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All locally convex topological vector spaces (LCTVS) are completely regular, since their topology is given by a family of semi-norms. I'm interested in conditions that imply that a LCTVS is paracompact or normal.

Some background: I have a particular space in mind. It is a directed colimit (union) over an uncountable family of nuclear Frechet spaces. It is complete but not metrisable nor separable nor nuclear. I have a very concrete description of it and can describe the bounded (and compact) subsets. This space is fairly badly behaved in other ways so I'm half anticipating a negative result, thus answers along the lines of "If you can find a subset that looks like X then it can't be normal" could be just what I'm looking for.

So in particular, I'm interested in the general question. To forestall a couple of "easy" answers: as my space is not Frechet, it is not metrisable so I can't directly use theorems on metrisability (however, as it is a colimit there may be some scope for indirect use). And, of course, paracompact would imply normality since it is completely regular.

To forestall another possible comment, I'm not going to tell you what the particular space is. Partly because I'm more interested in the general situation, this space is merely focussing my attention on the question, and partly because it's a more useful question if it's general. A bit of intelligent searching would reveal what space it is anyway so it's no great hardship.

Edit: There is a simple example of a space that would be very interesting to know about: the sum (i.e. coproduct) of an uncountable number of copies of $\mathbb{R}$, or even more specifically $\sum_{\mathbb{R}} \mathbb{R}$. This isn't the specific space that I'm interested in, but is close enough that I think that an answer either way for this space will tell me what to do for my space.

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...sum of an uncountable number of copies of R... with what topology? –  Gerald Edgar Nov 12 '09 at 13:48
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Colimit topology: strongest locally convex topology so that the inclusions of all finite subsums are continuous. –  Andrew Stacey Nov 13 '09 at 10:28
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2 Answers

Thanks to your other question, I was on a LCTVS kick. I did find one general criterion that implies that a locally convex space is paracompact. According to the Encyclopedia of Mathematics, if it is Montel (which means that it is barrelled and the Heine-Borel theorem holds true for it), then it is paracompact. Although this criterion is important, it is of no use to your specific question.

I thought I had a proof for half of your question, which I wrote up as the first version of this answer, but I made a mistake and proved something different. My thinking is based on the fact that the normality axiom for a topological space is equivalent to the Tietze extension theorem. (Tietze extension follows from normality. In the other direction, if $A$ and $B$ are the two closed sets, you obtain disjoint open neighborhoods from a continuous function that is 0 on $A$ and 1 on $B$.) However, in my argument I conflated the locally convex direct sum of spaces with the topological direct sum. For a countable direct sum of copies of $\mathbb{R}$, they are the same topology, and they agree with the box topology. But Waelbroeck, LNM 230 points out that they are different in the uncountable case.

Let $\alpha$ be an ordinal, for instance an ordinal of cardinality $2^{\aleph_0}$. Then $\mathbb{R}^\alpha$ in the topological direct sum topology satisfies Tietze extension. Let $A \subset \mathbb{R}^\alpha$ be a closed set and let $f:A \to \mathbb{R}$ be a continuous function. For $\beta < \alpha$, let $A_\beta$ be the intersection of $A$ and with $\mathbb{R}^\beta$. Suppose that $\alpha = \beta+1$ is a successor ordinal. If $\alpha$ is finite, then the conclusion is standard. Otherwise, by induction, there is an extension $f_\beta$ of $f$ to $\mathbb{R}^\beta$. Moreover, by induction in a different sense, we have already proved that $\mathbb{R}^{\beta+1}$ is normal, since $\beta$ and $\beta+1$ have the same cardinality. So there exists an extension $f_\alpha$ to $\mathbb{R}^\alpha$. If instead $\alpha$ is a limit ordinal, then the extensions all the way up to $\alpha$ work just because they work; that's the behavior of topological direct limits.

Having failed to normality for the locally convex direct sum, I can't say much about paracompactness either. :-) However, there is an interesting result called the Michael selection theorem which seems to do for paracompactness what the Tietze theorem does for normality. If the Tietze theorem is useful for your spaces, then maybe the Michael selection theorem is too.

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This is going to take me a while to parse. First question: you say that the direct limit for direct sums in LCTVS is the box topology. Do you have a reference for that? Is it peculiar to having a limit ordinal? I seem to be able to construct a neighbourhood using an infinite simplex that doesn't contain a box, but I may well be wrong. –  Andrew Stacey Dec 14 '09 at 18:56
    
Well, geez. I need to review whether or not it is finer than the box topology. If so, that is a mistake already, but not the essential point. What I really may need, whether or not it is the box topology, is that this l.c. inductive limit happens to equal the topological inductive limit. I construct a Tietze function by transfinite induction. I should also simplify and reword the argument a bit, but first let me review whether or not it works. –  Greg Kuperberg Dec 14 '09 at 19:34
    
It's been a while since I first worked on this problem and I'd forgotten some of the subtleties. In particular, reading your answer again I recall that I'm interested in both the inductive LCTVS topology and the inductive topology (over inclusions of finite dimensional subspaces) so your answer on the latter is still very useful even though it's not directly an answer of the stated question! –  Andrew Stacey Dec 16 '09 at 8:59
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A paper dealing with this (and other) questions for the weak topology of a Banach space ... H. H. Corson, "The weak toplology of a Banach space" Trans. Amer. Math. Soc. 101 (1961) 1--15.

In that case:

X is paracompact iff X is Lindelof.

If X^n is normal for all n, then X is real-compact.

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I'll take a look. My space is not Lindelof, but then it's not a Banach space with the weak topology either. Still, there may be ideas that still apply. Thanks. –  Andrew Stacey Oct 30 '09 at 8:45
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