Nonseparable example in dimension theory? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:53:28Z http://mathoverflow.net/feeds/question/43915 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43915/nonseparable-example-in-dimension-theory Nonseparable example in dimension theory? ε-δ 2010-10-28T00:14:42Z 2013-05-10T06:22:00Z <blockquote> <p>Could you give me an example of a complete metric space with covering dimension $> n$ all of which closed separable subsets have covering dimension $\le n$?</p> </blockquote> <p>The question closely related to <a href="http://mathoverflow.net/questions/43680" rel="nofollow">this one</a>.</p> http://mathoverflow.net/questions/43915/nonseparable-example-in-dimension-theory/45120#45120 Answer by Buschi Sergio for Nonseparable example in dimension theory? Buschi Sergio 2010-11-06T23:59:49Z 2010-11-07T08:54:27Z <p>Let $X$ the river metric on the positive plane $(R^+)^2$: $d((x,y),(u,v)) = |y-v|$ if $x=u$ and $|y| + |v| + |x-u|$ otherwise.</p> <p>(the x-axis is the river, there are othogonal paths in which we can move "as in R", while otherwise we have to go via the river first).</p> <p>Let $v_x$ the vertical line by orizonatal quote $x$, then the subspace topology of $v_x\setminus$ {$(x,0)$} is the euclidean topology, but any open that containing a point of $r_0$ the the bottom horizontal line is non-separable (involving a no countable vertical segments) we call a such open a "b-open" .</p> <p>THen any covering $\mathcal{U}$ of $X$ as order 2, i.e. there exixt almost 3 elemets by no-empty intersection:</p> <p>Considering a b-open $U\in \mathcal{U}$, exist a point $x\in r_o \cap (Cl(U)\setminus U)$ (i.e. in its boundary in $r_0$) then there exixt another b-open $V\in \mathcal{U}$ containing $x$, and $U\cap V$ is a b-open. The sets $U$, $V$ and $U\cap V$ are also open in the euclidean topology and we can assume also connected and containing its own projection on $r_0$, then because $R^2$ has dimension 2 follow that exist a point $(x', y)\notin U\cup V$ such that any open set containing $(x', y)$ intersect $U\cap V$ (otherwise we can make a cover of order 2, then any refinement has order 2). Then the element $F\in \mathcal{U}$ containing $(x', y)$ (like a open interval in the $v_{x'}$) meet also $U\cap V$ :</p> <p>IF NO,the $sup$-extrem of the quote that $U\cap V$ can reachedin in the $v_{x'}$ vertical is minor of $y$. Any (open ball) $B_\epsilon(x)$ with $x\in r_0$ is like a halph-square triangle then if this is included in $U\cap V$ it dont meet the follow open part $S\subset R^2$:</p> <p>Considering the two halph line by base in $x',y$ at right by $45°$ pendence and $-45°$ at left (like the graph of $y=|x|$ traslated from origin to $(x', y)$) and consider the superior part $S$ of plane $R^2$ these halph-line cut off, by these halph line included. </p> <p>but then $U\cap V$ being union of (open) balls dont meet $S$, then exixt a (euclidean) open neighbord of $(x', y)$ that dont meet $U\cap V$ .</p> http://mathoverflow.net/questions/43915/nonseparable-example-in-dimension-theory/45720#45720 Answer by Buschi Sergio for Nonseparable example in dimension theory? Buschi Sergio 2010-11-11T16:25:26Z 2010-11-11T16:25:26Z <p>@: $\varepsilon-\delta$: $X$ isn't the the topological sum of real lines. You consider this "more shorty paths" logic: </p> <p>For move in a vertical line you go on the line. But for go from $(x, y)$ to $x', y')$ ($x\neq x'$) you go (vertically) from $(x, y)$ to $(0, x)$ then go (horizontally) to $(x', 0)$ then go (vertically) to $(x', y').</p> <p>These "more short paths" distance describe the metric of$X$.</p> <p>Now a open ball centered in$(x, 0)$and radius$\epsilon$is a triangle isosceles, with the top right corner (the half of a square), its hight is$\epsilon$, and its base (intersection by the$x$-axis) is a open interval of lenght$2\epsilon$(centered in$x$), and the intersection by {$(x, y)\in R^2 | y>0 \$} is a open euclidean triangle.</p>