By Kenji Ueno, Koji Shiga, Shigeyuki Morita
This publication will deliver the wonder and enjoyable of arithmetic to the study room. It deals severe arithmetic in a full of life, reader-friendly type. incorporated are workouts and plenty of figures illustrating the most ideas.
The first bankruptcy provides the geometry and topology of surfaces. between different themes, the authors talk about the Poincaré-Hopf theorem on serious issues of vector fields on surfaces and the Gauss-Bonnet theorem at the relation among curvature and topology (the Euler characteristic). the second one bankruptcy addresses quite a few elements of the idea that of measurement, together with the Peano curve and the Poincaré strategy. additionally addressed is the constitution of three-d manifolds. specifically, it's proved that the 3-dimensional sphere is the union of 2 doughnuts.
This is the 1st of 3 volumes originating from a sequence of lectures given via the authors at Kyoto college (Japan).
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Additional info for A Mathematical Gift III: The Interplay Between Topology, Functions, Geometry, and Algebra (Mathematical World, Volume 23)
If G E I and G=UtF, with each F, closed, define F E I X o by F(x,t)ex~F, and notice that F is closed and G(x)w (3t)F(x, t ) . Thus G is Xi, and since G was arbitrary open, s. 2: E Hence Z: = 3” i 2: E s” i 2 : = 2; and inductively, 2 : E Z:+ 1. This establishes 2: E AZ+I for every n, so taking negations, II~E and the remaining inclusions in the diagram are trivial. 2. -X X , is a product space with at least one factor Xi = X and every Xieither o o r X , then I is homeomorphic with x. Hint. Construct homeomorphisms of o x X and X use induction on k.
The section above y. r A pointclass is Y-parametrized if for every product space X there is some G G Y X X which is universal for r l X . Let No, N1, N2Y. Thus 2; is X parametrized and it is trivial to prove from this that all the Bore1 pointclasses 2 : and their duals IIz are JV-parametrized. The next theorem establishes a little more. l. THEPARAMETRIZATION THEOREM FOR2 :. Let N ( Y , 0), N ( Y , 1),... and N(X, 0), N(X, 1),... enumerate bases for the topology of y and a fixed product space X respectively.
Thus G is Xi, and since G was arbitrary open, s. 2: E Hence Z: = 3” i 2: E s” i 2 : = 2; and inductively, 2 : E Z:+ 1. This establishes 2: E AZ+I for every n, so taking negations, II~E and the remaining inclusions in the diagram are trivial. 2. -X X , is a product space with at least one factor Xi = X and every Xieither o o r X , then I is homeomorphic with x. Hint. Construct homeomorphisms of o x X and X use induction on k. 3. Prove that if I = XI X X X, is a product space with at least one factor Xi not w, then I is a perfect Polish space.