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It is well known that there is no set of all sets. The argument goes as follows:

Define:

$$\mathcal{C}:=\{X: X\ \textrm{is a set and}\ X\not\in X\}.$$

If $\mathcal{C}$ was a set then $$\mathcal{C}\in \mathcal{C}\Leftrightarrow \mathcal{C}\not\in \mathcal{C}.$$

I am asking myself why this would show there is not a set of all sets.

Well, I think if there was a set of all sets it should be equal to $\mathcal{C}$, right? But this would rely on the fact that every set is NOT a member of itself, but what would justify that?

I am not even sure if I understand the definition of $\mathcal{C}$ anyway.

Can anyone give me further clarifications?

Thanks.

Andrés E. Caicedo
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2 Answers2

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The argument that Russell's paradox implies there is no universal set is rooted in the axiom that if $A$ is a set and $P$ is a predicate defined on $A$, then $\{ x \in A : P(x) \}$ is also a set. Using this axiom with $A$ being the universal set and $P(x)=x \not \in x$ results in a contradiction.

This axiom is built into ZF set theory. If you do not use this axiom, you can have a coherent set theory with a universal set. This is done in, for example, Quine's New Foundations.

Ian
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The argument goes that if there is a set of all sets $V$, then we can define your $\mathcal{C}$ as a subset. The existence of $\mathcal{C}$ is contradictory, so there was no such $V$ to start with.

James
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