Skip to main content
22 events
when toggle format what by license comment
Oct 12, 2021 at 13:32 history edited Paul Taylor
added newly created tag
Nov 13, 2017 at 10:32 answer added Christopher Townsend timeline score: 0
Mar 9, 2015 at 1:46 answer added Joseph Van Name timeline score: 1
Jun 30, 2014 at 11:35 answer added Paul Taylor timeline score: 16
May 16, 2014 at 2:45 answer added Joseph Van Name timeline score: 3
May 5, 2014 at 14:15 comment added Martin Argerami @AliTaghavi: if $X$ is not Hausdorff, then $C(X) $ does not detect the topology of $X$.
May 5, 2014 at 3:17 comment added Jonathan Beardsley @ToddTrimble Yeah, I find a lot of this stuff interesting and informative as well.
May 5, 2014 at 3:01 comment added Todd Trimble @JonBeardsley The "froth" generated by Andrej is genuinely interesting and informative, and I think perhaps you underestimate people's "abilities". (François did in fact address the question as asked.)
May 5, 2014 at 2:48 comment added Jonathan Beardsley This question gets an upvote because of all the froth it's generating, and because of people's general inability to just answer the question. =P
May 5, 2014 at 0:54 comment added Joseph Van Name Let's define the anti-compactness number of a space $X$ to be the smallest cardinal $\lambda$ such that every closed cover of $X$ has a subcover of cardinality less than $\lambda$. Then the anti-compactness number of a Hausdorff space is the least cardinal above its cardinality. However, the anti-compactness number of a space can be generalized to point-free topology. Therefore, the anti-compactness number seems to be a point-free generalization of the notion of the cardinality of a topological space.
May 4, 2014 at 19:36 answer added François G. Dorais timeline score: 4
May 4, 2014 at 13:28 comment added François G. Dorais At the risk of adding one more distracting terminological comment, let me point out that a space $X$ where every subset is 'anti-compact' in your sense is quite reasonably the dual idea of a Noetherian space. Though this would follow a nearly systematic naming tradition, I do not recommend calling such spaces 'Artinian spaces'!
May 4, 2014 at 11:59 comment added Henno Brandsma anti-compact as a property name is already taken: $X$ anti-compact means that the only compact subsets of $X$ are the finite ones. E.g. a co-countable space is an example.
May 4, 2014 at 10:05 comment added Martin Sleziak Similar question at MSE: Terminologies related to “compact?”. Some authors call spaces such that every closed cover has a finite subcover strongly S-closed. Some equivalent conditions are listed in my question here. (Sorry for the self-promotion, but link seemed better solution than posting the same list again here in a comment or an answer.)
May 4, 2014 at 9:59 answer added Andrej Bauer timeline score: 65
May 4, 2014 at 9:38 answer added Alexis Hazell timeline score: 4
May 4, 2014 at 9:27 comment added Martin Brandenburg This is not really a natural definition of dual compactness ...
S May 4, 2014 at 9:12 history suggested Ali Taghavi
I added a new tag
May 4, 2014 at 9:09 review Suggested edits
S May 4, 2014 at 9:12
May 4, 2014 at 9:07 comment added Ali Taghavi some particular questions arising from your question: Is the product of anti-compact space, anti-compact?what can be said about the continuous image of an anti compact space? assume that $X$ is a non Hausdorff compact space. Then $A=C(X)$, the algebra of all continuous complex functions on $X$, is a $C^{*}$ algebra. Now the question is that"what is an algebraic language for anti compactness, in term of $C^{*}$ algebras and non commutative topology?
May 4, 2014 at 9:02 review First posts
May 4, 2014 at 9:10
May 4, 2014 at 8:45 history asked p modabberi CC BY-SA 3.0