1617-257/TUT-R-8

On 11/3/16, we discussed some questions from the exam:

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 $x_{0}\in B$ 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 strictly closer to $z$ than is $f(x_{0})$.