Timeline for Limiting halves of a connected space
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31, 2018 at 19:19 | answer | added | MTyson | timeline score: 2 | |
| Jan 31, 2018 at 17:31 | comment | added | D.S. Lipham | @TarasBanakh Could you explain? The reason it holds for the compact boundary case is because given a $Y$-clopen set $A$, eventually, the "boundaries" of the halves (all homeomorphic to $\partial U$) must "match up" regarding their intersections with $A$ and its complement. And so $A$ will naturally correspond to a clopen subset of $X$. To MTyson: That is an interesting idea, but it's not immediately obvious to me that it works! | |
| Jan 31, 2018 at 6:55 | comment | added | Taras Banakh | The answer YES for the case of compact boundary implies the YES answer for a continuum-connected space. So, if a counterexample exists, it should not be continuum-connected. | |
| Jan 31, 2018 at 4:05 | comment | added | MTyson | I think there is a proof as follows, but I have not worked out the details. Suppose there were a partition of $Y$ into disjoint clopen sets $A_1,A_2$ both containing elements of $\{0\}\times\bar U$. For both $i$, let $A'_i=\{x\in X \mid \forall\epsilon>0\;\exists\delta\in[0,\epsilon)\text{ s.t. }(\delta,x)\in A_i\}$. Then one can show (I think) that $A'_1$ and $A'_2$ are disjoint, open, nonempty, and cover $X$. | |
| Jan 31, 2018 at 0:03 | history | edited | D.S. Lipham | CC BY-SA 3.0 |
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| Jan 30, 2018 at 22:24 | history | edited | D.S. Lipham | CC BY-SA 3.0 |
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| Jan 30, 2018 at 21:39 | history | asked | D.S. Lipham | CC BY-SA 3.0 |