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We have the interval $]-1,1[ \subset \mathbb{R}$. We have to say if this interval is homeomorphic with the open unit ball $U(0,1)$ of $\mathbb{R}^2$ and with the set $\mathbb{R}$.

I know that two sets are homeomorphic when it exists a continuous function between topological spaces that has a continuous inverse function. But how to prove that here ? I need an example to understand. For example, between $]-1,1[$ and $U(0,1)$. Someone could help me ?

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    $\begingroup$ If you delete some element from $(-1,1)$ then the result is obiously not path-connected. If you do the same with unit ball $U(0,1)\subset\mathbb R^2$ then the result is path-connected. $\endgroup$ – drhab Jun 18 '17 at 13:17
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    $\begingroup$ Why are you so sure this is true? Removing any point of $]-1,1[$ makes it disconnected, but the open unit ball $U(0,1)$ remains connected. $\endgroup$ – Demophilus Jun 18 '17 at 13:19
  • $\begingroup$ No, I'm not sure ! We have to say IF the two sets are homeomorphic. Thank you, I believe that I have understood. :) $\endgroup$ – Mélanie De la Cheminée Jun 18 '17 at 13:23
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This is the image you can have in mind when thinking about this solution. Other users have already given perfect explanations of solutions. I think this image will help with your intuition and aid you in wiring the proof yourself.

enter image description here

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We have the interval $]-1,1[ \subset \mathbb{R}$. We have to say if this interval is homeomorphic with the open unit ball $U(0,1)$ of $\mathbb{R}^2$ and with the set $\mathbb{R}$.

The unit ball case is already answered in comments. The standard homeomorhism $f$ between the interval $]-1,1[$ and $\mathbb{R}$ is given by the formula $f(x)=\tan\frac {\pi x}{2}$.

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