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It is a classic result that a separable metrizable space $X$ can be embeded into $\mathbb{R}^{2n+1}$, where $n=\dim X$; thus $D(X)\le2\dim X+1$. For the universal spaces of Menger and Nöbeling this is …

You can read a review of the paper in Zentralblatt, it contains a short description in German. The review on MathSciNet is a bit more extensive (but requires a subscription). There is indeed the condi …

This works best via the theorem on partitions (Theorem 7.2.15 in Engelking's General Topology): $\dim X\le n$ iff for every sequence $(A_1,B_1)$, ..., $(A_n,B_n)$, $(A_{n+1},B_{n+1})$ of pairs of disj …

References to Engelking's Dimension Theory (1978) ISBN 0-444-85176-3. The answer is yes for compact metrizable spaces, see Section 1.13. In general it is no in general, see Example 3.3.8 (Lokucievski …

Yes, the long ray $R$ works. If $\mathcal{U}$ is a finite open cover then $\bigcap\{R\setminus U:U\in\mathcal{U}\}=\emptyset$ and at least one of these closed sets must be bounded as in $R$ any two cl …

Using @Anonymous' second comment one can show that your property characterizes locally compact scattered spaces. For if $X$ is locally compact Hausdorff and not scattered then it contains a closed den …