Timeline for Maximal connected subtopologies
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 5, 2018 at 8:32 | comment | added | KP Hart | Split the real line into countably many disjoint dense sets $A_n$, e.g., $A_n=\mathbb{Q}+n\pi$ for $n\ge1$ and $A_0=\mathbb{R}\setminus\bigcup_{n\ge1}A_n$. Let $\tau_n$ be the topology generated by the usual topology together with $\{A_i:i\le n\}$. Each topology is connected, yet the union generates a topology with a pairwise disjoint cover by open sets. | |
| Dec 4, 2018 at 11:40 | comment | added | მამუკა ჯიბლაძე | Just in a way of additional relevant information - the (accepted) answer to the linked question refers to a paper with an example of a connected Hausdorff topology having no maximal connected topologies above it. | |
| Dec 4, 2018 at 7:56 | comment | added | Dominic van der Zypen | @ForeverMozart -- works fine if $\tau$ is connected, but if it isn't? | |
| Dec 4, 2018 at 3:29 | comment | added | Forever Mozart | Yes. Let $\tau_c=\tau$. ;-) | |
| Dec 3, 2018 at 20:46 | comment | added | user44191 | The straightforward application of Zorn's lemma doesn't work because one of the conditions on topologies uses infinitely many sets (in fact, arbitrarily many). An example: let $X = \mathbb{N}, \tau_n$ be the topology that doesn't distinguish between $m \geq n$, and otherwise is discrete. Then $\cup \tau_n$ is the set of "eventually constant" sets (and even is countable). But it does include each of the point-sets, which contradicts the "arbitrary unions" condition on topologies. Therefore, Zorn's lemma can't be directly applied. | |
| Dec 3, 2018 at 20:27 | comment | added | Dominic van der Zypen | Thanks for putting the effort into it... I also missed the fact that the union of a chain of topologies need not be a topology! Can you delete your answer and post as a comment, why a straightforward application of Zorn's lemma doesn't work? | |
| Dec 3, 2018 at 19:15 | comment | added | user44191 | The answer I gave below is incorrect; as pointed out by a comment to it, the union of a chain of topologies is not necessarily a topology. I think you should un-accept it. | |
| Dec 3, 2018 at 12:33 | vote | accept | Dominic van der Zypen | ||
| Dec 3, 2018 at 20:26 | |||||
| Dec 3, 2018 at 9:19 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |