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Exercise 1882Logic, sets and proofs › Sets and mapsL1Added 3 Nov 2026

Russell's paradox

#Russell paradox#set theory#self-reference

Statement

State Russell's paradox and explain why it forces restrictions on naive set theory.

Hint

Consider R={S:S∉S}R=\{S:S\notin S\}.

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Solution

Paradox. Let R={S:S∉S}R=\{S:S\notin S\} (the set of all sets that are not elements of themselves). Then:
- If R∈RR\in R, then R∉RR\notin R (by definition of RR).
- If R∉RR\notin R, then R∈RR\in R (since RR would be a set not containing itself).

Contradiction. Both cases lead to contradiction.

Resolution. Naive set theory (any property defines a set) is inconsistent. ZFC restricts comprehension to elements of an existing set: {x∈A:P(x)}\{x\in A:P(x)\}.

Conclusion. Russell's paradox motivates axiomatic set theory.