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## 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

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 $U\cap V$ .

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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
The sets U, V and U∩V are NOT open in the euclidean topology!!! – ε-δ Nov 10 2010 at 1:02
I'm sorry about the mistake of latex on the comment, add Another answer – Buschi Sergio Nov 11 2010 at 15:56

@: $\varepsilon-\delta$: $X$ 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