By Weil W.

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Extra info for A course on convex geometry

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1. 2. For K, L ∈ Kn , we have d(K, L) = hK − hL . Therefore, d is a metric on Kn and fulfills d(K + M, L + M ) = d(K, L), for all K, L, M ∈ Kn . Proof. 1 we obtain K ⊂ L + B(ε) ⇔ hL ≤ hK + εhB(1) and L ⊂ K + B(ε) ⇔ hK ≤ hL + εhB(1) . Since hB(1) ≡ 1 on S n−1 , this implies K ⊂ L + B(ε), L ⊂ K + B(ε) ⇔ hK − hL ≤ ε, and the assertions follow. In an arbitrary metric space (X, d), the class C(X) of nonempty compact subsets of ˜ which is defined by X can be supplied with the Hausdorff metric d, ˜ B) := max (max d(x, B), max d(y, A)).

3 to show that the origin 0 can be separated from A−B. 5. Let A, K ⊂ Rn be convex, A closed, K compact, and assume A ∩ K = ∅. Show that there is a hyperplane {f = α} with A ⊂ {f < α} and B ⊂ {f > α}. Show more generally that α can be chosen such that there is an > 0 with A ⊂ {f ≤ α− } and B ⊂ {f ≥ α + } (strong separation). 6. A bavarian farmer is happy owner of a large herd of happy cows, consisting of totally black and totally white animals. One day he finds them sleeping in the sun on his largest meadow.

1 we obtain K ⊂ L + B(ε) ⇔ hL ≤ hK + εhB(1) and L ⊂ K + B(ε) ⇔ hK ≤ hL + εhB(1) . Since hB(1) ≡ 1 on S n−1 , this implies K ⊂ L + B(ε), L ⊂ K + B(ε) ⇔ hK − hL ≤ ε, and the assertions follow. In an arbitrary metric space (X, d), the class C(X) of nonempty compact subsets of ˜ which is defined by X can be supplied with the Hausdorff metric d, ˜ B) := max (max d(x, B), max d(y, A)). d(A, x∈A y∈B Here A, B ∈ C(X), and we have used the abbreviation d(u, C) := min d(u, v), v∈C u ∈ X, C ∈ C(X), (the minimal and maximal values exist due to the compactness of the sets and the continuity of the metric).