Timeline for Can we express separability of a ray-remainder in terms of the function algebra?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2018 at 8:41 | comment | added | KP Hart | To get an idea of what you are looking for consider Boolean algebras and their Stone spaces: the Stone space of a Boolean algebra is separable iff the algebra itself is $\sigma$-centered (the union of countably many families with the finite intersection property). There is no getting around using subsets of the ring in the characterization. | |
| Feb 4, 2018 at 19:18 | comment | added | Daron | $g$ is just an arbitrary element of $A$. That should be clear now. | |
| Feb 4, 2018 at 19:17 | history | edited | Daron | CC BY-SA 3.0 |
added 14 characters in body
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| Feb 4, 2018 at 17:09 | history | edited | Daron | CC BY-SA 3.0 |
added 12 characters in body
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| Feb 4, 2018 at 17:09 | comment | added | Daron | Oh sorry! I should have noticed that! | |
| Feb 4, 2018 at 16:34 | comment | added | afton | There is a fatal flaw in the formulation---in order to get the Stone-Čech compactification as spectrum you need to consider the Banach algebra $C^b(X)$ of bounded continuous functions. The space $X$ is locally compact so that if we supply $C(X)$ with its natural topology (compact convergence), its spectrum is just $X$. I think you have to restate your problem to give it some content. | |
| Feb 4, 2018 at 15:51 | comment | added | Taras Banakh | What is $g$ in the definition of $A'$? | |
| Feb 4, 2018 at 12:28 | history | asked | Daron | CC BY-SA 3.0 |