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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.

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We can use Cantor's Theorem actually.

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.

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  • $\begingroup$ 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. $\endgroup$ Jun 11, 2019 at 15:35
  • $\begingroup$ Oh, sorry for that. I actually posted this answer to "are there other sets that we can easily prove to be excluded from every set?" question. I will think about some other logical sentences. $\endgroup$
    – ArsenBerk
    Jun 11, 2019 at 15:38
  • $\begingroup$ Well, thank you anyway! I’ve definitely learnt something from your answer. $\endgroup$ Jun 11, 2019 at 15:48
  • $\begingroup$ You're welcome :) After this question, I am also curious about some other simple condition. I will be following this question. $\endgroup$
    – ArsenBerk
    Jun 11, 2019 at 16:00

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