A subset $E$ contained in $\mathbb{R}^n$ is such that the function $x \mapsto \left\Vert x\right\Vert^2$ is uniformly continuous on $E$. For $r > 0$, let $E_r$ denote the union of all open balls of radius $r$ contained in $E$. Prove that $E_r$ is bounded for all $r > 0$. Find an example showing that $E$ itself does not have to be bounded.
I have been working on this one for a while and I seem to be stumped. I know what the definitions are, but I'm having trouble getting started on this problem.
Thanks
The function $f$ is uniformly continuous on the subset $E_r$ as well.
If contrariwise $E_r$ is unbounded for some $r>0$, then there is a sequence of vectors $x_n$ such that $\Vert x_n\Vert\to\infty$, and $B(x_n,r)\subseteq E$. For all $\delta\in(0,r)$ both $x_n$ and $x'_n(\delta)=x_n(1+\delta/(2\Vert x_n\Vert))$ are then in $E_r$, because $\Vert x_n-x'_n(\delta)\Vert<\delta<r$. But $$ f(x'_n(\delta))=f(x_n)\left(1+\frac{\delta}{2\Vert x_n\vert}\right)^2, $$ so $$ f(x'_n(\delta))-f(x_n)\ge f(x_n)\frac{\delta}{2\Vert x_n\Vert}=\frac{\delta}2 \Vert x_n\Vert. $$ This set of differences of values of $f$ is unbounded in spite of the arguments being within $\delta$ of each other violating the assumption that the restriction of $f$ to $E_r$ is uniformly continuous.
For an example of unbounded $E$ I proffer $E=\mathbb{Z}\subset\mathbb{R}$ (map this on the $x$-axis, if you want this to work for any $n$). There are no distinct points within distance $<1$ of each other, so uniform continuity of any function is automatic. Yet $E$ is unbounded.