0
$\begingroup$

I was wondering how to show that $\mathbb{C}\times\mathbb{R}^+$ is simple connected (every closed arc is continuously reducible to a dot).

The problem is more how can one write such paths in such space.

Can someone help ?

$\endgroup$
  • $\begingroup$ A path in this space is described by a path in $\mathbb C$ and a path in $\mathbb R^+$. Their carthesian product is a path in $\mathbb C \times \mathbb R^+$... $\endgroup$ – AlexR Dec 4 '13 at 17:25
  • $\begingroup$ Ok, I've got the point. Thank you ! $\endgroup$ – faero Dec 4 '13 at 17:33
  • $\begingroup$ Happy to help. Is your question still open or should I post this as an answer? $\endgroup$ – AlexR Dec 4 '13 at 17:36
  • $\begingroup$ I've found your comment complementary to Seirios' answer. $\endgroup$ – faero Dec 5 '13 at 20:01
0
$\begingroup$

Hint: $\mathbb{C} \times \mathbb{R}^+$ is star-convex. Therefore, it is sufficient to notice that any star-convex set is simply connected (see for example this question).

$\endgroup$
  • $\begingroup$ isn't it convex? $\endgroup$ – Stefan Hamcke Dec 4 '13 at 18:03
  • $\begingroup$ Yes, indeed. But it is not more complicated to use star-convexity instead of convexity. $\endgroup$ – Seirios Dec 4 '13 at 20:11

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.