MATH 5345H --- Week 12: The Tychonoff theorem

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1  The Tychonoff Theorem

1.1  (§37) The Tychonoff theorem

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    ★ Thm. (Tychonoff.) For any J and any compact spaces Xα, the product ∏α∈JXα is compact in the product topology.

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    Rmk. Proved above for finite J (tube lemma). The general case is omitted here; it needs the axiom of choice. Application below instead.

1.2  The profinite integers

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    Notation. a≡b⁢(mod⁢n) iff n∣b−a; classes [a]n=a+n⁢ℤ; ring ℤ/n with [a]n+[b]n=[a+b]n, [a]n⋅[b]n=[a⁢b]n; surjection φn:ℤ→ℤ/n.

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    Lem. Give each ℤ/n and ℤ the discrete topology. Then ∏n=1∞ℤ/n is compact Hausdorff, and Φ=(φ1,φ2,…):ℤ→∏nℤ/n is injective and continuous.

    Proof. Each ℤ/n is finite, hence compact; Tychonoff. Injective: if n∣b−a for all n, take n>|b−a|.

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    Def. For d∣m, reduction ρm,d:ℤ/m→ℤ/d, [a]m↦[a]d; note ρm,d∘φm=φd. Set

    ℤ^={(xn)n∈∏n=1∞ℤ/n|ρm,d⁢(xm)=xd⁢ for all ⁢d∣m},

    the ring of profinite integers (operations termwise; a topological ring).

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    ★ Lem. ℤ^ is a closed subspace of ∏nℤ/n, hence compact Hausdorff.

    Proof. ℤ^=⋂d∣mfm,d−1⁢([0]d) where fm,d⁢((xn))=ρm,d⁢(xm)−xd is continuous and ℤ/d is discrete.

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    Lem. Φ corestricts to an injective continuous ring map Ψ:ℤ→ℤ^, with dense image.

    Proof. Given p∈ℤ^ and a basis neighborhood constrained at finitely many n, let N be a common multiple of those n and choose a with [a]N=xN; then [a]n=xn for each constrained n.

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    Def. The subspace topology on ℤ from ℤ^ is the Fürstenberg topology. (Ψ is not an embedding of discrete ℤ.)

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    ★ Thm. (Euclid, ca. 300 BC.) There are infinitely many primes.

    Proof. Fürstenberg’s 1955 argument. In the Fürstenberg topology every nonempty open set is infinite, and each p⁢ℤ is closed. If there were finitely many primes, A=⋃pp⁢ℤ would be closed, so its complement {±1} would be open and finite — contradiction.