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

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time to close? nothing since November ... but still frequently bumped to the front page. –  Gerald Edgar Apr 29 '11 at 0:58
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@Gerald: Why?, maybe one day I will get an answer... –  ε-δ May 5 '11 at 19:09
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Nice question. I'd keep it. –  Wlodzimierz Holsztynski Feb 15 '13 at 2:28
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3 Answers 3

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+500

There are completely metrizable spaces $X$ with $\dim(X)=1$ and $\mathrm{ind}(X)=0$, where $\dim$ denotes the covering dimension and $\mathrm{ind}$ the small inductive dimension. The first such example is due to P. Roy (see "nonequality of dimensions for metric spaces", TAMS 1968), but there are many others. My favourite is due to E.K. van Douwen (see "The small inductive dimension can be raised by the adjunction of a single point", Indag.Math. 1973).

Note that if $Y$ is a separable (not necessarily closed) subspace of such an $X$ then $\dim(Y)=0$ since $\dim$ and $\mathrm{ind}$ coincide for separable metrizable spaces. So any such $X$ satisfies what you want for $n=0$.

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

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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 '10 at 16:58
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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 '10 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 '10 at 19:39
    
It is nearly impossible to read your comments... –  ε-δ Nov 10 '10 at 1:02
    
The sets U, V and U∩V are NOT open in the euclidean topology!!! –  ε-δ Nov 10 '10 at 1:02
    
I'm sorry about the mistake of latex on the comment, add Another answer –  Buschi Sergio Nov 11 '10 at 15:56
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