Download A Mathematical Gift I: The Interplay Between Topology, by Kenji Ueno, Koji Shiga, Shigeyuki Morita PDF

By Kenji Ueno, Koji Shiga, Shigeyuki Morita

This e-book will convey the sweetness and enjoyable of arithmetic to the study room. It bargains critical arithmetic in a full of life, reader-friendly type. incorporated are workouts and plenty of figures illustrating the most innovations.
The first bankruptcy offers 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 features of the concept that of measurement, together with the Peano curve and the Poincaré method. additionally addressed is the constitution of 3-dimensional manifolds. particularly, 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 by means of the authors at Kyoto collage (Japan).

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Additional info for A Mathematical Gift I: The Interplay Between Topology, Functions, Geometry, and Algebra (Mathematical World, Volume 19)

Example text

No one else uses the word and hence no confusion should result. 48 IL Topologica/ Spaces Exercise 1. The passage from set to derived set possesses the following properties: (a) (b) (c) (d) d(dM) d(MvN) = d(M)ud(N). d(dM) ç Mud(M). d(o) = 0. For all p e Ry we have p $ d({p}). We can not in general sharpen (b) to c d(M). 2. Conversely, let a mapping d: ty(R) -> ty(R) verify conditions (a)-(d). Then M = M u d(M) defines a closure operator on R, and the derived set of M defined by this closure operator ({/> : p E M n Gp}) coincides with d(M).

An e 91, constitute a basis of b. Moreover, (2^) implies D e 91, whence 1—1 m 91 < b . 9 77*£ lattice of filters, F(R), is distributive. In fact, we even have the sharper equality Û v Λ b< = Λ (a v bf) Proof: For all indices t0 , we have a v /\ b( ^ a v bt . This implies a v Λ b, < Λ (a v bf) (*) In order to prove equality in (*), we must show that every set C belonging to the left-hand filter of (*) also belongs to the right-hand filter. 6 we have C = A υ (B1 n ... n Bn) = (A u Bx) n ... , «.

If the topology τ defined on the set R has previously been mentioned and no confusion can occur, we shall often write R rather than (R, r ) and call R itself the topological space. Since in this book we shall deal almost exclusively with topological spaces, the adjective "topological" will often also be dropped and the word space used alone. Note also that we shall always consider the empty set to be a topological space—this will allow us to state certain results without special provisos to cover exceptional cases.

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