Download A Course of Differential Geometry and Topology by Aleksandr Sergeevich Mishchenko PDF

By Aleksandr Sergeevich Mishchenko

This is often basically a textbook for a contemporary direction on differential geometry and topology, that is a lot wider than the normal classes on classical differential geometry, and it covers many branches of arithmetic a data of which has now develop into crucial for a latest mathematical schooling. we are hoping reader who has mastered this fabric should be capable of do self sufficient examine either in geometry and in different comparable fields. to achieve a deeper figuring out of the fabric of this e-book, we suggest the reader may still remedy the questions in A.S. Mishchenko, Yu.P. Solovyev, and A.T. Fomenko, difficulties in Differential Geometry and Topology (Mir Publishers, Moscow, 1985) which used to be especially compiled to accompany this path.

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A d(A,B) U S. Show that B d(x,B) x 1. It follows from d(x, B) = inf{d(x, y) | y ∈ B}, that d(x, B) = d({x}, B) ≥ d(A, B) for every x ∈ A, and the claim is proved. Furthermore, d(x, B) = inf{d(x, y) | y ∈ B} ≤ d(x, y0 ), for y0 ∈ B, inf{d(x , B) | x ∈ A} ≤ d(x, B) ≤ d(x, y) for x ∈ A and y ∈ B. com 49 7. o. Global Analysis It follows from these two inequalities that inf{d(x , B) | x ∈ A} ≤ inf{d(x, y) | x ∈ A, y ∈ B} = d(A, B) ≤ inf{d(x, B) | x ∈ A}, hence we have equality d(A, B) = inf{d(x, B) | x ∈ A}.

Xp ∈ K such that K Bε (x1 ) ∪ · · · ∪ Bε (xp ). 1. Show that a subset K if it is bounded. Rn in the space Rn (with Euclidean metric) is precompact if and only X 2. Let f : X → Y be a mapping between the metric spaces (X, dX ) and (Y, dY ), and let K be a precompact subset in X. Show that if f is uniformly continuous in K, then the image set f (K) Y is precompact in Y . 1. Let K be precompact, and put ε = 1. There are points x1 , . . , xp , such that K B1 (x1 ) ∪ · · · ∪ B1 (xp ). Defining R = max{d(x1 , xj ) | j = 1, .

Then we have the estimate f 1 1 = 0 |f (x)| dx ≥ J |f (x)| dx ≥ c · ε > 0, and the claim follows. This proof should be well-known to the reader, so it is only given here for completeness in a remark. ♦ Furthermore, α·f 1 1 = 0 |α · f (x)| dx = |α| 1 0 |f (x)| dx = |α| · f 1, and f +g 1 1 = 0 |f (x) + g(x)| dx ≤ and we have proved that · 1 1 0 |f (x)| dx + 1 0 |g(x)| dx = f 1 + g 1, is a norm. com 57 9. Normed vector spaces and integral operators Global Analysis 2. It follows from 1 |I(f )| = 0 1 f (x) dx ≤ 0 |f (x)| dx = f that I is continuous.

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