulae, and $x \in x$ is not stratified. Oct 1 at 15:43
• @DanChristensen, Thank you for the proof. I was thinking to apply a rule to forbid any set to contain itself so the universal set also would contain everything but itself, but that still doesn't work, and the proof you stated is still valid since we're only using the subset axiom to form $\mathbb M$, and the same contradiction would still arise. Thank you for your guidance.
– Aria
Oct 2 at 17:31