# Partitioning $\mathbb{R}^2$ into disjoint path-connected dense subsets

Does there exist a partition of the plane into $n=3$ (or more generally $n\ge 3$) disjoint path-connected dense subsets?

Note that the answer is yes if "path-connected" is replaced by "connected", as shown here. The linked question also shows that the answer is yes for $n=1$ (trivial) and $n=2$ (not quite trivial, but nice and explicit.)

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I don't have an answer now nor do I have time now to give this much thought, but perhaps some of the ways that ${\mathbb Q}^{2} \cup \left({\mathbb R}\;-\;{\mathbb Q}\right)^{2}$ can be proved path-connected could be of help. See my 23 June 2009 sci.math post and the earlier posts I cite in it (two from 2002 and one from 2005). –  Dave L. Renfro Dec 4 '12 at 22:01
Thanks Dave, that sounds very interesting, I'll have a look at it later. –  Lukas Geyer Dec 5 '12 at 17:13
@DaveL.Renfro: Tell me if I'm wrong, but isn't there a straight-forward proof that $\mathbb{Q}^2 \cup (\mathbb{R} - \mathbb{Q})^2$ is path-connected, using the fact that any line segment with rational endpoints which is not vertical or horizontal is contained in that set? Then for any two points in your set, approximate them with points with rational coordinates, connect them with line segments, just making sure that you don't get horizontal or vertical line segments in the approximation (which is pretty easy). Or am I missing something? –  Lukas Geyer Dec 5 '12 at 17:30
This is the proof I gave in my 14 May 2002 sci.math post, where I wrote: I asked my Ph.D. advisor whether he thought the result was worthy of publication. He said he'd think about it for a day or two and see if he could come up with a more straightforward proof. The next day he said there's a fairly easy way to do it, so it's probably not publishable. I gave a more detailed version of this proof in my 19 October 2005 sci.math post, where I also said it can be found on p. 187 (Example 2 in Chapter 6.3) of Allan J. Sieradski, An Introduction to Topology and Homotopy (1992). –  Dave L. Renfro Dec 5 '12 at 19:14
@DaveL.Renfro: Oh yeah, I missed that short paragraph with the easy proof when I skimmed through the post. –  Lukas Geyer Dec 5 '12 at 20:06

First we partition $\mathbb Q^2$ into $n$ dense subsets $A_1, \ldots , A_n$ and fix an enumeration $a_{i,1}, a_{i,2}, a_{i,3}, \ldots$ for each of set $A_i$.

As a starter, assign colour $i$ to $a_{i,1}$, that is let $S_i^{(1)}=\{a_{i,1}\}$.

Assume we have the following situation:

$m\in\mathbb N$ and we have pairwise disjoints sets $S_1^{(m)}, \ldots , S_n^{(m)}$, where each $S_i^{(m)}$ is a polygonal curve from $a_{i,1}$ to $a_{i,m}$ with $S_i^{(m)}\cap \mathbb Q^2 = \{a_{i,k}\mid k\le m\}$.

Then the complement $\mathbb R^2\setminus \bigcup S_i^{(m)}$ is connected and open. For $i=1, \ldots , n$ we will append a path to $a_{i,m+1}$ as follows: First observe that the set $$X=\mathbb R^2\setminus\left(\bigcup_{j<i}S_j^{(m+1)}\cup \bigcup_{j\ge i}S_j^{(m)}\right)$$ is open and connected. There is a path in $X\cup\{a_{i,m}\}$ from $a_{i,m}$ to $a_{i,m+1}$. Since $X$ is open, we can replace the path with a polygonal path. Also, we can slightly move the nodes of the path (apart from start and end node) such that all line segments avoid rational points (apart from $a_{i,m}$ and $a_{i,m+1}$); in short, this is possible because we have uncountably many choides to position the nodes in an open disc, but there are only countably many points to avoid. Then let $S_i^{(m+1)}$ be the polyline $S_i^{(m)}$ together with the polyline just found. Especially, $S_i^{(m)}\subset S_i^{(m+1)}$.

When we have done this for all $i$, we have obtained the situation described above, but with $m$ replaced by $m+1$.

Now let $$S_i=\bigcup_{m\in\mathbb N}S_i^{(m)}.$$ Then $S_i$ is pathwise connected (any two points are connected in some $S_i^{(m)}$) and is dense in $\mathbb R^2$ because $A_i\subset S_i$. However, I must admit that we do not have $\mathbb R^2=\bigcup S_i$ yet, i.e. the set $$Y=\mathbb R^2\setminus\bigcup S_i$$ need not be empty.

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This is similar to the construction for the connected case, and yes, the main difficulty is how to attach the set $Y$ in a path-connected way to the pieces $S_i$. (You can attach all of it to $S_1$ and end up with $n-1$ path-connected and one connected set. However, I have no idea how to make this one set path-connected.) –  Lukas Geyer Nov 28 '12 at 22:52
@LukasGeyer You don't need to "make" it path-connected. It is path-connected because it is connected (in $\mathbb R^2$). –  Phira Dec 5 '12 at 10:06
@Phira Actually I don't see myself how e.g. $S_1\cup Y$ would still be path-connected –  Hagen von Eitzen Dec 5 '12 at 16:02
@HagenvonEitzen You are right, I am sorry, I just read that $X$ was open, but $Y$ isn't. –  Phira Dec 5 '12 at 16:22