# Are there other simple conditions we can use to demonstrate the non-existence of a universal set?

Here is (what I think is) a standard proof of the non-existence of a universal set in ZFC:

Let $$A$$ be an arbitrary set and, by the axiom of specification, let $$B = \{x \in A \mid x \not \in x\}$$. Is it true that $$B \in A$$?

Suppose $$B \in A$$. We have two cases: $$B \in B$$ or $$B \not \in B$$. In the first case, we see from the definition of $$B$$ and the assumption that $$B \in A$$ that $$B \not \in B$$, and in the second we see that $$B \in B$$. In either case we reach a contradiction, so it must be true that $$B \not \in A$$. Since $$A$$ is arbitrary we can conclude that $$B \not \in A$$ for any set $$A$$. Therefore, there is no set of all sets (since $$B$$ is always missing).

In this argument we showed that the subset of any set satisfying the condition $$x \not \in x$$ is never an element of that set.

Are there other conditions (simple expressions like $$x \not \in x$$) that can be used to achieve the same result? In other words, are there other sets that we can easily prove to be excluded from every set? I can think of trivial examples like $$B \cup A$$ or indeed $$B \cup C$$ for any set $$C \neq B$$ (taking $$B$$ as above), but I’m wondering if there are any other well-known logical sentences that define subsets through the axiom of specification as in the argument above.

Cantor's Theorem states that for any set $$X$$, cardinality of power set $$|P(X)| > |X|$$.
Now, suppose for a contradiction that set of all sets $$S$$ exists. But then $$P(S) \subseteq S$$ because every set in $$P(S)$$ also included in $$S$$ by definition of $$S$$. But then $$|P(S)| \le |S|$$ and by Cantor's Theorem $$|P(S)| > |S|$$, a contradiction.
• Cool, but I wasn’t exactly asking for another whole proof of this; just another logical sentence we can insert in place of $x \not \in x$ in my proof. Jun 11, 2019 at 15:35