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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.

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    $\begingroup$ What exactly do you mean by a tiling? $\endgroup$
    – Servaes
    Feb 10, 2019 at 22:41
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    $\begingroup$ 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. $\endgroup$ Feb 10, 2019 at 22:45
  • $\begingroup$ I mean to divide up space into distinct regions. Each region must be connected and have a nonzero measure. $\endgroup$
    – Halbort
    Feb 10, 2019 at 22:45
  • $\begingroup$ I forgot to add continuous. $\endgroup$
    – Halbort
    Feb 10, 2019 at 22:45

1 Answer 1

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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 $f(Q_i)=P_i$.

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  • $\begingroup$ 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. $\endgroup$
    – Halbort
    Feb 10, 2019 at 22:55
  • $\begingroup$ 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 $\Bbb{R}^m$. $\endgroup$
    – Servaes
    Feb 14, 2019 at 22:00

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