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Let $U$ be the set of all sets. Define a partial ordering on $U$ by inclusion: $A \leq B$ iff $A \subseteq B$ for $A, B \in U$. Consider a chain $C$ of $U$ under this partial ordering: $$ C : A_1 ≤ A_2 ≤ A_3 ≤ \cdots $$ Define $B = \bigcup_{i\geq1}A_i$. Clearly, $B \in U$ and it is an upper bound of the chain $C$. Hence, Zorn’s Lemma implies that $U$ has a maximal element, say $M$. The argument is clearly wrong since $M$ is not a maximal element: $$ M \subsetneq \{M, \{M\}\} \in U. $$ I can't identify which step in the argument is wrong and why.

Nick Matteo
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Jaswant
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  • Trivially, the set of all sets is a set, and it is the maximum of $U$. (Of course, there is no set of all sets, but even in a scenario that it does exist...) – Asaf Karagila Dec 02 '21 at 21:18
  • $M$ is not actually a subset of the set ${M, {M}}$. I think you meant $M \cup {M}$. But it's only a strict subset of the latter if $M$ is not an element of itself (leading to Asaf's resolution). – Nick Matteo Dec 02 '21 at 21:32

1 Answers1

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The first step is wrong: There is no set of all sets. Objects like this are not allowed in ZFC set theory in order to prevent contradictions like Russel's paradox.

You could define a class that contains all sets; however that class is no longer a set itself and cannot be treated as such.

Andreas Tsevas
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