MATH 5345H --- Week 5: Continuous functions and the general product topology

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0.1  (§18) Continuous functions

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    ★ Def. f:X→Y is continuous if f−1⁢(V) is open in X for every open V⊂Y. A continuous function is called a map.

    Note. Refer back to §2: preimages respect ∪, ∩, −. Images do not. That is the entire reason the definition takes this form.

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    Lem. Composites of maps are maps.

    Proof. (g∘f)−1⁢(W)=f−1⁢(g−1⁢(W)).

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    ★ Lem. It is enough to test a basis: f continuous ⇔f−1⁢(B) open for all B∈ℬ. And enough to test a subbasis: ⇔f−1⁢(S) open for all S∈𝒮.

    Proof. Because f−1 commutes with unions (basis) and with finite intersections (subbasis).

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    Ex. Metric spaces: continuity ⇔ for all x and ε>0 there is δ>0 with d⁢(x,y)<δ⇒d′⁢(f⁢(x),f⁢(y))<ε. The ε–δ definition, recovered.

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    ★ Ex. id:ℝ→ℝℓ is not continuous ([a,b) is not open in ℝ); id:ℝℓ→ℝ is continuous.

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    Def. f is continuous at x if for each neighborhood V of f⁢(x) there is a neighborhood U of x with f⁢(U)⊂V.

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    Thm. f continuous ⇔ f continuous at every x∈X.

    Proof. ⇐ is the gluing trick again: f−1⁢(V)=⋃x∈f−1⁢(V)Ux.

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    ★ Thm. TFAE:

    1. (1)

      f is continuous;

    2. (2)

      f⁢(A¯)⊂f⁢(A)¯ for every A⊂X;

    3. (3)

      f−1⁢(B) is closed for every closed B⊂Y.

    Proof. (1)⇒(2) via the neighborhood criterion for closure; (2)⇒(3) apply to A=f−1⁢(B); (3)⇒(1) complement.

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    ★ Def. f:X→Y is a homeomorphism if it is bijective and both f and f−1 are continuous. Then X≅Y, topologically equivalent.

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    Lem. ≅ is an equivalence relation (identity, inverse, composite).

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    Lem. For f bijective, TFAE: f homeomorphism; U open in X⇔f⁢(U) open in Y; V open in Y⇔f−1⁢(V) open in X.

    Note. So a homeomorphism is a bijection 𝒯X→𝒯Y of the topologies, not just of the points.

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    Def. A topological property is one expressible in terms of points and open sets; such properties are preserved by homeomorphism. E.g. finiteness, discreteness, Hausdorff, and later compactness and connectedness.

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    Ex. (0,1)≅(a,b) via f⁢(x)=(1−x)⁢a+x⁢b.

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    Ex. (−1,1)≅ℝ via f⁢(x)=x/(1−x2),   f−1⁢(y)=2⁢y/(1+1+4⁢y2). Hence every open interval ≅ℝ.

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    ★ Caution. Continuous bijection ≠ homeomorphism.

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      id:ℝℓ→ℝ.

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      f:[0,1)→S1, f⁢(t)=(cos⁡2⁢π⁢t,sin⁡2⁢π⁢t). Here U=[0,12) is open in [0,1) but f⁢(U) is not open in S1: no neighborhood of (1,0) lies inside it.

Constructing maps.

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    Thm. The subspace topology on A⊂X is the coarsest making the inclusion i:A→X continuous.

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    Cor. Restrictions f|A of maps are maps. Lem. Corestrictions X→B (with f⁢(X)⊂B) are maps.

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    Def. f:X→Y is an embedding if the corestriction X→f⁢(X) is a homeomorphism onto the subspace f⁢(X).

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    Lem. f is an embedding ⇔ f=j∘h with h a homeomorphism onto a subspace and j the inclusion. Embeddings are injective.

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    Ex. [0,1)→ℝ2, t↦(cos⁡2⁢π⁢t,sin⁡2⁢π⁢t): injective and continuous, not an embedding.

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    Lem. Maps to a one-point space are continuous; hence constant maps are continuous.

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    ★ Thm. (Pasting lemma.)

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      X=⋃α∈JUα with all Uα open: f continuous ⇔ every f|Uα continuous.

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      X=A1∪⋯∪An with all Ai closed (finitely many): f continuous ⇔ every f|Ai continuous.

    Proof. Open case: f−1⁢(V)=⋃(f|Uα)−1⁢(V), a union of open sets. Closed case: same with closed sets — and now finiteness is essential.

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    ★ Thm. (Maps into a product.) f:A→X×Y is continuous ⇔ both components f1=π1⁢f and f2=π2⁢f are continuous.

    Proof. Subbasis criterion: f−1⁢(U×Y)=f1−1⁢(U), f−1⁢(X×V)=f2−1⁢(V).

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    Cor. The product topology is the coarsest topology making both projections continuous.

0.2  (§19) The product topology (general)

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    ★ Def. On ∏α∈JXα the product topology is generated by the subbasis 𝒮={πβ−1⁢(Uβ)∣β∈J,Uβ⊂Xβ⁢ open}.

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    ★ Lem. Concretely:

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      subbasis elements are ∏αUα with Uα≠Xα for at most one α;

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      basis elements are ∏αUα with Uα≠Xα for finitely many α.

    Note. Emphasise “all but finitely many coordinates unrestricted”. This is the point where students expect the naive definition and get a different one.

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    Thm. f:A→∏Xα continuous ⇔ every component fβ=πβ⁢f is continuous.

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    Cor. The product topology is the coarsest making all πβ continuous.

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    Thm. Aα⊂Xα subspaces: on ∏Aα, product topology = subspace topology from ∏Xα.

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    Thm. All Xα Hausdorff ⇒∏Xα Hausdorff.

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    Thm. ∏αAα¯=∏αAα¯.