Talk:Paracompact space

Relation to compactness[edit]

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)[reply]

Two statements removed[edit]

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)[reply]

The definitions section can now be cut back, by referring to the open cover page. Charles Matthews 14:50, 13 Sep 2004 (UTC)

"Every metric space (hence, every metrisable space) is paracompact. In fact, every metric space is hereditarily paracompact. The proof of this is hard and requires the Axiom of choice for the non-separable case." Shouldn't the second and third sentences be reversed? What's hard is proving metric spaces are paracompact. Once you know that, since subspaces of metric spaces are metric spaces, once you know metric spaces are paracompact you automatically know they are hereditarity paracompact.--Syd Henderson 04:06, 28 February 2007 (UTC)[reply]

Long line not paracompact[edit]

The article states: "The long line is locally compact, but not second countable." I am having some trouble with this. Firstly, what difference does it make that it's not second countable? How does that keep it from being paracompact? And secondly, it seems paracompact to me: choose an open cover that is an uncountable collection of open intervals. This will be locally finite. What am I missing? -Lethe | Talk 23:04, May 18, 2005 (UTC)

Your open cover isn't locally finite. The long line can't be paracompact, because it's countably compact yet not compact. I agree that "locally compact but not second countable" doesn't appear to be relevant. --Zundark 10:56, 19 May 2005 (UTC)[reply]

The locally compact but not second countable refers, I think, to the statement above it that second countable, locally compact spaces are paracompact. It shows we cannot remove the second countability there. Also, the lower limit topology is given which shows that local compactness nor second countability are necessary for paracompactness. Hennobrandsma 15:54, 21 February 2007 (UTC)[reply]

Is the property of being locally finite a local property of a space? Isn't the long line locally homeomorphic to R? -Lethe | Talk 23:25, May 19, 2005 (UTC)

Yes, the long line is locally homeomorphic to R (except at its initial point, of course). But being locally finite isn't a property of a space, it's a property of a set of subsets of a space, so I'm not sure what it would mean for it to be a local property. Note that in a countably compact space a set of subsets is locally finite if and only if it's finite, and I wouldn't consider finiteness to be a local property. --Zundark 07:58, 20 May 2005 (UTC)[reply]

Fully normal vs. paracompact[edit]

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)[reply]

Just to weigh in here, since I wrote most of Separation axiom (and thanks for fixing that, Fropuff!): I don't know. —Toby Bartels 11:54, 31 October 2007 (UTC)[reply]

Since fully normal spaces are normal, any compact space that's not normal is a counterexample -- Walt Pohl (talk) 08:57, 16 May 2020 (UTC)[reply]

Second Property - F-sigma Case[edit]

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)[reply]

"Counterexamples" to definition[edit]

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)[reply]

Paracompactness[edit]

This article could do with a seperate section on paracompactness, as that redirects here. Brad7777 (talk) 12:32, 6 February 2012 (UTC)[reply]

The entire article is about paracompactness. What would you expect to see in a separate section? --Zundark (talk) 12:38, 6 February 2012 (UTC)[reply]

Simpler language and no mention of additional properties of other topological spaces. Brad7777 (talk) 20:26, 7 February 2012 (UTC)[reply]

Faulty Proof?[edit]

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)[reply]

The proof of the Theorem also seems to be wrong. It assumes the open cover has a locally finite sub-cover O*, when all we can assume for paracompact Hausdorff spaces is that it has a locally finite refinement. — Preceding unsigned comment added by 193.188.47.46 (talk) 13:23, 15 November 2017 (UTC)[reply]