Timeline for If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Current License: CC BY-SA 3.0
15 events
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| Apr 16, 2014 at 14:05 | comment | added | Salvo Tringali | Here's a "natural" follow up: mathoverflow.net/questions/163559. | |
| Apr 16, 2014 at 8:29 | answer | added | Chris Schommer-Pries | timeline score: 1 | |
| Apr 16, 2014 at 8:02 | answer | added | Salvo Tringali | timeline score: 1 | |
| Apr 13, 2014 at 14:26 | comment | added | Salvo Tringali | Eric: Thanks for mentioning the example with the Alexandrov topology. As a minor addendum to what you said in your 1st comment, the canonical topology of a semimetric $d$ is T1 if and only if $d(x,y)\ne 0$ for all distinct $x,y$. @Wlodzimierz Holsztynski: I will give a look at Archangielski's work (thanks for the hint). And I agree with the splitting that you suggest. | |
| Apr 13, 2014 at 1:35 | comment | added | Eric Wofsey | To elaborate on my last comment, every poset with the Alexandrov topology is canonically semimetrizable (let $d(x,y)=0$ if $x\leq y$ and $d(x,y)=1$ otherwise). Incidentally, every semilattice is also a topological monoid and this semimetric is subinvariant. | |
| Apr 13, 2014 at 1:24 | comment | added | Włodzimierz Holsztyński | I seem to remember that Soviet topologists, in particular A.Archangielski, considered some notions of asymmetric (not necessarily symmetric) metrics. Perhaps in the context of metrization. | |
| Apr 13, 2014 at 1:19 | comment | added | Włodzimierz Holsztyński | The notion of left/right semi-invariance is very nice. Nevertheless I'd split the given problem into two questions: 1.Under what conditions on a topological monoid does there exist a semi-metric which induces the given topology (without worrying about the semi-metric being semi-invariant); and 2.Under what conditions is a semi-metric equivalent to a (left/right) semi-invariant semi-metric? | |
| Apr 13, 2014 at 1:18 | comment | added | Eric Wofsey | Do you know of a characterization of what spaces admit any semimetric structure at all? If I'm not mistaken, these are a lot more general than metrizable spaces, since the possibility of having $d(x,y)=0$ but $d(y,x)>0$ lets you get interesting non-Hausdorff topologies. | |
| Apr 13, 2014 at 0:09 | comment | added | Salvo Tringali | I don't expect this to be true in general: let me clarify this point in the OP. (For the record, I've in mind some topological monoids considered by J. Snellman in the context of factorization theory.) | |
| Apr 13, 2014 at 0:07 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Some clarifications after comments of Mariano Suárez-Alvarez
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| Apr 12, 2014 at 23:49 | comment | added | Mariano Suárez-Álvarez | Why do you expect this to be true? | |
| Apr 12, 2014 at 23:46 | comment | added | Salvo Tringali | True. So let us restrict to "sufficiently small" topological monoids. What happens? | |
| Apr 12, 2014 at 23:43 | comment | added | Mariano Suárez-Álvarez | The topology deternined by a semimetric always has countable bases of neighborhoods at each point, and there are topological monoids which don't have that property; for example, the product of uncountably many discrete non-trivial finite groups. | |
| Apr 12, 2014 at 23:38 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
deleted 66 characters in body; edited tags
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| Apr 12, 2014 at 23:33 | history | asked | Salvo Tringali | CC BY-SA 3.0 |