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A standard characterization of the Baire space is that is the only nonempty, zero dimensional, Polish space in which every compact set has empty interior (up to homeomorphism of course).

I'm interested in what happens when the zero-dimensional hypothesis is dropped: does there exist, for every $n\in\{1,2,3,\ldots,\infty\}$, an $n$-dimensional Polish space in which all compact sets have empty interior? How many such spaces up to homeomorphism are there if I insist that I only want spaces where the local dimension at every point is $n$ (this is to avoid producing many boring examples by taking disjoint unions with smaller dimensional spaces)?

I know that every infinite dimensional, separable Banach space is an example for $n=\infty$, and that we have the complete Erdős space for $n=1$, but I'm already having troubles finding more examples for finite $n\geq 2$.

Edit: Following discussion in the comments, for every $n$ the space $X_n=\Bbb R^n\setminus \Bbb Q^n$ satisfies $\dim X_n=n-1$ and all of its compact subspaces have empty interior (and it is clearly $G_\delta$ in $\Bbb R^n$ hence Polish). Note that for $n=1$ we recover the Baire space, while for $n=2$ we get an example distinct from the complete Erdős space ($X_2$ is connected, unlike the complete Erdős space). I now suspect that there is an easy construction of infinite families in all dimensions starting from $X_n$.

Alessandro Codenotti
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  • Could you do a trick similar to the Cantor Leaky Tent? Like, take complete Erdos space in $\mathcal{H}$ and plant copies of $\mathbb{R}^n \setminus \mathbb{Q}^n$ at irrational points along some coordinate? – John Samples Feb 14 '21 at 05:48
  • @JohnSamples it's not very clear to me what the construction you are suggesting looks like, but what is the dimension of $\Bbb R^n\setminus\Bbb Q^n$? – Alessandro Codenotti Feb 14 '21 at 09:52
  • @AlessandroCodenotti it’s probably $n-1$ dimensional (we’re removing a zero-dimensional set). No full proof of it yet, though it’s fairly clear for the plane. – Henno Brandsma Feb 14 '21 at 21:49
  • @HennoBrandsma Actually thinking again isn't it at least $n-1$ always because $\Bbb R^{n-1}\times{\pi}\subseteq\Bbb R^n$ is a closed subspace of dimension $n-1$? (and $\dim A\leq\dim X$ for $A\subseteq X$ closed is clear) – Alessandro Codenotti Feb 14 '21 at 22:03
  • @AlessandroCodenotti you’re right, the hyperplane gives a lower bound. The upperbound is also clear from other dimension theory theorems. Of course $\dim A \le\dim X$ for any subspace (in normal spaces). – Henno Brandsma Feb 14 '21 at 22:06
  • @HennoBrandsma I believe you need extra assumptions, Roy's example of a metrizable space with $\mathrm{ind}P=0$ and $\mathrm{dim}P=1$ embeds in a totally disconnected compact Hausdorff space. In Pears book the subspace inequality is proved for totally normal spaces for example (and is stated as open for completely normal spaces). In any case the spaces $\Bbb R^n\setminus\Bbb Q^n$ are examples in all dimensions for my question. – Alessandro Codenotti Feb 14 '21 at 22:16
  • @AlessandroCodenotti engelking in his most recent dimension theory book proves it for all strongly hereditary normal spaces. In particular for separable metric ones, which is what you’re working in. – Henno Brandsma Feb 14 '21 at 22:23
  • Let us continue this discussion in chat. – Alessandro Codenotti Feb 14 '21 at 22:27

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