Timeline for Why are there no wild arcs in the plane?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2011 at 11:29 | comment | added | Jim Conant | @Bill: That's a clever way to do it! | |
| Mar 8, 2011 at 4:46 | comment | added | Bill Thurston | Actually, if the arc goes from $0$ to $\infty$ in $\hat{\mathbb C}$, then just take the preimage under $z \mapsto z^2$: it's a circle. It's nice, but it reduces the problem to another problem that ab initio is just as hard, just better known. | |
| Mar 8, 2011 at 4:38 | comment | added | Ryan Budney | But a fixable flaw. I like this argument. | |
| Mar 8, 2011 at 4:36 | history | edited | John Stillwell | CC BY-SA 2.5 |
added 87 characters in body; added 45 characters in body
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| Mar 8, 2011 at 4:33 | comment | added | Ryan Budney | Oh, I think I see how one can tweak Stillwell's argument to make it work -- the lifted arc is in $(-1,1) \times \mathbb R$ which you could compactify to $[-1,1] \times [-\infty,\infty]$. Crush $\{-1\}\times [-\infty,\infty]$ and $\{1\}\times [-\infty,\infty]$ to points and identify the quotient space with a compact disc -- the arc has ends in this quotient space. | |
| Mar 8, 2011 at 4:24 | comment | added | Ryan Budney | @Richard: in case your response is aimed at me, what I mean is in Stillwell's argument "our arc on a strip in the plane" is an arc in the universal cover of a twice punctured sphere -- this cover is homeomorphic to $\mathbb R^2$, so showing the arc has endpoints on the boundary is a non-trivial problem (it's easy to construct ones that don't, even if they come from embedded intervals in $S^2$ via this construction). | |
| Mar 8, 2011 at 4:19 | comment | added | Autumn Kent | Based on the stackexchange question, I also assumed the arc is a compact one. | |
| Mar 8, 2011 at 4:14 | comment | added | Ryan Budney | How do you know the arc has ends? | |
| Mar 8, 2011 at 4:12 | history | answered | John Stillwell | CC BY-SA 2.5 |