# 1617-257/TUT-R-8

Problem 1. Let $E$ be an infinite subset of a compact metric space $X$. Show that $E$ has a limit point.
Proof. If $E$ has no limit points, then $E$ is a closed subset of a compact space and is therefore compact in itself. Since each point of $E$ is isolated, we may find for each point $e \in E$ a neighborhood $U_e$ such that $E \cap U_e = \{ e \}$. The collection $\{ U_e\}_{e \in E}$ is an open cover of E which clearly has no finite subcover.
Problem 2. Let $f: B_1(0) \to \mathbb{R}^2$ be a function which is "jelly-rigid": for all $x,y \in B_1(0)$: $|f(x) - f(y) - (x - y)| \leq 0.1 |x - y|$. Prove that $f$ maps onto $B_{0.4}(0)$.
Proof. Since $f$ is Lipschitz, it has a unique extension to its boundary and so we regard it as a function on $B:= \overline{B_1(0)}$. A simple estimate shows if $f(x) \in B_{0.4}(0)$, then $x \in B_1(0)$. Suppose now that there is some point $z \in B_{0.4}(0)$ which is not in the image of $f$. Let $x_0 \in B_1(0)$ be a closest element to $z$ in the image of $f$. Consider now the point $x_1 := x_0 + \delta (z - f(x_0))$ (jelly-rigidity says that the function is almost like the identity, so moving closer to the point $z$ from $f(x_0)$ in the codomain side can be obtained by moving in that same direction on the domain side first) where $\delta > 0$ is chosen to be small enough so that $x_1 \in B_1(0)$ and $0< \delta < 1$. Then $f(x_1)$ is closer to $z$ than is $f(x_0)$.