Locally complete space is topologically equivalent to a complete space - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T04:55:54Z http://mathoverflow.net/feeds/question/21954 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21954/locally-complete-space-is-topologically-equivalent-to-a-complete-space Locally complete space is topologically equivalent to a complete space Tom Ellis 2010-04-20T11:43:14Z 2010-04-20T16:28:07Z <p>Can someone please tell me where I can find a citeable reference for the following result:</p> <p>Call the metric space $(X, d)$ "locally complete" if for every $x \in X$ there a neighbourhood of $x$ which is complete under $d$.</p> <p>If $(X,d)$ is locally complete and separable then there exists a metric $d'$ on $X$ such that $(X,d) \to (X,d')$ is a homeomorphism and $(X, d')$ is complete.</p> <p>This result follows immediately from Alexandrov's theorem that a $G_\delta$ subset of a Polish space is Polish, but I'd rather find a statement in the literature of the above more straightforward result if there is one.</p> http://mathoverflow.net/questions/21954/locally-complete-space-is-topologically-equivalent-to-a-complete-space/21956#21956 Answer by Carl Mummert for Locally complete space is topologically equivalent to a complete space Carl Mummert 2010-04-20T12:12:25Z 2010-04-20T12:12:25Z <p>The result you want to prove also follows directly from Choquet's characterization of completely metrizable metric spaces as those for which the second player has a winning strategy in the "strong Choquet game". See Kechris, <i>Classical descriptive set theory</i>. The usual proof of this characterization uses Alexandrov's theorem, though, so it may not be what you are looking for. </p> http://mathoverflow.net/questions/21954/locally-complete-space-is-topologically-equivalent-to-a-complete-space/21957#21957 Answer by Sergei Ivanov for Locally complete space is topologically equivalent to a complete space Sergei Ivanov 2010-04-20T12:16:18Z 2010-04-20T12:16:18Z <p>This is not a reference but a short direct proof.</p> <p>Let $\bar X$ be the completion of $X$. Define $f:X\to\mathbb R$ by $f(x)=dist(x,\bar X\setminus X)$. Obviously $f$ is continuous, and the local completeness implies that $f$ is strictly positive.</p> <p>Let $X'\subset X\times\mathbb R$ be the graph of $1/f$. Then $X'$ is homeomorphic to $X$ and complete for the following reason:</p> <p>If <code>$\{p_n\}$</code> is a Cauchy sequence in $X'$, then the first coordinates of $p_n$ are a Cauchy sequence in $X$. Hence they converge to a point of $\bar X$. This point cannot be in $\bar X\setminus X$ because the second coordinates of $p_n$ are bounded.</p> <p>The metric on $X\times\mathbb R$ can be defined as the sum of two coordinate distances (any other common definition will work too).</p> http://mathoverflow.net/questions/21954/locally-complete-space-is-topologically-equivalent-to-a-complete-space/21983#21983 Answer by Henno Brandsma for Locally complete space is topologically equivalent to a complete space Henno Brandsma 2010-04-20T16:28:07Z 2010-04-20T16:28:07Z <p>For a reference: <a href="http://www.chimiefs.ulg.ac.be/SRSL/newSRSL/modules/FCKeditor/upload/File/73_1/Bella%20TSIRULNIKOV%20p%209-19.pdf" rel="nofollow">this paper</a> has a reference [30] that has a proof. The author cites your result and refers to it. I don't have access to these papers, so I cannot verify exactly.</p>