By Igor Pak
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Extra resources for Lectures on Discrete and Polyhedral Geometry (Draft)
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Let us remark here that not all pairs of points at the same height may necessarily appear as (u1 , u2 ). 12. Line L⊥ℓ through midpoints of an interval (u1 , u2 ) ℓ. For every t ∈ [0, 1], consider a vertical line L through the midpoints of (u1 (t), u2 (t)) and plot the height h = ϕ(x) of the intersection points x ∈ L ∩ X in a graph Γ = {(t, h), 0 ≤ t ≤ 1}. By construction, the resulting graph Γ is a union of non-intersecting piecewise linear curves. There are at least two curve endpoints at t = 0 and at least two curve endpoints at t = 1.
One of the four possible local changes of A. Third proof. Consider the subset A ⊂ [0, 1]2 defined in the first proof. Think of A as a finite graph with straight edges. Observe that except for (0, 0) and (1, 1), every vertex has degree 2 or 4, where the vertices of degree 4 appear when two corresponding peaks of two valleys are at the same level. This immediately implies that (0, 0) and (1, 1) lies in the same connected component of A. 6. Let us mention that the first proof is misleadingly simple and suggests that the mountain climbing lemma holds for all continuous functions.
Thus, [z, z + 12 ] is the desired inscribed chord of length 21 . Now that we know that some distances always occur as lengths of inscribed chords, let us restate the problem again. Denote by D(f ) the set of distances between points with equal values: D(f ) = y − x | f (x) = f (y), 0 ≤ x ≤ y ≤ 1 . Observe that for a given f , the set D(f ) contains all distances small enough. 2). In fact the only distances guaranteed to be in D(f ) are given by following theorem. 5 (Inscribed chord theorem). For every continuous function f : [0, 1] → R with f (0) = f (1) = 0, we have n1 ∈ D(f ), for all n ∈ N.