Mathematical logic and foundations Proof

Proof of Russell's paradox

Proof

We argue by contradiction.

Suppose that there is a set $R$ such that, for each set $x$ , $x \in R$ if and only if $x \notin x$ .
The object $R$ is itself a set. So we may take $x = R$ in step 1. This gives: $R \in R$ if and only if $R \notin R$ .
Suppose $R \in R$ . By step 2, $R \notin R$ . This contradicts the assumption $R \in R$ .
So $R \in R$ is false, that is, $R$ is not an element of $R$ . In symbols, $R \notin R$ .
By step 2 and step 4, $R \in R$ .
Steps 4 and 5 give both $R \notin R$ and $R \in R$ , a contradiction.
So the assumption of step 1 is false. There is no set $R$ such that, for each set $x$ , $x \in R$ if and only if $x \notin x$ .

Idea

Ask whether $R$ contains itself. The defining property of $R$ says that $R \in R$ exactly when $R \notin R$ . No statement can be equivalent to its own negation, so no such $R$ exists.

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Element definition
Set definition
Proof of Russell's paradox proof

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Tag	`4CVPENYR0S`
URI	`m2://4CVPENYR0S`
Kind	proof
Writing	W0 stub
MSC	03E30
Proves	Russell's paradox
Uses	Set, Element
Methods	Proof by contradiction
Changed	4 November 2026

Proof of Russell's paradox

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