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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?

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    $\begingroup$ Yes. This follows from the Riemann mapping theorem. $\endgroup$
    – Mark
    Aug 2, 2011 at 11:27
  • $\begingroup$ I'm asking more general question. $\endgroup$ Aug 2, 2011 at 11:45
  • $\begingroup$ You have changed the question after correct answers have been posted. I think it'd be better to ask a separate question for general $n$. $\endgroup$
    – lhf
    Aug 2, 2011 at 11:47
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    $\begingroup$ 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?) $\endgroup$
    – GEdgar
    Aug 2, 2011 at 14:39
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    $\begingroup$ @GEdgar: Proving this without the Riemann mapping theorem was the subject of this MathOverflow question. $\endgroup$ Aug 2, 2011 at 16:26

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.

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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!

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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, $U\subsetneq \mathbb{R}^2$.

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    $\begingroup$ $\mathbb{R}^2$ is also homeomorphic to an open ball. $\endgroup$ Aug 2, 2011 at 11:37
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    $\begingroup$ That's because the asker changed the question... $\endgroup$
    – user864940
    Aug 2, 2011 at 13:19

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