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The Claim

Every infinite set contains a countably infinite subset.

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The Short Version

As stated, the theorem is too broad. In ordinary ZFC-based mathematics, every infinite set does contain a countably infinite subset, but without some form of the Axiom of Choice this is not generally true. In ZF, there can be infinite Dedekind-finite sets with no countably infinite subset, so the omitted assumption materially changes the claim.

Caveats

  • The claim omits a load-bearing hypothesis: it is valid in ZFC, but not in bare ZF set theory.
  • Without choice, an infinite set need not contain a countably infinite subset; Dedekind-finite counterexamples can exist.
  • Many textbook or homework-style sources state the result without making their background axioms explicit.

The Receipts

  1. set theory - Is it possible to show that an infinite set has a countable infinite subset without using Choice?

    MathOverflow

  2. Is my proof correct? => Prove that any infinite set contains a countably infinite subset

    Reddit

  3. Arithmetic of infinite Dedekind-finite sets

    LSA (University of Michigan)

  4. Infinite set

    Wikipedia

  5. Infinite set always has a countably infinite subset

    Math StackExchange

  6. Infinite Set has Countably Infinite Subset

    ProofWiki

  7. 9.5: Countable sets

    Mathematics LibreTexts

  8. Countability 1 - CS173 Lectures

    University of Illinois Urbana-Champaign

  9. Prove that every infinite set has a countable subset. [duplicate]

    Math StackExchange

  10. CHAPTER FIVE: INFINITIES

    Ohio State University

+ 8 more sources — see the full list on Lenz

Filed Under

Axiom Of ChoiceSet Theory

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