Talk:Paracompact space

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

---

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

Long line not paracompact

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)

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)