This question was asked in my real analysis quiz and I was not able to solve it.
Let $U$ be a non-empty open subset of $\mathbb{R} $. Suppose that there exists a uniformly continuous homeomorphism $ h: U \longrightarrow \mathbb{R} $ . Show that $U=\mathbb{ R} $.
I thought of assuming that $U$ is a subset of $\mathbb{R} $ and then trying to find a contradiction. But I am confused on what property I should use and unable to find a contradiction.
Best Answer
Since $U(\subset\Bbb R)$ and $\Bbb R$ are homeomorphic, $U$ is an open interval. Suppose that it is an interval of the form $(a,b)$, with $a\in\Bbb R$ and $a<b$ ($b\in\Bbb R$ or $b=\infty$). there is some $\delta>0$ such that$$x,y\in(a,b)\wedge|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<1.$$But it follows from that $f$ maps bounded sets into bounded sets. So, if $c\in(a,b)$, $f\bigl((a,c)\bigr)$ is bounded. On the other hand, $f$ is injective and continuous, and therefore it is monotonic. So, you can have $\lim_{x\to b}f(x)=\infty$ or $\lim_{xto b}f(x)=-\infty$, but you can't have both. Therefore, either $f$ cannot take arbitrarily large values or it cannot take arbitrarily small values. In either case, it cannot be a bijection between $(a,b)$ and $\Bbb R$.