Exercise 2098Logic, sets and proofs › Countability and cardinalityM1Added 3 Nov 2026

Trichotomy of cardinals (statement)

#trichotomy#cardinals#choice

Statement

State the trichotomy theorem for cardinalities (assuming the Axiom of Choice).

Hint

For any two sets $A,B$ , exactly one of $|A|<|B|$ , $|A|=|B|$ , $|A|>|B|$ holds.

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Theorem (with AC). For any sets $A,B$ , exactly one of $|A|<|B|$ , $|A|=|B|$ , $|A|>|B|$ holds.

Equivalent. Every set can be well-ordered; cardinals are linearly ordered.

Without AC. The statement can fail: some cardinalities are incomparable.

Conclusion. With AC, cardinalities form a total order.