Talk:Paracompact space

message: Every open cover of R has a locally finite subcover, which contradicts For paracompactness, "subcover" is replaced by "open refinement" and "finite" by is replaced by "locally finite". Both of these changes are significant: if we take the definition of paracompact and change "open refinement" back to "subcover", or "locally finite" back to "finite", we end up with the compact spaces in both cases.
:That doesn't seem to be true; (-\infty, n),\ n \in \mathbb{Z} doesn't have locally finite subcover (nor point-finite for that matter).  — Preceding unsigned comment added by 89.68.149.185 (talk) 00:49, 1 February 2017 (UTC)
thread_title:  Relation to compactness 

message: I removed the following two statements:
*Of course, a fully normal space is normal.
That doesn't seem to be true, since fully normal spaces don't have to be Hausdorff.
:Normal spaces don't have to be Hausdorff either. 
::Ok, Topology Glossary agrees, but normal space links to Separation axioms where "normal" appears in a table which suggests that normal spaces are T1 and hence Hausdorff, but then the glossary further down says that that's usually no longer true. This is a mess. AxelBoldt
:::The text following that table in Separation axioms and which explains that table describes how "normal" originally meant that a space satisfies both T4 and T1 but now only means that it satisfies T4. Of course, this is still a mess, but that is unavoidable; it's a mess in real life. The Topology Glossary, like the glossary in at the end of Separation axioms, explains what the term means in Wikipedia: the modern interpretation. The article Normal space itself should do so as well; right now it redirects only because nothing has been written on normal spaces as such. (There is much to write on that subject, however, and I'll be sure to do it someday.) Given the easy possibility for confusion, however, it'd be best to have an explanation right in Normal space itself, even if that makes it a pitiful stub; I'll do that now for anything that redirects to Separation axioms and is one of these controversial terms. — Toby Bartels, Wednesday, May 22, 2002 — Actually, on second that, I won't do that stuff, because you already did!!!
:Steen & Seebach say that "fully T4" spaces are "T4"; since they use the older terminology for separation axioms, this means that fully normal spaces are normal. Since I trust Steen & Seebach's facts, if not their terminology, I'll restore this.
*Thus, a fully T4 space is the same thing as a paracompact Hausdorff space (see Separation axioms).
The term "fully T4" hasn't been defined. Or is this sentence supposed to be a definition?
:The reference to Separation axioms is supposed to give the reader enough information to come up with a definition (that is, fully normal and T1). But this could be made clearer, which I will now do.
:-- Toby Bartels, Monday, May 20, 2002
Fully normal, as defined here, does NOT imply regular. For instance, the Sierpinski space is fully normal, but not regular.
Manta 19:36, 19 December 2006 (UTC)
thread_title:  Two statements removed 

message: This page use to claim that every fully normal space is paracompact. I believe this is in error. What is true is that every fully T4 space (= fully normal and T1) is paracompact (and Hausdorff). Steen and Seebach (who interchange the meanings of fully normal and fully T4), do not claim that full normality (our definition) implies paracompactness, however there are not any counterexamples in their book. Are there any counterexamples, or is this an open question? -- Fropuff 18:48, 23 October 2007 (UTC)
thread_title:  Fully normal vs. paracompact 

message: In the article, the second property is "Paracompactness is weakly hereditary, i.e. every closed subspace of a paracompact space is paracompact. This can be extended to F-sigma subspaces as well."
I think for the F-sigma case, the paracompact space is required to be Hausdorff.  —Preceding unsigned comment added by 219.236.198.32 (talk) 15:39, 29 January 2011 (UTC)
thread_title:  Second Property - F-sigma Case 

message: Isn't it uterly illogilal to speak about a counterexample to a definition? Counterexample can show up when you say "every object with property X also has property Y" and then someone says, "hey, wait here is something that has property X but doesn't have property Y". But since definition basically says only "lets refer to every object with property X as x, because its shorter that way". So since paracompactness is a property, you can't really have a counterexample to it - what you can have are objects that don't have this property.
Matejo (talk) 15:27, 10 May 2011 (UTC)
thread_title:  "Counterexamples" to definition 

message: This article could do with a seperate section on paracompactness, as that redirects here. Brad7777 (talk) 12:32, 6 February 2012 (UTC)
replies: text: The entire article is about paracompactness. What would you expect to see in a separate section? --Zundark (talk) 12:38, 6 February 2012 (UTC)
replies: text: Simpler language and no mention of additional properties of other topological spaces. Brad7777 (talk) 20:26, 7 February 2012 (UTC)
thread_title:  Paracompactness 

message: The proof of lemma 1 seems wrong. First the proof that x\in U\setminus\bar{W_{U}}\, gives a contradiction only shows that U\subseteq\bar{W_{U}}, not what is intended. Secondly, W_U=U since, by regularity and local finiteness, every point of U belongs to some set A\in\mathcal{V} with \bar{A}\subseteq U.  — Preceding unsigned comment added by 193.188.47.48 (talk) 10:40, 11 July 2014 (UTC)
thread_title:  Faulty Proof?