An example of a non-paracompact tvs (over the reals, say)

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
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

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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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.
$\omega$ 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
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