## Nonseparable example in dimension theory?

Could you give me an example of a complete metric space with covering dimension $> n$ all of which closed separable subsets have covering dimension $\le n$?

The question closely related to this one.

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time to close? nothing since November ... but still frequently bumped to the front page. – Gerald Edgar Apr 29 2011 at 0:58
@Gerald: Why?, maybe one day I will get an answer... – ε-δ May 5 2011 at 19:09
Nice question. I'd keep it. – Wlodzimierz Holsztynski Feb 15 at 2:28

@: $\varepsilon-\delta$: $X$ isn't the the topological sum of real lines. You consider this "more shorty paths" logic:
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 $(x', y'). These "more short paths" distance describe the metric of$X$. Now a open ball centered in$(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 $} is a open euclidean triangle. -  Sorry, the metric you construct, gives$\mathbb R_+$with$\mathbb R_+$attached to each point. And I know this is 1-dimensional... Your statement about$\epsilon$-ball is correct, but for opens sets it is not longer true. – ε-δ Nov 11 2010 at 16:58 Let$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|$otherwise. (the x-axis is the river, there are othogonal paths in which we can move "as in R", while otherwise we have to go via the river first). Let$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$the the bottom horizontal line is non-separable (involving a no countable vertical segments) we call a such open a "b-open" . THen any covering$\mathcal{U}$of$X$as order 2, i.e. there exixt almost 3 elemets by no-empty intersection: Considering a b-open$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$: IF NO,the$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$: Considering the two halph line by base in$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. 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$U\cap V$. - If I understand you correctly,$X$is$\mathbb R_+$with attached$\mathbb R_+$to each point. In this case$X$is 1-dimensional; i.e. your argument should be wrong somewhere... – ε-δ Nov 7 2010 at 4:30 @$\varepsilon-\delta$: No the distance is euclidean on vertical line, but for go from a pont$(x,y)$to a point$(x’, y’)$with$x\neq x’$you have to go vertically from$(x, y) $to$(x, 0)$then horizontally to$(x’,0)$then go vertically to$(x’,y’)$. These path’s define the distance in$X$. PS. We can chose copy af a finite close interval instead the vertical lines (copies of$R^+\$) then any subspace compact is separable. This answere to another question in this forum. Please I wish delete my bad formatted comment above. – Buschi Sergio Nov 8 2010 at 19:39