6
$\begingroup$

Let $f:U\to\mathbb{R}^n$ be a homeomorphism from an open subset of $\mathbb{R}^n$ onto $\mathbb{R}^n$, where $f$ is also uniformly continuous. Show that $U=\mathbb{R}^n$.

The solution I found here Uniformly continuous homeomorphism from open set to $\mathbb{R}^n$. but I don't understand the last part of the accepted answer: Since $f$ is uniformly continuous with image in a complete metric space, $f$ can be continuously extended $F:\overline{U}\to\mathbb{R}^n$. If $U\not=\mathbb{R}^n$, then $\overline{U}\not=U$ (otherwise, $U$ is a proper clopen subset of the connected space $\mathbb{R}^n$). Since $f$ maps $U$ onto $\mathbb{R}^n$, $F$ can't be injective, which contradicts $f$ being a homeomorphism.

I'm confused about the last line. Why does this contradict $f$ being a homeomorphism. Does it have something to do with connectedness? I know $U$ and hence $\overline{U}$ must be connected, but what is the contradiction?

$\endgroup$
  • $\begingroup$ Homeomorphisms are bijective by definition. $\endgroup$ – Ethan Alwaise Mar 28 '17 at 3:12
  • 1
    $\begingroup$ $f$ is a homeomorphism. How do we know $F$ is as well? $\endgroup$ – user124910 Mar 28 '17 at 3:19
  • $\begingroup$ See also MSE 590128. $\endgroup$ – Benjamin Dickman Mar 28 '17 at 3:28
  • $\begingroup$ @BenjaminDickman that's where I actually found this problem. The proposed solution I refer to is actually the one given by Daniel Fischer, but I don't understand the last line $\endgroup$ – user124910 Mar 28 '17 at 3:30
  • $\begingroup$ @user124910 It would be best to include that in the original question (although you could try commenting there to ask for clarification!)... $\endgroup$ – Benjamin Dickman Mar 28 '17 at 3:31
4
$\begingroup$

Under the supposed conditions, there are points $x \in U$ and $y \in \overline{U} \setminus U$ such that $F(x) = F(y) = z$. Then there is a sequence $\{y_n\} \subset U$ with $y_n \to y$. By the continuity of $F$, we have $F(y_n) \to F(y) = z$. By the continuity of $f^{-1}$, we have $$y_n = f^{-1}(F(y_n)) \to f^{-1}(z) = x$$ Limits in a metric space are unique, so this could only happen if $x=y$ which is absurd.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.