Timeline for A fixed point problem
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5, 2013 at 16:41 | comment | added | Sergei Akbarov | Oh, I missed this. | |
| Aug 5, 2013 at 16:09 | vote | accept | dimo | ||
| Aug 5, 2013 at 16:00 | comment | added | dimo | @SergeiAkbarov No ! $r$ is not fixed as it is clear from the statement of the problem above. | |
| Aug 5, 2013 at 13:27 | comment | added | Sergei Akbarov | @dimo Something is wrong, we do not understand each other. You consider a map from the closed interval $[0,1]$ into the plane $\mathbb R^2$ acting by formula $t\mapsto (tr,1-t)$ -- is this correct? And $A$ is the image of this map. If this is correct (and $[0,1]$ and $\mathbb R^2$ have usual topology), then this map is a homeomorphism between $[0,1]$ and $A$. (By the way, in this case it doesn't matter whether $r$ belongs to $\mathbb Q$ or to $\mathbb R$.) | |
| Aug 5, 2013 at 9:57 | comment | added | dimo | @SergeiAkbarov Actually you claim is not true. Since if you delete any point of $A$ which lays on the x-axes then the reminder is still connected. But $[0,1]$ has not this property | |
| Aug 5, 2013 at 9:16 | comment | added | Sergei Akbarov | Of course, this works only if I understand you correctly: the topology on $\mathbb R^2$ must be the usual topology of direct product of two $\mathbb R$, and each of those $\mathbb R$ is endowed with the usual topology of $\mathbb R$ (i.e. intervals $(a,b)$ form a base of open sets). | |
| Aug 5, 2013 at 9:08 | comment | added | Sergei Akbarov | This is a general theorem: every bijective and continuous map from a compact space to a Hausdorff space is a homeomorphism, see R.Engelking, General topology, 3.1.13, see also Wikipedia en.wikipedia.org/wiki/Compact_space#cite_note-10 (the properties of Hausdorff spaces). | |
| Aug 5, 2013 at 9:01 | comment | added | dimo | @SergeiAkbarov Would you please tell why $A\cong [0,1]$ ? | |
| Jul 29, 2013 at 11:24 | comment | added | Sergei Akbarov | I don't understand the problem. $A$ is homeomorphic to the interval $[0,1]$. So everything follows from the Brouwer fixed-point theorem (it's variant for the dimension 1). The statement of the problem looks strange. | |
| Jul 28, 2013 at 21:01 | answer | added | Pietro Majer | timeline score: 3 | |
| Jul 28, 2013 at 21:01 | comment | added | dimo | Yes, the usual topology | |
| Jul 28, 2013 at 20:59 | comment | added | Igor Rivin | Is the topology induced from $\mathbb{R}^2?$ | |
| Jul 28, 2013 at 20:37 | review | First posts | |||
| Jul 28, 2013 at 21:14 | |||||
| Jul 28, 2013 at 20:27 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced tags
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| Jul 28, 2013 at 20:21 | history | asked | dimo | CC BY-SA 3.0 |