Maths WorkshopMaths problems with full solutions
Exercise 1560Logic, sets and proofs › Countability and cardinalityL1Added 3 Nov 2026

Cantor's theorem

#Cantor theorem#power set#larger cardinality

Statement

Prove ∣A∣<∣P(A)∣|A|<|\mathcal{P}(A)| for any set AA.

Hint

Suppose f:A→P(A)f:A\to\mathcal{P}(A) is a surjection, consider S={x∈A:x∉f(x)}S=\{x\in A:x\notin f(x)\}.

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Solution

Injection. a↦{a}a\mapsto\{a\} injects AA into P(A)\mathcal{P}(A).

No surjection. If ff surjective, let S={x∈A:x∉f(x)}S=\{x\in A:x\notin f(x)\}. Since ff surjective, S=f(a)S=f(a) for some aa. Then a∈S  ⟺  a∉f(a)=Sa\in S\iff a\notin f(a)=S, contradiction.

Conclusion. ∣A∣<∣P(A)∣|A|<|\mathcal{P}(A)|; power sets strictly increase cardinality.