Timeline for Simple closed curves in a simply connected domain
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7, 2023 at 11:27 | comment | added | Pietro Majer | Yes, exactly, that’s what I was also thinking. I think it is OK because any connected component $A$ has a closure that meets $\partial U$. Yes, $V=h^{-1}(A)$. Thank you! | |
| Dec 7, 2023 at 11:15 | comment | added | Sam Nead | Since $y_\infty$ is not a point of $U$ it does not lie in the intersection $U \cap B_{\epsilon_0/2}(y_\infty)$. So $A$ is not well-defined. Perhaps the following works: "let $A$ be any connected component of $U \cap B_{\epsilon_0/2}(y_\infty)$." Also, is $V = h^{-1}(A)$, on the last line? | |
| Dec 7, 2023 at 9:51 | comment | added | Pietro Majer | In the notations above, let $A$ be the connected component of $y_\infty$ in $U\cap B_{\epsilon_0/2}(y_\infty)$. Then $h^{-1}(A)$ is a connected open subset of $B_1$ whose closure meets $\partial B_1$. It follows that the set of $r<1$ such that $\partial B_r\cap V\neq \emptyset$ is a non-empty interval $(a,1)$, which gives the last sentence applying $h$. | |
| Dec 7, 2023 at 8:21 | comment | added | Sam Nead | Why is the last sentence of the second bullet true? That is, why are all points of $\partial U$ approached by $\Gamma_r$ as $r$ tends to one? I see this holds when $h$ extends continuously to $\partial B_1$. But there are examples where that does not happen... | |
| Dec 7, 2023 at 2:53 | vote | accept | D.S. Lipham | ||
| Dec 6, 2023 at 23:31 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Dec 6, 2023 at 23:22 | history | answered | Pietro Majer | CC BY-SA 4.0 |