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 ?
Asked
Active
Viewed 48 times
0
-
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 $\mathbb C \times \mathbb R^+$... – AlexR Dec 04 '13 at 17:25
-
Ok, I've got the point. Thank you ! – faero Dec 04 '13 at 17:33
-
Happy to help. Is your question still open or should I post this as an answer? – AlexR Dec 04 '13 at 17:36
-
I've found your comment complementary to Seirios' answer. – faero Dec 05 '13 at 20:01
1 Answers
0
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).
-
-
Yes, indeed. But it is not more complicated to use star-convexity instead of convexity. – Seirios Dec 04 '13 at 20:11