MHB The set of all sets does not exist.

[evinda]

Hey! (Wave)

Theorem (Russell's paradox is not a paradox in axiomatic set theory)

The set of all sets does not exist.

Proof

We suppose that the set of all sets exist, let $V$. So, for each set $x$, $x \in V$.
We define the type $\phi: \text{ a set does not belong to itself, so } x \notin x$.

So, from the Axiom schema of specification, the set $\{ x \in V: x \notin x \}$ exists.

Since $V$ is the set of all sets,

$$\{x \in V: x \notin x \} \subset V$$

So, $V'=\{ x: x \notin x \}$ is a set.

Therefore: $V' \in V' \leftrightarrow V' \notin V'$, contradiction.How do we know that $V'$ is a set? Also, could you explain me why, having found this contradiction, we have proven that the set of all sets do not exist? (Thinking)

 

[Siron]

The proof starts with the assumption: suppose that the set of all sets exists. At the end the proof concludes that $V' \in V' \Leftrightarrow V' \notin V'$ which is a contradiction since a statement can not be equivalent with it's negation. Because of the contradiction the assumption (this is the only thing what was assumed) is incorrect hence the set of all sets does not exist.

 

[evinda]

We want to prove that the set of all sets do not exist.

We suppose that the set of all sets, let $V$, exists.
So, for each set $x, x \in V$.
We define the type $\phi: \text{ a set does not belong to itself , so } x \notin x$.

From the axiom schema of specification, we conclude that there is the set $B=\{ x \in V: x \notin x \}$

$$\forall y(y \in B) \Leftrightarrow y \notin y$$

How can we continue? (Thinking)