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Russell’s paradox

Let A be any set. We define B such as:
B = \{x \in A : x \notin x\}

Can it be that B \in A? No.

Proof by contradiction.

Assume B \in A.

  • If B \in B then by the definition of B:
    B \notin B. Contradiction.
  • If B \notin B then by the definition of B:
    B \in A \text{ and } B \notin B \implies B \in B. Contradiction.

Therefore B \notin A.

Credits for the succinct proof.

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