Is the long line paracompact? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:10:59Z http://mathoverflow.net/feeds/question/229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/229/is-the-long-line-paracompact Is the long line paracompact? Anton Geraschenko 2009-10-09T17:12:28Z 2009-11-04T17:54:04Z <p>A manifold is usually defined as a <a href="http://en.wikipedia.org/wiki/Second-countable%5Fspace" rel="nofollow">second-countable</a> <a href="http://en.wikipedia.org/wiki/Hausdorff%5Fspace" rel="nofollow">hausdorff</a> topological space which is locally homeomorphic to <b>R</b><sup>n</sup>. My understanding is that the reason "second-countable" is part of the definition is to make sure that the space is <a href="http://en.wikipedia.org/wiki/Paracompact%5Fspace" rel="nofollow">paracompact</a>, which you want so that you get locally finite partitions of unity. Once you have locally finite partitions of unity, basically anything you can multiply by a function can be constructed locally (any presheaf that is a module over the sheaf of functions is automatically a sheaf), a property you want manifolds to have.</p> <p>But do we unnecessarily throw out some paracompact topological spaces which "should" be manifolds by requiring second-countability. A boring example is an uncountable disjoint union of manifolds, but there are other more interesting spaces that kind of look like they should be manifolds.</p> <p>In particular, is the <a href="http://en.wikipedia.org/wiki/Long%5Fline%5F%28topology%29" rel="nofollow">long line</a> paracompact? Should I consider it a manifold?</p> http://mathoverflow.net/questions/229/is-the-long-line-paracompact/233#233 Answer by Richard Dore for Is the long line paracompact? Richard Dore 2009-10-09T18:56:50Z 2009-10-09T18:56:50Z <p>The wikipedia article states that the long line is not paracompact. Here is a proof that the long ray is not paracompact (so neither is the long line):</p> <p>Start with the open covering [0, α) for every ordinal α &lt; ω<sub>1</sub>. Let X be some refinement of that covering. Let S be the limit ordinals below ω<sub>1</sub>. S is a stationary subset of ω<sub>1</sub>. For each β in S, pick a Y<sub>β</sub> in X so that β is in Y<sub>β</sub>. Consider the function f from S to ω<sub>1</sub> which sends β to the least ordinal in Y<sub>β</sub>. For any β in S, f(β) &lt; β since β is a limit ordinal. So by <a href="http://en.wikipedia.org/wiki/Fodor%27s%5Flemma" rel="nofollow">Fodor's Lemma</a>, There is a stationary subset of S, call it S', so that f is constant on S'. Let γ be the value of f on S'. Then γ is in Y<sub>β</sub> for all β in S'. So the refinement is not finite (or even countable) at γ.</p> http://mathoverflow.net/questions/229/is-the-long-line-paracompact/235#235 Answer by Andrew Stacey for Is the long line paracompact? Andrew Stacey 2009-10-09T19:56:52Z 2009-10-09T19:56:52Z <p>For <em>connected</em> locally Euclidean spaces, paracompactness is equivalent to second countable. Therefore if we truly insisted that our manifolds be second countable then the only thing that we would lose would be arbitrary coproducts. However, we don't insist that. Although it's often stated in the definition early on in a differential topology course or book, in my experience that's just because it's easier to explain than paracompactness (I tend to pick metrisable, myself). Of course, that early in a course we're probably not too concerned about arbitrary coproducts either.</p> <p>Moreover, because manifolds split into a coproduct of their connected components, we tend to deal with connected manifolds unless we really can't avoid it. And even when they aren't connected, they most often have countably many components. So in practise the distinction doesn't arise.</p> <p>I don't have it in front of me here, but I believe that an appendix to the first volume of Spivak's "Introduction to differential geometry" contains a proof of four equivalent conditions for locally Euclidean spaces (perhaps requiring Hausdorff). If I remember aright, the conditions are: paracompact, second countable, metrisable, and &sigma;-compact.</p> <p>There was a paper on the arxiv on Monday, <a href="http://www.arxiv.org/abs/0910.0885" rel="nofollow">0910.0885</a>, which lists 107 conditions for a <em>connected</em> locally Euclidean Hausdorff space equivalent to that it be metrisable. Amongst them are paracompactness and second countable.</p> http://mathoverflow.net/questions/229/is-the-long-line-paracompact/237#237 Answer by Eric Wofsey for Is the long line paracompact? Eric Wofsey 2009-10-09T20:04:58Z 2009-10-09T20:15:03Z <p>Here's another proof, which shows that any connected paracompact locally Euclidean space X is second-countable. Cover X by Euclidean charts and take a locally finite refinement. Say an open set is good if it only intersects finitely many of the charts. Now take any point x and take a good neighborhood of it. The charts that intersect that good neighborhood can then themselves be covered by countably many good open sets. There are then only countably many charts intersecting those good open sets, and those charts can be covered by countably many good sets. Iterating this countably many times, you get an open set U associated to x which is covered by countably many charts such that if a chart intersects U, it is contained in U. It follows that the complement of U is also a union of charts, so by connectedness U is all of X. Thus X can be covered by countably many charts and is second-countable.</p> http://mathoverflow.net/questions/229/is-the-long-line-paracompact/4127#4127 Answer by Apollo for Is the long line paracompact? Apollo 2009-11-04T17:54:04Z 2009-11-04T17:54:04Z <p>The <em>Handbook of Set-Theoretic Topology</em> is the bible for this kind of question. It has a chapter on the long-line and a chapter on questions of paracompactness of manifolds.</p>