By Walter D. Neumann, B. N. Apanasov, B. N. Apanasov, Alan W. Reid, Ohio State University International Mathematical Research Institute
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Unfortunately, as experiment will show, the whole thing gets hopelessly tangled. The point is, that this sort of model making is impossible in R3 —an extra dimension is needed. ) S2 \ D D1 b b D1 E M E Fig. 8 The question is: can we anyway say what we mean by this stitching process without having to produce the result as a subset of R3 ? One of the properties of the model we should like is that if in Fig. 4] 17 S2 is identified with b in M, then the curve shown should be continuous. This can be arranged by defining neighbourhoods suitably.
For each λ ∈ U there is a basic neighbourhood M × N of λ such that M × N ⊆ U. Let Uλ = Int M, Vλ = Int N. Then Uλ , Vλ are open and U = λ∈U Uλ × Vλ . ✷ E XAMPLE Let α = (a, b) ∈ R2 , and let r > 0. The open ball about α of radius r is the set B(α, r) = {(x, y) ∈ R2 : (x − a)2 + (y − b)2 < r2 }. This open ball is an open set: For, let α = (a , b ) ∈ B(α, √r) and let s = (a − a)2 + (b − b)2 . Then s < r. Let 0 < δ < (r − s)/ 2, M = ]a − δ, a + δ[, N = ]b − δ, b + δ[. Then M × N ⊆ B(α, r) and so B(α, r) is a neighbourhood of α .
6 Let f : Z → X, g : Z → Y be maps. Then (f, g) : Z → X × Y is a map. Proof Let h = (f, g), so that h sends z → (f(z), g(z)). Let P be a neighbourhood of h(z), and let M × N be a basic neighbourhood of h(z) contained in P. Then h−1 [P] contains the set h−1 [M × N] = {z ∈ Z : f(z) ∈ M, g(z) ∈ N} = f−1 [M] ∩ g−1 [N]. It follows that h−1 [P] is a neighbourhood of z. This result can also be expressed: a function h : Z → X × Y is continuous ⇔ p1 h, p2 h are continuous. 6 since p1 h, p2 h are the components f, g of h.