By Volker Runde

ISBN-10: 0387283870

ISBN-13: 9780387283876

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 continuously interesting workout one has to move via prior to you can actually 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 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 obtainable for college students who've a heritage in calculus and trouble-free algebra, yet now not inevitably in actual or advanced 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 broadly, particularly for an intuitive evidence of Tychonoff's theorem;

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

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Then it is immediate that B (x) ⊂ U1 ∩ U2 . This proves (iii). 5(i) may seem odd at the ﬁrst glance. 5(i). 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.

Analogously, (iii) =⇒ (ii) is proved. 7(iii) for each x ∈ X. We now give an example which shows that continuous maps between general metric spaces can be quite diﬀerent from what we may intuitively expect. 11. Let (X, dX ) and (Y, dY ) be metric spaces such that (X, dX ) is discrete, and let f : X → Y be arbitrary. Let U ⊂ Y be open. Since in a discrete space every set is open, it follows that f −1 (U ) is open. Consequently, f must be continuous. As we have seen, there can be diﬀerent metrics on one set.

Hence, there is > 0 such that B (x0 ) ⊂ N . By (iii), there is δ > 0 such that Bδ (x0 ) ⊂ f −1 (B (f (x0 ))) ⊂ f −1 (N ). This implies that f −1 (N ) ∈ Nx0 . (iv) =⇒ (i): Let (xn )∞ n=1 be a sequence in X with xn → x0 . Let N ∈ Nf (x0 ) , so that f −1 (N ) ∈ Nx0 . Since xn → x0 , there is nN ∈ N such that xn ∈ f −1 (N ) for n ≥ nN ; that is, f (xn ) ∈ N for n ≥ nN . Since N ∈ Nf (x0 ) was arbitrary, this yields f (xn ) → f (x0 ). The following deﬁnition should also look familiar. 8. Let (X, dX ) and (Y, dY ) be metric spaces.

### A Taste of Topology (Universitext) by Volker Runde

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