An example of a non-paracompact tvs (over the reals, say) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:17:58Z http://mathoverflow.net/feeds/question/44907 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44907/an-example-of-a-non-paracompact-tvs-over-the-reals-say An example of a non-paracompact tvs (over the reals, say) David Roberts 2010-11-05T01:25:37Z 2013-02-11T07:54:48Z <p>What is an example of a non-paracompact topological vector space?</p> <p>I'm aware of <a href="http://mathoverflow.net/questions/3241/when-is-a-locally-convex-topological-vector-space-normal-or-paracompact" rel="nofollow">this question</a>, 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.</p> <p>This is in order to answer <a href="http://mathoverflow.net/questions/44859/cw-complexes-and-paracompactness" rel="nofollow">this other question</a>.</p> http://mathoverflow.net/questions/44907/an-example-of-a-non-paracompact-tvs-over-the-reals-say/44929#44929 Answer by anon for An example of a non-paracompact tvs (over the reals, say) anon 2010-11-05T07:37:30Z 2010-11-05T08:09:25Z <p>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 $y &lt; x$ 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.</p> http://mathoverflow.net/questions/44907/an-example-of-a-non-paracompact-tvs-over-the-reals-say/45283#45283 Answer by Buschi Sergio for An example of a non-paracompact tvs (over the reals, say) Buschi Sergio 2010-11-08T09:53:31Z 2010-11-08T09:53:31Z <p>I find it: <a href="http://www.jstor.org/pss/2040014" rel="nofollow">http://www.jstor.org/pss/2040014</a></p> http://mathoverflow.net/questions/44907/an-example-of-a-non-paracompact-tvs-over-the-reals-say/121377#121377 Answer by jbc for An example of a non-paracompact tvs (over the reals, say) jbc 2013-02-10T06:18:08Z 2013-02-10T06:18:08Z <p>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.</p> http://mathoverflow.net/questions/44907/an-example-of-a-non-paracompact-tvs-over-the-reals-say/121450#121450 Answer by David Roberts for An example of a non-paracompact tvs (over the reals, say) David Roberts 2013-02-11T07:54:48Z 2013-02-11T07:54:48Z <p>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.</p> <p>Consider an uncountable set $\mathfrak{n}$ and the space $\mathbb{R}^\mathfrak{n}$ in the product topology. This is <a href="http://www.ams.org/journals/bull/1948-54-10/S0002-9904-1948-09118-2/home.html" rel="nofollow">not normal</a>, but still Hausdorff, hence not paracompact.</p>