Non-regular Connected Hausdorff Banach Manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:49:31Z http://mathoverflow.net/feeds/question/104965 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104965/non-regular-connected-hausdorff-banach-manifold Non-regular Connected Hausdorff Banach Manifold Benjamin Dickman 2012-08-18T03:02:17Z 2012-09-02T19:10:28Z <p>After reading <a href="http://mathoverflow.net/questions/104575/under-exactly-what-extra-conditions-if-any-is-a-connected-hausdorff-manifold" rel="nofollow">this MO post</a>, I am wondering: </p> <p><strong>Is every (connected) Hausdorff Banach manifold a regular space?</strong></p> <p>Though unjustified, page 53 of <a href="http://www.maik.ru/full/rusmath/97/10/rusmath10_97p49full.pdf" rel="nofollow">this paper</a> nonchalantly states: "Note that a Hausdorff Banach manifold X is a regular space."</p> <p>But does anyone know of a proof of this statement (or a counterexample)?</p> <p>Of course, the real difficulty arises in proving the statement for the infinite-dimensional version, since such a Banach manifold will <em>not</em> be locally compact.</p> <hr> <p><strong>Follow-up:</strong> Now that Theo Buehler has kindly pointed to a counterexample (i.e. a connected Hausdorff Banach manifold which is <em>not</em> regular) perhaps it will give someone an idea about how to tackle <a href="http://mathoverflow.net/questions/104575/under-exactly-what-extra-conditions-if-any-is-a-connected-hausdorff-manifold" rel="nofollow">the question</a> that provided the inspiration for this one.</p> http://mathoverflow.net/questions/104965/non-regular-connected-hausdorff-banach-manifold/105075#105075 Answer by unknown (google) for Non-regular Connected Hausdorff Banach Manifold unknown (google) 2012-08-20T05:03:51Z 2012-08-20T16:21:03Z <p>I'm confused as to why (failure of) local compactness is an issue. I think one can just prove this directly as follows. It's rather trivial, though, so maybe I'm missing something.</p> <p>Let $x\in X$ be a point and $K\subseteq X$ a closed subset with $x\notin K$. Find some coordinate chart near $x$, that is, a map $f:E\to X$ which is a homeomorphism onto some open set containing $x$, where $E$ is a Banach space and $f(0)=x$. Now $f^{-1}(K)$ is a closed subset of $E$ which does not contain $0$. Thus there exists $\epsilon>0$ such that $\|k\|\geq\epsilon$ for all $k\in f^{-1}(K)$. Let $B(0,a)$ denote the open ball in $E$ centered at $0$ with radius $a$. Then we simply observe that the following two open sets are disjoint and "separate" $x$ and $K$: $$x\in f(B(0,\epsilon/3))$$ $$K\subseteq X\setminus f(\overline{B(0,2\epsilon/3)})$$ I'm using Hausdorffness to conclude that $f(\overline{B(0,2\epsilon/3)})$ is closed in $X$.</p> <p><strong>EDIT</strong>: As the comments point out, the point here is exactly to show that $f(\overline{B(0,2\epsilon/3)})$ is closed, which in finite dimensions (i.e. locally compact) follows easily from Hausdorfness. So, as of now this argument is incomplete.</p> http://mathoverflow.net/questions/104965/non-regular-connected-hausdorff-banach-manifold/105595#105595 Answer by Theo Buehler for Non-regular Connected Hausdorff Banach Manifold Theo Buehler 2012-08-27T02:34:31Z 2012-08-27T20:37:15Z <p>Apparently the answer is <strong>no</strong>, not every connected Hausdorff Banach manifold is regular, not even when it is modeled on a separable Hilbert space.</p> <p>I quote (verbatim) from J.&nbsp;Margalef-Roig, E.&nbsp;Outerelo-Dominguez, <em>Differential Topology</em>, North Holland Mathematics Studies 173, 1992, page&nbsp;44f.</p> <blockquote> <p>It is well known the result of General Topology that every Hausdorff locally compact topological space satisfies the Tychonoff axiom [M-O-P, V.2, pg. 231]. By this and the Riesz's theorem every Hausdorff locally finite dimensional differentiable manifold satisfies the Tychonoff axiom. This last affirmation is not true for arbitrary Hausdorff differentiable manifolds. In [M.O.1] there is an example of a Hausdorff connected differentiable manifold $X$ of class $\infty$, such that $\partial X = \emptyset$, $X$ is not regular and $X$ admits an atlas whose charts are modelled over an infinite dimensional separable real Hilbert space.</p> </blockquote> <p>They continue to add the regularity hypothesis in their results whenever it is needed.</p> <p>The cited references are:</p> <ul> <li><p>[M.O.P.] MARGALEF, J.-OUTERELO, E.-PINILLA, J.L.: Topologia, I-V. Alhambra, Madrid 1975, 79, 79, 80 and 1982.</p></li> <li><p>[M.O.1] MARGALEF, J.-OUTERELO, E.: Una variedad diferenciable de dimension infinita, separada y no regular. Rev. Mat. Hisp.-Am, IV, V.42, 1982, 51-55. <a href="http://tinyurl.com/8kb3m45" rel="nofollow">(QuickView link)</a>.</p></li> </ul> <hr> <p><strong>Edit:</strong> As was pointed out by Benjamin Dickman in the comments, the example also appears in English in A. Kriegl, P.W.&nbsp;Michor, <em><a href="http://books.google.com/books?id=s7fPYRqhXEUC" rel="nofollow">The convenient setting of global analysis</a></em>, AMS (1997), <a href="http://books.google.com/books?id=s7fPYRqhXEUC&amp;pg=PA266" rel="nofollow"><strong>27.6</strong> Non-regular manifold</a>, page&nbsp;266. The book is available as <a href="http://www.mat.univie.ac.at/~kriegl/Skripten/apbook.pdf" rel="nofollow">a pdf file</a> on Kriegl's homepage.</p>