Timeline for How do the compact Hausdorff topologies sit in the lattice of all topologies on a set?
Current License: CC BY-SA 2.5
29 events
| when toggle format | what | by | license | comment | |
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| S Nov 10, 2023 at 0:32 | history | suggested | Mark Schultz-Wu |
wrong lattices tag
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| Nov 10, 2023 at 0:20 | review | Suggested edits | |||
| S Nov 10, 2023 at 0:32 | |||||
| Apr 17, 2015 at 20:25 | answer | added | Will Brian | timeline score: 9 | |
| Oct 12, 2014 at 17:07 | comment | added | Todd Trimble | @QiaochuYuan For any nonempty set $X$, pick an element $x$ and topologize $X$ as a one-point compactification of the discrete space $X \backslash \{x\}$ (I think this is what dominiczypen was trying to say). | |
| Sep 2, 2014 at 6:48 | comment | added | Dominic van der Zypen | About Qiaochu's question: >> Is it obvious that there exists a compact Hausdorff topology on every set? << Let $X\neq \emptyset$ be a set, fix $x_0\in X$. Let $\tau = \mathcal{P}(X\setminus\{x_0\}) \cup \{U\subseteq X : X\setminus U \textrm{ is finite }\}$. Then $\tau$ is a compact Hausdorff topology on $X$. | |
| Jun 27, 2013 at 8:24 | answer | added | Włodzimierz Holsztyński | timeline score: 13 | |
| May 23, 2013 at 0:35 | answer | added | Cameron Buie | timeline score: 17 | |
| Mar 26, 2011 at 7:50 | answer | added | Dan | timeline score: 9 | |
| Mar 16, 2010 at 21:42 | vote | accept | Joel David Hamkins | ||
| Feb 22, 2010 at 14:13 | comment | added | Joel David Hamkins | Gerald, you are completely right about the uncountable case. About AC, I think everyone would want to define compact via the finite subcover definition, even if we lose the equivalence with other formulations. Finally, I don't think that "finite" loses its meaning if AC fails. A set is finite if it is bijective with a natural number, and without AC this can be different from being Dedekind finite (not being bijective with any proper subset), but I believe most people are careful to distinguish these notions. | |
| Feb 20, 2010 at 13:41 | comment | added | Gerald Edgar | Another strange comment... without AC, you have to say what you mean by "compact" ... and if you choose something about finite subcovers you have to say what you mean by "finite". | |
| Feb 20, 2010 at 13:39 | comment | added | Gerald Edgar | @Joel ... I'm not sure I understand this. Why not split off a countable subset, do your example there, and put a compact Hausdorff topology on the rest? | |
| Feb 20, 2010 at 6:38 | comment | added | Joel David Hamkins | Also, the question about maximal compact but non-Hausdorff topologies and minimal Hausdorff but incompact topologies appears to still be unsettled when the underlying space is uncountable. | |
| Feb 20, 2010 at 4:44 | comment | added | Joel David Hamkins | The questions about uniqueness are not yet answered, and neither are the questions about other lattice features. | |
| Feb 20, 2010 at 4:19 | comment | added | Joel David Hamkins | About Qiaochu's question: is it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology? | |
| Feb 20, 2010 at 1:09 | answer | added | François G. Dorais | timeline score: 42 | |
| Feb 20, 2010 at 0:41 | comment | added | François G. Dorais | Steen & Seebach 100 (Minimal Hausdorff Topology) answers the other question. | |
| Feb 20, 2010 at 0:35 | comment | added | François G. Dorais | For those who don't have the book, Steen & Seebach #99 is on Google books: books.google.com/… | |
| Feb 20, 2010 at 0:21 | comment | added | Gerald Edgar | Steen & Seebach 99. | |
| Feb 20, 2010 at 0:19 | comment | added | Gerald Edgar | every compact topology is refined by a compact Hausdorff topology No, there exist maximal compact topologies that are not Hausdorff and vice versa. Look for "maximal compact" in the title of the paper... | |
| Feb 20, 2010 at 0:17 | comment | added | Gerald Edgar | Is it obvious that there exists a compact Hausdorff topology on every set? Yes (using the well-ordering theorem) ... the order topology on the set of ordinals up to and including a given ordinal. | |
| Feb 20, 2010 at 0:17 | comment | added | François G. Dorais | @Qiaochu: The order topology on a successor ordinal is compact Hausdorff. | |
| Feb 20, 2010 at 0:11 | comment | added | Qiaochu Yuan | Is it obvious that there exists a compact Hausdorff topology on every set? | |
| Feb 19, 2010 at 23:16 | comment | added | Qiaochu Yuan | You probably already know this, but Terence Tao wrote a nice post on the "compact/Hausdorff duality" here: terrytao.wordpress.com/2009/02/09/… | |
| Feb 19, 2010 at 22:11 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
added 76 characters in body; edited title
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| Feb 19, 2010 at 22:03 | comment | added | Joel David Hamkins | Yes, of course you are right, Francois; the first question is too easy. Please proceed to answer the rest of my questions so beautifully! | |
| Feb 19, 2010 at 21:52 | comment | added | François G. Dorais | The maximal antichain question has a negative answer for infinite X. Split X into two infinite halves put the discrete topology on one half and the indiscrete topology on the other. | |
| Feb 19, 2010 at 21:28 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
added 127 characters in body
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| Feb 19, 2010 at 21:09 | history | asked | Joel David Hamkins | CC BY-SA 2.5 |