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What is an example of a non-paracompact topological vector space?

I'm aware of this question, but I don't care if my tvs is locally convex. In fact the wilder the better. The only criterion is that it should be contractible, which it would be, assuming $\mathbb{R} \times V \to V$ is continuous.

This is in order to answer this other question.

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(In your haste, I think you forgot to ask the question.) –  Harry Gindi Nov 5 '10 at 1:30
    
@Harry: Is there <title>? –  Andres Caicedo Nov 5 '10 at 1:38
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tvs = topological vector space –  Anton Geraschenko Nov 5 '10 at 1:59
    
And if any functional analysts are reading this: I'm still interested in any thoughts anyone might have on the first question that David cites. –  Andrew Stacey Nov 5 '10 at 7:33
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4 Answers

up vote 1 down vote accepted

Henno Bransma provided this answer in a comment to anon's answer, which I think is the easiest and best, hence I'm adding it as a community wiki answer.

Consider an uncountable set $\mathfrak{n}$ and the space $\mathbb{R}^\mathfrak{n}$ in the product topology. This is not normal, but still Hausdorff, hence not paracompact.

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This is an old question which has been answered satisfactorily but the following remark might shed some light on it and put it in a broader context: we can embed any completely regular space as a closed subspace of a separated locally convex space in a canonical manner. We simply provide the free vector space over the set with the finest locally convex structure so that the natural inclusion is continuous. Hence if any topological property is stable under closed subspaces (e.g., normality, paracompactness), then whenever we can find a completely regular space which fails it, we can find a locally convex space which fails it.

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If I have not goofed on some detail, here is an example. Let $w$ be the first uncountable ordinal (= the set of countable ordinals). Regarded as a topological space with its usual order topology, it is not paracompact. The space $\mathbb R^w$ with the usual product topology is a topological vector space over $\mathbb R$ in an obvious way. Let $V$ denote the subspace $\mathbb R^w$ consisting of those functions from $w$ to $\mathbb R$ whose support is at most countable. Then $V$ is a topological vector space over R in an obvious way. $V$ contains a closed subset homeomorphic to $w$ (namely the set of functions $g_x$, where $g_x(y) = 1$ if $y < x$ and $0$ otherwise; the map sending $x$ in $w$ to $g_x$ is a homeomorphism onto its range). A closed subspace of a paracompact space must be paracompact, so $V$ is not paracompact.

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$\omega$ usually refers to the first infinite ordinal, not the first uncountable ordinal, which could be denoted by $\omega_1$ or $\aleph_1$ or, outside of set theory, by $\Omega$. –  Joel David Hamkins Nov 5 '10 at 12:54
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You could also use that $\bf{R}^\kappa$ in the product topology is not normal for any uncountable index set $\kappa$, so in particular not paracompact. The latter is turns out to be the typical way to fail: $C_p(X)$ (space of continuous functions on a Tychonov space X in the pointwise topology) is paracompact iff it is normal. And the standard product is of course a $C_p(X)$ space for a discrete space. –  Henno Brandsma Nov 8 '10 at 20:15
    
@Henno - if you add this as an answer, I will accept it. –  David Roberts Nov 8 '10 at 23:01
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