By Kenji Ueno, Koji Shiga, Shigeyuki Morita, Toshikazu Sunada

This publication will carry the sweetness and enjoyable of arithmetic to the study room. It deals critical arithmetic in a full of life, reader-friendly kind. integrated are workouts and plenty of figures illustrating the most suggestions.

The first bankruptcy offers the geometry and topology of surfaces. between different subject matters, the authors speak 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 a variety of features of the idea that of size, together with the Peano curve and the Poincaré technique. additionally addressed is the constitution of third-dimensional manifolds. specifically, it truly is 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 way of the authors at Kyoto collage (Japan).

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**Extra resources for A Mathematical Gift, 1: The Interplay Between Topology, Functions, Geometry, and Algebra**

**Sample text**

In the proof of the following theorem, note that the “Case 1” part shows that δ b = εb for each b ∈ A implies δ = ε. In other words, when points δ and ε in N (A) project onto the same “values” on each of the |A | arms of a Cantor star SC (A ), then those points are equal. 2 Theorem (characterization of δ ∼ ε in N (A)) Let |A| ≥ 2, z ∈ A be ﬁxed, A = A \ {z}, and δ, ε ∈ N (A). Then δ ∼ ε if and only if πb (δ) ∼ πb (ε) for every b ∈ A . L. 1007/978-0-387-85494-6 4, c Springer Science+Business Media, LLC 2009 33 34 IMBEDDING JA IN HILBERT SPACE CHAPTER 4 Proof.

And each θβ bijectively corresponds to a number θb ∈ [0, 1] on the bth arm [0, ub ] of the star in l2 (A ). The numbers θb , b ∈ A , are then used as coordinates of (θb ) ∈ ω A ⊂ 2 l (A ), which is the image f (θ) = (θb ) of θ. From this viewpoint, f is the “l2 (A ) synthesis” of the |A |-mappings θ → θβ → θb . 1 Lemma (f : JA → ω A is bijective) Let f be given by θ → f (θ) = (θb ). Then f is a bijection. Proof. By deﬁnition (9) of ω A , the map f is surjective. , θb = ρb for each b ∈ A . It suﬃces to show that δ ∈ p−1 (θ) and ε ∈ p−1 (ρ) imply δ ∼ ε: Since each ψβ : I(ζ, β) → [0, 1] is bijective, we have θb = ρb ∈ [0, ub ] if and only if θβ = ρβ ∈ I(ζ, β).

Each πb is continuous: Given the open-in-C(z, b) set x1 , . . , xn , deﬁne Xi = A \ {b} if xi = z {b} if xi = b, (i = 1, . . , n) and then note that πb−1 ( x1 , . . , xn ) = X1 × · · · × Xn × A × A × · · · . Whether each projection πb is closed or not closed depends on |A|: When A is ﬁnite, then N (A) is compact and each πb is necessarily closed. When A is inﬁnite, then N (A) is not compact and each πb is not a closed mapping — let a1 , a2 , . . be a sequence in A such that i = j implies ai = aj , then (1) F = {a1 ba1 , a2 a2 ba2 , a3 a3 a3 ba3 , .