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.


3 Answers 3


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.

  • $\begingroup$ Sir, I can't understand why it suffices to show that $U$ closed, given that we know $\Bbb R^n$ connected. $\endgroup$
    – Silent
    May 7, 2019 at 12:37
  • 2
    $\begingroup$ @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. $\endgroup$ May 7, 2019 at 13:54
  • $\begingroup$ What if $U$ is the empty set? $\endgroup$
    Feb 1, 2021 at 4:44
  • $\begingroup$ @user439126: There is no homeomorphism from the empty set onto $\mathbb{R}^n$. $\endgroup$ 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.

  • 1
    $\begingroup$ Hi Daniel. Would you be able to clarify why this would contradict $h$ being a homeomorphism? $\endgroup$
    – user124910
    Mar 28, 2017 at 3:49
  • $\begingroup$ @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. $\endgroup$ 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 $(hp_n)$.


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