Read e-book online A Taste of Topology (Universitext) PDF

By Volker Runde

ISBN-10: 0387283870

ISBN-13: 9780387283876

If arithmetic is a language, then taking a topology direction on the undergraduate point is cramming vocabulary and memorizing abnormal verbs: an important, yet no longer regularly fascinating workout one has to move via prior to you could learn nice works of literature within the unique language.

The current publication grew out of notes for an introductory topology direction on the college of Alberta. It presents a concise advent to set-theoretic topology (and to a tiny bit of algebraic topology). it's obtainable to undergraduates from the second one 12 months on, yet even starting graduate scholars can reap the benefits of a few parts.

Great care has been dedicated to the choice of examples that aren't self-serving, yet already available for college students who've a historical past in calculus and straightforward algebra, yet no longer inevitably in actual or complicated analysis.

In a few issues, the publication treats its fabric otherwise than different texts at the subject:
* Baire's theorem is derived from Bourbaki's Mittag-Leffler theorem;
* Nets are used largely, particularly for an intuitive evidence of Tychonoff's theorem;
* a brief and chic, yet little recognized evidence for the Stone-Weierstrass theorem is given.

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Additional info for A Taste of Topology (Universitext)

Sample text

Show that a finite subset of a metric space is closed. 2. Let (E, · ) be a normed space, let U ⊂ E be open, and let S ⊂ E be any set. Show that S + U := {x + y : x ∈ S, y ∈ U } is open in E. 3. Let U ⊂ R be open. (a) For each x ∈ U , let Ix be the union of all open intervals contained in U and containing x. Show that Ix is an open (possibly unbounded) interval. (b) For x, y ∈ U , show that Ix = Iy or Ix ∩ Iy = ∅. (c) Conclude that U is a union of countably many, pairwise disjoint open intervals.

C) Let S = ∅ be a set, and let (Y, d) be a complete metric space. 2(d) is complete. Let > 0, and choose n > 0 (fn )∞ n=1 be a Cauchy sequence in B(S, Y ). Let such that D(fn , fm ) < for all n, m ≥ n . For x ∈ S, we then have d(fn (x), fm (x)) ≤ D(fn , fm ) < (n, m ≥ n ). Consequently, (fn (x))∞ n=1 is a Cauchy sequence in Y for each x ∈ S. Since Y is complete, we can therefore define f : S → Y by letting f (x) := lim fn (x) n→∞ (x ∈ S). We first claim that f lies in B(S, Y ) and is, in fact, the limit of (fn )∞ n=1 with respect to D.

Let X be equipped with a sequence (dn )∞ n=1 of semimetrics such that, for any x, y ∈ X with x = y, there is n ∈ N with dn (x, y) > 0. Then d : X × X → R defined by ∞ d(x, y) := 1 dn (x, y) 2n 1 + dn (x, y) n=1 (x, y ∈ X) is a metric. Clearly, d is symmetric and satisfies the triangle inequality, and if x, y ∈ X are such that x = y, there is n ∈ N with dn (x, y) > 0, so dn (x,y) that d(x, y) ≥ 21n 1+d > 0. n (x,y) (h) The previous example can be used, for instance, to turn a Cartesian product X of countably many metric spaces ((Xn , dn ))∞ n=1 into a metric space again.

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A Taste of Topology (Universitext) by Volker Runde


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