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Let $U$ be a simply connected open set in $\mathbb{R}^2$. Is it true that $U$ is homeomorphic to an open ball?

  • 4
    Yes. This follows from the Riemann mapping theorem.2011-08-02
  • 0
    I'm asking more general question.2011-08-02
  • 0
    You have changed the question after correct answers have been posted. I think it'd be better to ask a separate question for general $n$.2011-08-02
  • 0
    Since you now re-asked your follow-up question I rolled back to the previous version.2011-08-02
  • 3
    BUT, isn't the fact in the question **MUCH EASIER TO PROVE** than the Riemann mapping theorem? (A snipe: is the empty set simply connected?)2011-08-02
  • 3
    @GEdgar: Proving this without the Riemann mapping theorem was the subject of [this MathOverflow question](http://mathoverflow.net/questions/66048/riemann-mapping-theorem-for-homeomorphisms).2011-08-02

3 Answers 3

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Yes, this is the Riemann mapping theorem. You get much more than a homeomorphism: you get a biholomorphic map.

2

Yes. In fact, more can be said... The Riemann Mapping Theorem states that the homeomorphism can be taken to be biholomorphic (as a complex map), if $U \neq \mathbb{C}$. See this link for a much more detailed treatment and proof.

Hope this helps!

0

According to the Riemann mapping theorem that's true iff U is a simply connected nonempty open set in $\mathbb{R}^2$ which is a strict subset. That is, $U\subsetneq \mathbb{R}^2$.

  • 3
    $\mathbb{R}^2$ is also homeomorphic to an open ball.2011-08-02
  • 1
    That's because the asker changed the question...2011-08-02