Talk:Compact space

message: I'm concerned that the "Hyperreal definition" subsubsection is not stated as being specific to metric spaces.  It's not clear to me what being "infinitely close to a point of X" might mean, if there is no metric around.  Should this section be merged into the "Metric spaces" section?  What about the paragraph just above it, the second paragraph of "Characterization by continuous functions", which by the way also seems to be redundant with "Hyperreal definition"? --Trovatore (talk) 07:02, 12 July 2018 (UTC)
thread_title:  Is "hyperreal definition" specific to metric spaces? 

message: I think it is not too bold to say that the nearly universal agreement among mathematicians is that the word "compact" by itself, in the most general context and without further qualification, means that every open cover has a finite subcover.  I am not saying that definition should appear in the first sentence; we've discussed this before and I agree with opening more gently with the example of closed and bounded subsets of a Euclidean space.
But I am saying that the third sentence of the six-sentence fourth paragraph is way too deeply buried.
I know that Sławomir has strong feelings about the importance of sequential compactness, and that's fine; I don't care to argue that point.  Just the same, when the distinction is made explicitly, sequential compactness is the one that takes the adjective; compactness simpliciter is the cover definition.  This is entirely standard in the literature.
I don't have an immediate proposal but I do think the standard general definition needs to be treated earlier and more prominently. --Trovatore (talk) 07:17, 12 July 2018 (UTC)
thread_title:  The most standard definition is not prominent enough 

message: The introduction to the Definitions section mentions "Dirichlet's theorem". It seems to me that this is actually meant to refer to the Heine–Cantor theorem, but I don't feel competent enough to edit the article.  — Preceding unsigned comment added by 2A02:8070:A191:7D00:AB:87BA:3A49:1844 (talk) 12:23, 20 January 2019 (UTC)
thread_title:  Heine-Cantor? 

message: I think this part is wrong or vague:
"typical examples of compact spaces include spaces consisting not of geometrical points but of functions."
Mojtabakd (talk) 09:22, 8 November 2019 (UTC)
replies: text: I agree, and I have completely rewritten the paragraph. D.Lazard (talk) 10:13, 8 November 2019 (UTC)
thread_title:  What does it mean?! 

message: That is not grammatical English, at the very least, a verb must be adduced, e.g. the extreme value theorem follows, or since... we have... etc. 2A01:CB0C:CD:D800:C12E:287B:1E8A:E532 (talk) 20:44, 1 February 2021 (UTC)
thread_title:  "Since a continuous image of a compact space is compact, the extreme value theorem" 

message: The introduction states:
: One such generalization is that a topological space is sequentially compact if every infinite sequence of points sampled from the space has an infinite subsequence that converges to some point of the space.
While this is true for subsequences (s_{i_n})_n of (s_n)_n with \forall n, i_n < i_{n+1}, the article subsequence specifically states that subsequences are defined in terms of \leq instead of {<}:
: Subsequences can contain consecutive elements which were not consecutive in the original sequence.
That does not make sense here, since it would allow one to prove that every subset of a metric space is compact via the constant "subsequence" (or, equivalently, every topological space is sequentially compact).
The non-strict comparison is also in contrast to the linked reference that explicitely defines the index sequence via {<} as well as, for example, .
In my opinion, it should be made clear - either here or in subsequence - that (non-)strictness matters depending on the context. --217.251.39.96 (talk) 23:23, 9 May 2021 (UTC)
replies: text: Apparently you confuse "consecutive" and "equal", or you confuse elements of the sequence with their indexes. Said otherwise, the definition of "subsequence" implies \forall n, i_{n+1} > i_n, and your second quotation means that one may have i_{n +1} > 1+ i_{n}. For example, the positive even integers form a subsequence of the positive integers; the integers 6 and 8 are consecutive in the subsequence of even integers, but not in the integers, since 7 is between them. D.Lazard (talk) 08:49, 10 May 2021 (UTC)
thread_title:  `Subsequence' in this article is different from the linked [[subsequence]] 

message: In a compact Hausdorff space, each closed subset is compact and each compact subset is closed.  (Here, "subset" means proper or improper.) Is there an example of a compact non-Hausdorff space where each subset is closed if and only if it is compact?  If so, let's add it to the article.  —Quantling (talk | contribs) 19:08, 1 June 2023 (UTC)
thread_title:  Each subset is closed if and only if it is compact 

message: This doesn't seem to be correct. In the cocountable topology, the only compact sets are the finite sets. These are closed, as the cocountable topology is T1. But the cocountable topology on an uncountable set is not Hausdorff. 2A02:A03F:8CEC:D00:CC18:E805:F3B7:DD52 (talk) 16:56, 3 January 2024 (UTC)
replies: text: Good observation, but I think the cocountable topology is not a TVS topology because addition is not continuous (any T1 topological group is hausdorff).  A reference would be nice though. Tito Omburo (talk) 17:38, 3 January 2024 (UTC)

text: After further reflection, the singleton \{0\} is compact, and is closed iff the topology is hausdorff.  Tito Omburo (talk) 17:47, 3 January 2024 (UTC)
thread_title:  "every non-Hausdorff TVS contains compact (and thus complete) subsets that are not closed"