- #1
evinda
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MHB
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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)
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)