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 no longer consistently fascinating workout one has to move via ahead of you will 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 advent to set-theoretic topology (and to a tiny bit of algebraic topology). it truly is available to undergraduates from the second one yr 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 historical past in calculus and basic algebra, yet now not unavoidably in actual or complicated analysis.
In a few issues, the ebook treats its fabric otherwise than different texts at the subject:
* Baire's theorem is derived from Bourbaki's Mittag-Leffler theorem;
* Nets are used commonly, specifically for an intuitive facts of Tychonoff's theorem;
* a quick and stylish, yet little recognized facts for the Stone-Weierstrass theorem is given.
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Additional info for A Taste of Topology
If U = X, we have dU = d, so that the claim is trivially true. Hence, suppose that U X. ∞ Let (xn )∞ n=1 be a Cauchy sequence in (U, dU ). Then (xn )n=1 is easily seen to be a Cauchy sequence in (X, d) as well. Let x ∈ X be its limit in (X, d). We ﬁrst claim that x ∈ U . Assume towards a contradiction that x ∈ X \ U . 3, we conclude that dist(xn , X \ U ) → 0. Since (xn )∞ n=1 is a Cauchy sequence in (U, dU ), there is n1 ∈ N such that 1 1 − ≤ dU (xn , xm ) ≤ 1 dist(xn , X \ U ) dist(xm , X \ U ) (n, m ≥ n1 ).
G2 ε_ 2 g3 g1 0 2 1 t Fig. 3: Sawtooth functions Then gk is continuous with gk ∞ = 2 , but |gk (t + h) − gk (t)| = k h 2 h∈(0,1) sup (†) holds for any t ∈ [0, 1]. Since f + gk ∈ B (f ) ⊂ Fn0 , there is t ∈ [0, 1] such that |(f + gk )(t + h) − (f + gk )(t)| sup ≤ n0 . 4 Completeness sup h∈(0,1) 51 |gk (t + h) − gk (t)| h |(f + gk )(t + h) − (f + gk )(t)| |f (t + h) − f (t)| + sup h h h∈(0,1) h∈(0,1) ≤ sup = n0 + f ∞, which contradicts (†) if we choose k ∈ N so large that 2 k > n0 + f ∞ . ∞ Hence, the sets F1 , F2 , .
Let (X, d) be a metric space, and let ∅ = S ⊂ X. Show that the function X → R, x → dist(x, S) is continuous. 4. Let E and F be normed spaces, and let T : E → F be linear. Show that the following are equivalent. (i) T is continuous; (ii) T is continuous at 0; (iii) There is C ≥ 0 such that T (x) ≤ C x for all x ∈ E. 5. Let E and F be normed spaces, let T : E → F be linear, and suppose that dim E < ∞. Show that T is continuous. ) 6. 2(c). Show that the metrics induced by these two norms are not equivalent.