# Can a tiling on $\mathbb{R}^n$ continuously map to a tiling on $\mathbb{R}^m$ if $n > m$

Let $$n, m$$ be naturals such that $$n > m$$. Suppose $$\mathbb{R}^n$$ is tiled into finitely many regions. Each region must be disjoint, connected, continuous and have a nonzero measure.Let $$f: \mathbb{R}^n \to \mathbb{R^m}$$ be continuous. Is it possible for the images of regions in $$\mathbb{R^m}$$ to be non-intersecting. What if the number of regions is countably infinite? Are there any interesting restrictions on the tiling or the mapping that make this impossible?

Here is one idea I had. Suppose a region in $$\mathbb{R}^2$$ is $$A$$ and $$f(A) = B$$. We know that $$B$$. $$A$$ can border multiple regions while $$B$$ can only border two.

• What exactly do you mean by a tiling? Feb 10, 2019 at 22:41
• I suspect that you have restrictions on $f$ in mind that have not been specified. Is it supposed to be linear? continuous? or what? I can certainly chop $\Bbb R^m$ into the same number of pieces and make a function that takes one piece to one piece. Cardinality says so. Feb 10, 2019 at 22:45
• I mean to divide up space into distinct regions. Each region must be connected and have a nonzero measure. Feb 10, 2019 at 22:45
• I forgot to add continuous. Feb 10, 2019 at 22:45

Consider the trivial tilings of $$\Bbb{R}^n$$ and $$\Bbb{R}^m$$ consisting of one tile. The map $$f:\ \Bbb{R}^n\ \longrightarrow\ \Bbb{R}^m:\ (x_1,\ldots,x_n)\ \longmapsto\ (x_1,\ldots,x_m),$$ is continuous and maps the one tile of $$\Bbb{R}^n$$ onto the one tile of $$\Bbb{R}^m$$.
More generally, assuming that a tiling is a partition: Given any partition $$\mathcal{P}$$ of $$\Bbb{R}^m$$, we can construct a partition $$\mathcal{Q}$$of $$\Bbb{R}^n$$ such that the map above has the desired properties. Describing the partition $$\mathcal{P}$$ as a collection $$\{P_i\}_{i\in I}$$ of subsets $$P_i\subset\Bbb{R}^m$$ such that $$\bigcup_{i\in I}P_i=\Bbb{R}^m$$ and $$P_i\cap P_j=\varnothing$$ whenever $$i\neq j$$, define $$Q_i:=P_i\times\Bbb{R}^{n-m}=\{(x_1,\ldots,x_n)\in\Bbb{R}^n:\ (x_1,\ldots,x_m)\in P_i\}\subset\Bbb{R}^n.$$ It is not hard to verify that $$\mathcal{Q}:=\{Q_i\}_{i\in I}$$ is a partition of $$\Bbb{R}^n$$, and that $$f(Q_i)=P_i$$.
• This shows that every partition of $\mathbb{R}^m$ corresponds to a partition of $\mathbb{R}^n$. I want to see the other way around. Feb 10, 2019 at 22:55
• Then you should have asked that; in stead you asked whether it is possible for a tiling of $\Bbb{R}^n$ to map continuously into a tiling of $\Bbb{R}^m$. My answer shows that this is possible, and even gives an example for every tiling of $\Bbb{R}^m$. Feb 14, 2019 at 22:00