A Taste of Topology by Volker Runde (auth.), S Axler, K.A. Ribet (eds.)

By Volker Runde (auth.), S Axler, K.A. Ribet (eds.)

If arithmetic is a language, then taking a topology direction on the undergraduate point is cramming vocabulary and memorizing abnormal verbs: an important, yet now not consistently interesting workout one has to head via sooner than you could learn nice works of literature within the unique language.

The current ebook grew out of notes for an introductory topology path on the college of Alberta. It presents a concise creation to set-theoretic topology (and to a tiny bit of algebraic topology). it really is obtainable to undergraduates from the second one 12 months on, yet even starting graduate scholars can make the most of a few parts.

Great care has been dedicated to the choice of examples that aren't self-serving, yet already available for college kids who've a historical past in calculus and effortless algebra, yet now not inevitably in genuine or complicated analysis.

In a few issues, the e-book treats its fabric otherwise than different texts at the subject:

* Baire's theorem is derived from Bourbaki's Mittag-Leffler theorem;

* Nets are used widely, particularly for an intuitive facts of Tychonoff's theorem;

* a brief and stylish, yet little identified facts for the Stone-Weierstrass theorem is given.

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Yn }, we obtain that B (x) ∩ Y ⊂ B (x) ∩ Bδy1 (y1 ) ∪ · · · ∪ Bδyn (yn ) = ∅ and thus B (x) ⊂ X \ Y . Since x ∈ X \ Y was arbitrary, this means that X \ Y is open. 5. Let (K, dK ) be a compact metric space, let (Y, dY ) be any metric space, and let f : K → Y be continuous. Then f (K) is compact. Proof. Let U be an open cover for f (K). 10. Hence, there are U1 , . . , Un ∈ U with K = f −1 (U1 ) ∪ · · · ∪ f −1 (Un ) and thus f (K) ⊂ U1 ∪ · · · ∪ Un . This proves the claim. 6. Let (K, d) be a non-empty, compact metric space, and let f : K → R be continuous.

There are U1 , . . , Un ∈ U such that X = U1 ∪ · · · ∪ Un ∪ X \ Y. Taking the intersection with Y , we see that Y ⊂ U1 ∪ · · · ∪ Un . For (ii), let x ∈ X \ Y . For each y ∈ Y , there are y , δy > 0 such that B y (x) ∩ Bδy (y) = ∅. Since {Bδy (y) : y ∈ Y } is an open cover for Y , there are y1 , . . , y1 ∈ Y such that Y ⊂ Bδy1 (y1 ) ∪ · · · ∪ Bδyn (yn ). Letting := min{ y1 , . . , yn }, we obtain that B (x) ∩ Y ⊂ B (x) ∩ Bδy1 (y1 ) ∪ · · · ∪ Bδyn (yn ) = ∅ and thus B (x) ⊂ X \ Y . Since x ∈ X \ Y was arbitrary, this means that X \ Y is open.

20. Let (X, d) be a metric space, and let S ⊂ X. Then the boundary of S is defined as ∂S := {x ∈ X : B (x) ∩ S = ∅ and B (x) ∩ (X \ S) = ∅ for all > 0}. 13 yields immediately that ∂S = {x ∈ X : N ∩ S = ∅ and N ∩ (X \ S) = ∅ for all N ∈ Nx } for each subset S of a metric space X. 21. Let (X, d) be a metric space, and let S ⊂ X. Then: (i) ∂S = ∂(X \ S); (ii) ∂S is closed; (iii) S = S ∪ ∂S. Proof. (i) is a triviality. For (ii), let x ∈ X \ ∂S; that is, there is N ∈ Nx such that N ∩ S = ∅ or N ∩ (X \ S) = ∅.

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