IMC 1996 · Day 1 · P6

(a) The choice $E_1 = L$ gives $\mathcal{C}(L) \le \operatorname{lenght}(L)$. If $E \subset \bigcup\limits_{i=1}^{n} E_i$ then $\sum\limits_{i=1}^{n} \operatorname{diam}(E_i) \ge \operatorname{lenght}(L)$: By induction, $n = 1$ obvious, and assuming that $E_{n+1}$ contains the end point $a$ of $L$, define the segment $L_\varepsilon = \{ x \in L : \operatorname{dist}(x,a) \ge \operatorname{diam}(E_{n+1}) + \varepsilon \}$ and use induction assumption to get $\sum\limits_{i=1}^{n+1} \operatorname{diam}(E_i) \ge \operatorname{lenght}(L_\varepsilon) + \operatorname{diam}(E_{n+1}) \ge \operatorname{lenght}(L) - \varepsilon$; but $\varepsilon > 0$ is arbitrary.

(b) If $f$ is a contraction of $E$ onto $L$ and $E \subset \bigcup\limits_{n=1}^{n} E_i$, then $L \subset \bigcup\limits_{i=1}^{n} f(E_i)$ and $\operatorname{lenght}(L) \le \sum\limits_{i=1}^{n} \operatorname{diam}(f(E_i)) \le \sum\limits_{i=1}^{n} \operatorname{diam}(E_i)$.

(c1) Let $E = T \cup T'$ where $T$ is the triangle with vertices $(-2,2)$, $(2,2)$ and $(0,4)$, and $T'$ is its reflexion about the $x$-axis. Suppose $E \subset \bigcup\limits_{i=1}^{n} E_i$. If no set among $E_i$ meets both $T$ and $T'$, then $E_i$ may be partitioned into covers of segments $[(-2,2),(2,2)]$ and $[(-2,-2),(2,-2)]$, both of length 4, so $\sum\limits_{i=1}^{n} \operatorname{diam}(E_i) \ge 8$. If at least one set among $E_i$, say $E_k$, meets both $T$ and $T'$, choose $a \in E_k \cap T$ and $b \in E_k \cap T'$ and note that the sets $E_i' = E_i$ for $i \ne k$, $E_k' = E_k \cup [a,b]$ cover $T \cup T' \cup [a,b]$, which is a set of upper content at least $8$, since its orthogonal projection onto $y$-axis is a segment of length $8$. Since $\operatorname{diam}(E_j) = \operatorname{diam}(E_j')$, we get $\sum\limits_{i=1}^{n} \operatorname{diam}(E_i) \ge 8$.

(c2) Let $f$ be a contraction of $E$ onto $L = [a', b']$. Choose $a = (a_1, a_2)$, $b = (b_1, b_2) \in E$ such that $f(a) = a'$ and $f(b) = b'$. Since $\operatorname{lenght}(L) = \operatorname{dist}(a', b') \le \operatorname{dist}(a, b)$ and since the triangles have diameter only 4, we may assume that $a \in T$ and $b \in T'$. Observe that if $a_2 \le 3$ then $a$ lies on one of the segments joining some of the points $(-2,2)$, $(2,2)$, $(-1,3)$, $(1,3)$; since all these points have distances from vertices, and so from points, of $T'$ at most $\sqrt{50}$, we get that $\operatorname{lenght}(L) \le \operatorname{dist}(a,b) \le \sqrt{50}$. Similarly if $b_2 \ge -3$. Finally, if $a_2 > 3$ and $b_2 < -3$, we note that every vertex, and so every point of $T'$ is in the distance at most $\sqrt{10}$ for $a$ and every vertex, and so every point, of $T$ is in the distance at most $\sqrt{10}$ of $b$. Since $f$ is a contraction, the image of $T'$ lies in a segment containing $a'$ of length at most $\sqrt{10}$ and the image of $T$ lies in a segment containing $b'$ of length at most $\sqrt{10}$. Since the union of these two images is $L$, we get $\operatorname{lenght}(L) \le 2\sqrt{10} \le \sqrt{50}$. Thus $\mathcal{K}(E) \le \sqrt{50} < 8$.