Uniformly continuous homeomorphism from open set to $\mathbb{R}^n$.

Let $U \subset \mathbb{R}^n$ be an open set. Suppose that the map $h:U \to \mathbb{R}^n$ is a homeomorphism from $U$ onto $\mathbb{R}^n$, which is uniformly continuous. Prove $U = \mathbb{R}^n$.

My first attempt guided by intuition was to look at covering $\mathbb{R}^{n}$ by balls of radius $=\frac{1}{n}$ and conclude something forwarding to contradiction by looking at inverse-image of such covering, but I cannot see it for now.

Here's a sketch of another argument along the line of Daniel Fischer's.

By the connectedness of $\mathbb{R}^n$, it suffices to show $U$ is closed. So suppose $x_n \in U$ and $x_n \to x$. Then $\{x_n\}$ is Cauchy, and it follows from the uniform continuity that $\{f(x_n)\}$ is Cauchy as well. $\mathbb{R}^n$ is complete, so $f(x_n)$ converges to some $y \in \mathbb{R}^n$. By the continuity of $f^{-1}$, we have $x_n = f^{-1}(f(x_n)) \to f^{-1}(y)$ so $x = f^{-1}(y)$. In particular, $x \in U$.

This would work just as well if we replaced $\mathbb{R}^n$ by another connected complete metric space.

• Sir, I can't understand why it suffices to show that $U$ closed, given that we know $\Bbb R^n$ connected. May 7, 2019 at 12:37
• @Silent: Theorem: If $X$ is a connected topological space and $U \subset X$ is both open and closed, then either $U = X$ or $U = \emptyset$. Proof: If not, then $U$ and $U^c$ are two disjoint nonempty open sets, and we have $X = U \cup U^c$, contradicting the assumption that $X$ was connected. May 7, 2019 at 13:54
• What if $U$ is the empty set? Feb 1, 2021 at 4:44
• @user439126: There is no homeomorphism from the empty set onto $\mathbb{R}^n$. Feb 1, 2021 at 4:46

$\mathbb{R}^n$ is a complete metric space. A uniformly continuous map from a metric space $X$ to a complete metric space can be extended to a uniformly continuous map from the completion of $X$ to $Y$.

If $U \neq \mathbb{R}^n$, the extension $\overline{h} \colon \overline{U} \to \mathbb{R}^n$ could not be injective, and that would contradict the assumption that $h\colon U \to \mathbb{R}^n$ is a homeomorphism.

• Hi Daniel. Would you be able to clarify why this would contradict $h$ being a homeomorphism? Mar 28, 2017 at 3:49
• @user124910 For $y \in \overline{U}\setminus U$, there is an $x\in U$ with $\overline{h}(y) = h(x)$. Pick disjoint neighbourhoods (in $\mathbb{R}^n$) $V$ of $y$ and $W$ of $x$. Since $h$ is assumed to be a homeomorphism, $h(W)$ is a neighbourhood of $h(x)$. By continuity of $\overline{h}$, we however have $\varnothing \neq h(V \cap U) \cap h(W)$ - if we chose $V$ small enough, $h(V\cap U) \subset h(W)$ - and thus $h$ can't have been injective. Apr 13, 2017 at 11:52

Suppose $$(p_n) \to p$$ for $$p_n \in U$$. We have by uniform continuity that $$\forall \epsilon > 0$$, $$\exists\delta>0$$ $$|x-y|<\delta \implies |hx-hy|<\epsilon$$ Thus, given $$\epsilon>0$$, we have for sufficiently large $$n,m$$ that $$|p_n-p_m|<\delta$$ (since $$(p_n)$$ is Cauchy) and thus $$|hp_n-hp_m|<\epsilon$$. So $$(hp_n)$$ is Cauchy and thus converges in $$\mathbb{R}^m$$ say $$(hp_n) \to q$$. Since $$h^{-1}$$ is continuous, we have $$(h^{-1}(hp_n))=(p_n) \to h^{-1}q$$ which shows $$p=h^{-1}q \in U$$. Thus, $$U$$ is clopen and since $$U \ne \emptyset$$, $$U=\mathbb{R}^m$$ by connectedness of $$\mathbb{R}^m$$.

Note the necessity of uniform continuity since regular continuity does not give a globally applicable $$\delta$$ to show Cauchyness of $$(hp_n)$$.