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 continually interesting workout one has to head via sooner than you can still learn nice works of literature within the unique language.

The current booklet grew out of notes for an introductory topology direction on the college of Alberta. It offers a concise creation to set-theoretic topology (and to a tiny bit of algebraic topology). it's available to undergraduates from the second one yr on, yet even starting graduate scholars can take advantage 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 heritage in calculus and easy algebra, yet no longer unavoidably in actual or complicated analysis.

In a few issues, the e-book 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 quick and chic, yet little recognized evidence for the Stone-Weierstrass theorem is given.

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Example text

But, of course, we know that [0, 1] is not open. How is this possible? The answer is that openness (as well as all the notions that are derived from it) depends on the context of a given metric space. Thus, [0, 1] is open in [0, 1], but not open in R. 6. Let (X, d) be a discrete metric space, and let S ⊂ X. Then {x} = S= x∈S B1 (x) x∈S is open; that is, all subsets of X are open. A notion closely related to open sets is that of a neighborhood of a point. 7. Let (X, d) be a metric space, and let x ∈ X.

Hence, all we have to show is that the triangle inequality holds. First note that the function [0, ∞) → R, t→ t 1+t (∗) is increasing (this can be verified, for instance, through differentiation). Let x, y, z ∈ X, and observe that ˜ z) = d(x, ≤ = ≤ = d(x, z) 1 + d(x, z) d(x, y) + d(y, z) , because (∗) is increasing, 1 + d(x, y) + d(y, z) d(x, y) d(y, z) + 1 + d(x, y) + d(y, z) 1 + d(x, y) + d(y, z) d(x, y) d(y, z) + 1 + d(x, y) 1 + d(y, z) ˜ y) + d(y, ˜ z). d(x, Consequently, d˜ is indeed a metric on X.

For x, y ∈ E, define d(x, y) := x − y . This turns E into a metric space. For example, let E be C([0, 1], F), the space of all continuous F-valued functions on [0, 1]. Then there are several norms on E, for example, · 1 defined by 1 f 1 := |f (t)| dt (f ∈ E) 0 or · ∞ given by f ∞ := sup{|f (t)| : t ∈ [0, 1]} (f ∈ E). Each of them turns E into a normed space. (d) Let S = ∅ be a set, and let (Y, d) be a metric space. A function f : S → Y is said to be bounded if sup d(f (x), f (y)) < ∞ x,y∈S The set B(S, Y ) := {f : S → Y : f is bounded} becomes a metric space through D defined by D(f, g) := sup d(f (x), g(x)) x∈S (f, g ∈ B(S, Y )).

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