By Volker Runde

If arithmetic is a language, then taking a topology path on the undergraduate point is cramming vocabulary and memorizing abnormal verbs: an important, yet now not continually intriguing workout one has to move via ahead of you'll learn nice works of literature within the unique language.

The current publication grew out of notes for an introductory topology path on the collage 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 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 easy algebra, yet no longer unavoidably in genuine or advanced analysis.

In a few issues, the ebook treats its fabric in a different way than different texts at the subject:

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

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

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

**Read or Download A Taste of Topology (Universitext) PDF**

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**Extra resources for A Taste of Topology (Universitext)**

**Sample text**

Let (X, d) be a metric space, and let U1 , . . , Un ⊂ X be dense open subsets of X. Then U1 ∩ · · · ∩ Un is dense in X. Proof. By induction, it is clear that we may limit ourselves to the case where n = 2. Let x ∈ X, and let > 0. Since U1 is dense in X, we have B (x) ∩ U1 = ∅. Since B (x)∩U1 is open—and thus a neighborhood of each of its points—it follows from the denseness of U2 that B (x) ∩ U1 ∩ U2 = ∅. Since > 0 was arbitrary, we conclude that x ∈ U1 ∩ U2 . 16 (Baire’s theorem). Let (X, d) be a complete metric space, and let (Un )∞ n=1 be a sequence of dense open subsets of X.

Fix h ∈ (0, 1) and > 0, and choose m ∈ N so large that ⎧ ⎫ ⎨ |f (t + h) − f (tm + h)| ⎬ f − fm ∞ (m ≥ m ). < h ⎩ ⎭ 4 |f (tm ) − f (t)| For m ≥ m , this implies |f (t + h) − f (t)| ≤ |f (t + h) − f (tm + h)| + |f (tm + h) − fm (tm + h)| <4h <4h + |fm (tm + h) − fm (tm )| + |fm (tm ) − f (tm )| + |f (tm ) − f (t)| ≤nh ≤ nh + h, <4h <4h 50 2 Metric Spaces so that |f (t + h) − f (t)| ≤n+ . h Since h and were arbitrary, this means that f ∈ Fn . Hence, Fn is closed. Assume towards a contradiction that every f ∈ C([0, 2], R) is diﬀerentiable ∞ at some point in [0, 1], so that C([0, 2], R) = n=1 Fn .

Conversely, let x ∈ S, and suppose that x ∈ / S. 13 yields that N ∩ S = ∅ as well. The closure of a given set is, by deﬁnition, the smallest closed set containing it. Analogously, one deﬁnes the largest open set contained in a given set. 22. Let (X, d) be a metric space. For each S ⊂ X, the interior of S is deﬁned as ◦ S := {U : U ⊂ X is open and contained in S}. 23. Let (X, d) be a metric space, and let S ⊂ X. Then we have: ◦ S = {x ∈ X : S ∈ Nx } = S \ ∂S. ◦ Proof. Let x ∈S . Then there is an open subset U of S with x ∈ U , so that S ∈ Nx .