Timeline for What are the algebras for the ultrafilter monad on topological spaces?
Current License: CC BY-SA 4.0
13 events
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| Dec 21, 2019 at 19:30 | history | edited | Tim Campion | CC BY-SA 4.0 |
Tidying up -- I think the original answer had some very interesting insights and I don't want them to be lost amid the errors.
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| Dec 18, 2019 at 18:20 | comment | added | Tim Campion | Let us continue this discussion in chat. | |
| Dec 18, 2019 at 18:15 | comment | added | Andrea Marino | You are so damn right man. It didn't work! Maybe we can join forces and try to prove some other condition. I feel like we are close to! Sorry for the misunderstanding about indiscrete topology, in my mind I read "discrete" topology. | |
| Dec 18, 2019 at 18:11 | history | edited | LSpice | CC BY-SA 4.0 |
Alexander's theorem; \operatorname
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| Dec 18, 2019 at 18:10 | comment | added | Tim Campion | As another sanity check, Remark 2 shows that any $\tau^\xi$-open set has compact complement -- not just that it's in the topology generated by the sets with compact complement. So if $\tau^\xi = cp(\tau)$ is a $\beta$-structure, then every set in the topology generated by the sets with compact complement actually has compact complement, i.e. the sets with compact complement already form a topology and don't just generate one. | |
| Dec 18, 2019 at 18:04 | comment | added | Tim Campion | Indeed, let $(X,\tau)$ be an infinite indiscrete space. This is compact -- every open cover has a singleton subcover! It's also locally compact, and compactly-separated. Then $K \subseteq X$ is $\tau$-compact for every subset $K$, so $cp(\tau)$ is the discrete topology, which is not compact. | |
| Dec 18, 2019 at 18:04 | comment | added | Tim Campion | If, as now, you define $cp(\tau)$ to be the topology generated by the complements of compact sets, then I think there's an error in the proof that $cp(\tau)$ is compact. The famous lemma (essentially Cantor's intersection theorem) says that a codirected intersection of nonempty compact closed sets is nonempty -- but here the $K_i$ are not necessarily $\tau$-closed (and on the other hand, they are not necessarily $cp(\tau)$-compact). | |
| Dec 18, 2019 at 17:59 | comment | added | Andrea Marino | By locally compact I mean that - at each point - you have a local basis of compact neighbourhoods. I am quoting for Wikipedia: this is a possible definition of the local compactness, which is equivalent to the classical one if the space is Hausdorff. The definition that I just quoted is indeed equivalent to: for any point x and open $V$ containing x, there exist an open $U$ containing x and a compact $K$ such that $U \subset K \subset V$. | |
| Dec 18, 2019 at 17:40 | comment | added | Andrea Marino | Fiuu! You were right! I hope I fixed this on the edge :) however, the infinite indiscrete topology is not compact, so I don't see any contraddiction! | |
| Dec 18, 2019 at 17:39 | history | edited | Andrea Marino | CC BY-SA 4.0 |
added 398 characters in body
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| Dec 18, 2019 at 17:02 | comment | added | Tim Campion | Note also that if $cp(\tau)$ is a compact Hausdorff topology, then for any $\beta$-structure $\tau^\xi$, the map $id: (X, cp(\tau)) \to (X,\tau^\xi)$ is a continuous bijection of compact Hausdorff spaces and therefore a homeomorphism. So whenever $cp(\tau)$ is a compact Hausdorff topology, it is the unique $\beta$ structure on $(X,\tau)$. So I think $cp(\tau)$ must not be a topology when $X$ is an infinite indiscrete space. | |
| Dec 18, 2019 at 16:57 | comment | added | Tim Campion | Thanks! Remark 2 seems especially insightful to me. But I'm not sure that $cp(\tau)$ is a topology -- it's closed under finite intersections but is it closed under even finite unions? Also, in the absence of Hausdorffness, one needs to be careful about the definition of "locally compact". I see how condition (3) and remark (2) imply that for every $x$ and every open neighborhood $U$ of $x$, there exists an open neighborhood $V$ of $x$ such that $V \subseteq K \subseteq U$ for some compact $K$ -- is this equivalent to a standard definition? | |
| Dec 18, 2019 at 16:24 | history | answered | Andrea Marino | CC BY-SA 4.0 |