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I will proceed by assuming the negation. So, let $V$ be set of all sets. So, $V$ exists. Now, I will use axiom schema of specification

$$ \forall w_1,\cdots, w_n \, \forall A \,\exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in A \wedge \varphi(x,w_1, \cdots , w_n, A)] \right) $$

I will let $A = V$ and $\varphi(x, A) = x \notin x $. So, using the universal instantiation, I get that

$$ \exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in V \wedge \varphi(x, V)] \right) $$

And then using existential instantiation, there exists a set $B$ such that

$$ \forall x \left( x \in B \Longleftrightarrow [ x \in V \wedge \varphi(x, V)] \right) $$

$$ \forall x \left( x \in B \Longleftrightarrow [ x \in V \wedge x \notin x] \right) $$

$$ \forall x \left( x \in B \Longleftrightarrow x \in \{ x \in V \,| x \notin x\} \right) $$

Using, axiom of extensionality, it follows that

$$ B = \{ x \in V \,| x \notin x\} $$

Now, I want to use other ZF axioms to reach my contradiction. How do I proceed ?

user9026
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    Once you've shown the existence of set $B$ as you defined it, you have Russell's paradox in hand. No more axioms are needed. Do you have a more precise question in mind? – hardmath Aug 26 '20 at 14:23
  • I see the point. If $B \in B$, then $B \in V$ and $B \notin B$. On the other hand, if $B \notin B$ and $B$ will always be in $V$ since its set of all sets. Then $B \in B$. So, we reach contradiction in both directions. Which means $V$ can not exist. – user9026 Aug 26 '20 at 15:17

1 Answers1

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You don't need any more ZF axioms to continue, just excluded middle. Ask whether or not $B$ is an element of $B$ to reach the desired contradiction.

halrankard2
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  • Thanks. I was thinking of axiom of foundation here. But what you say suffices it seems. – user9026 Aug 26 '20 at 15:18
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    The axiom of foundation can be used to prove that no set is an element of itself. For example, this appears in the wikipedia article: https://en.wikipedia.org/wiki/Axiom_of_regularity#No_set_is_an_element_of_itself. So this is a different proof that $V$ is not a set. If $V$ is a set then $V\in V$ by definition of $V$. – halrankard2 Aug 26 '20 at 15:22