MATH 5345H --- Week 12: The Tychonoff theorem and compactness of products

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1  The Tychonoff Theorem

1.1  (§37) The Tychonoff theorem

  •  Thm. (Tychonoff.) For any J and any compact spaces Xα, the product αJXα is compact in the product topology.

  • 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

  • Notation. ab(modn) iff nba; classes [a]n=a+n; ring /n with [a]n+[b]n=[a+b]n, [a]n[b]n=[ab]n; surjection φn:/n.

  • Lem. Give each /n and the discrete topology. Then n=1/n is compact Hausdorff, and Φ=(φ1,φ2,):n/n is injective and continuous.

    Proof. Each /n is finite, hence compact; Tychonoff. Injective: if nba for all n, take n>|ba|.

  • Def. For dm, reduction ρm,d:/m/d, [a]m[a]d; note ρm,dφm=φd. Set

    ^={(xn)nn=1/n|ρm,d(xm)=xd for all dm},

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

  •  Lem. ^ is a closed subspace of n/n, hence compact Hausdorff.

    Proof.^=dmfm,d1([0]d) where fm,d((xn))=ρm,d(xm)xd is continuous and /d is discrete.

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

    Proof. Given p^ 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.

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

  •  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 is closed. If there were finitely many primes, A=pp would be closed, so its complement {±1} would be open and finite — contradiction.