Does there exist pairwise disjoint, connected, dense subsets $U_1,\dots, U_n \subset \mathbb{R}^2$ such that $U_1\cup \cdots \cup U_n =\mathbb{R}^2$?

If $n=1$, then we can take $U_1 = \mathbb{R}^2$.

If $n=2$, then we can take $$U_1= \{(x,y)\in\mathbb{R}^2| y\neq 0~\text{and} \sqrt{x^2+y^2}\in\mathbb{Q}\}\cup\{(x,0)\in\mathbb{R}^2|x\geq 0\}$$ and $$U_2=\{(x,y)\in\mathbb{R}^2|y\neq 0~\text{and}~\sqrt{x^2+y^2}\notin\mathbb{Q}\}\cup\{(x,0)\in\mathbb{R}^2|x<0\}.$$

I do not know how to construct such sets $U_1,\dots, U_n$ for $n\geq 3$, nor do I know a proof that it is impossible.

Edit: Lukas showed below how to construct such sets $U_1,\dots, U_n$. What if instead of insisting that each $U_i$ is connected, we insist that each $U_i$ is path connected?

  • $\begingroup$ In your example for $n = 2$, how do you know the $U_i$ are connected? $\endgroup$ – Hew Wolff Nov 3 '12 at 2:50
  • $\begingroup$ $U_1$ and $U_2$ are path connected. The set $U_1$ can be formed by taking all circles centered at the origin with rational radius, then removing the left-most point of each circle, and then adding the non-negative part of the $x$-axis. Each of the individual circles in $U_1$ are path connected, and we forced the whole space $U_1$ to be path connected by adding a ray that connects all of the circles. $\endgroup$ – Adam Lowrance Nov 3 '12 at 3:09
  • $\begingroup$ Not sure if it helps: If you assume the $U_i$ to be pathconnected, then you can find continuous $\gamma_i : \mathbb R \to \mathbb R^2$ such that $\gamma_i(\mathbb R) \subset U_i$ and $\overline{\gamma_i(\mathbb R)} = \mathbb R^2$ for all $i$. $\endgroup$ – Alexander Thumm Nov 3 '12 at 15:56

Here is a construction in the $2$-sphere $S^2$, equipped with any reasonable metric. By removing one point it becomes homeomorphic to the plane, so it gives an example in $\mathbb{R}^2$. (You have to be a little careful which point to remove, but it is not that hard to figure out that there exists one that works. Alternatively, equip $\mathbb{R}^2$ with a bounded metric and run the same construction.) The construction is similar to the standard "Lakes of Wada" construction in spirit.

Let $U_1^1$ be a simple path which is $1$-dense in $S^2$, i.e., such that every point on the sphere has distance $\le 1$ to a point on $U_1^1$. Now let $U_2^1$ be a simple path (i.e., a homeomorphic image of $[0,1]$) in $S^2 \setminus U_1^1$ which is $1$-dense in $S^2$. Proceed to get disjoint $1$-dense simple paths $U_1^1,\ldots U_n^1$. Now extend $U_1^1$ to obtain a $1/2$-dense simple path $U_1^2$ in $S^2 \setminus (\bigcup_k U_k^1)$. Inductively construct a sequence of mutually disjoint simple paths $U_1^2,\ldots U_n^2$ which are $1/2$-dense extensions of $U_1^1,\ldots,U_n^1$. Now keep extending those inductively to get mutually disjoint paths $U_1^m,\ldots U_n^m$ which are $1/m$-dense in $S^2$. This construction is possible because at any step the complement of the already constructed paths is connected, since it is the complement in $S^2$ of a finite set of disjoint homeomorphic images of $[0,1]$.

Now let $U_k^\infty = \bigcup_m U_k^m$ for $k=1,\ldots,n$. This is a collection of mutually disjoint open paths (continuous images of $[0,1)$ or $(0,1)$, depending on how exactly the extensions are chosen), each of them dense in the plane. Their union is not necessarily all of $\mathbb{R}^2$, so let $T=S^2 \setminus \bigcup_k U_k^\infty$, and let $U_1 = U_1^\infty \cup T$ and $U_k = U_k^\infty$ for $k\ge 2$. Then $S^2 = \bigcup_k U_k$ is a disjoint partition, and since $U_2,\ldots,U_n$ are continuous images of an interval, they are connected, even path-connected. The set $U_1$ is not necessarily path-connected, so in order to show connectedness assume that $U_1 = A \cup B$ with relatively open disjoint sets $A$ and $B$. Since $U_1^\infty$ is path-connected, it has to be contained in either $A$ or $B$. We may assume $U_1^\infty \subseteq A$. Assume $t \in T \cap B$. Since $U_1^\infty$ is dense and $B$ is relatively open, there has to exist $u \in U_1^\infty \cap B$. However, this contradicts $A \cap B = \emptyset$.

The last argument is probably some standard topology result, that if $U$ is connected, and $V\supseteq U$ is contained in the closure of $U$, then $V$ is connected. The crucial point is to find disjoint connected dense subsets in the first place.

This construction does not guarantee that $U_1$ is path-connected, and I am not sure whether the similar question about a path-connected partition has a positive answer.

  • $\begingroup$ Very nice, but now I am even more interested in the path connected case. $\endgroup$ – Adam Lowrance Nov 3 '12 at 3:42

Here's an explicit construction that I think will work. This is for the connected, not the path connected, case.

Take $X_1$, ..., $X_n$ to be a partition of $\mathbb{R}$ into dense subsets. (For example, for $i > 1$ choose a prime $p_i$ and take $X_i$ be all rational numbers with denominator of the form ${p_i}^j$, and finally take $X_1$ to be the complement.) View this as a function $f: \mathbb{R} \rightarrow \{1, ..., n\}$. Take $Y_1$, ..., $Y_n$ to be a partition of $\mathbb{R}$ into intervals. Then define $U_i = \bigcup_{x \in \mathbb{R}} \{x\} \times Y_{f(x)}$.

It's clear that the $U_i$ form a partition and that each one is dense. I am not sure how to show that each one is connected, but here's a sketch. A disconnection of $U_i$ would have the form $A_1$, $A_2$, where the $A_i$ are disjoint open sets which cover $U_i$. In this case I would expect there is an embedded circle $C \subseteq \mathbb{R}^2 - A_1 \cup A_2$. (The circle might pass through the point at infinity.) In particular $C$ lies in the complement of $U_i$. But the complement of $U_i$ has the same form as $U_i$ itself. Its path components are straight lines, and there is no room there for the circle $C$.

Here are crude pictures with each $U_i$ in a different color. The original path-connected construction with $n = 2$ looks like original construction The new construction with $n = 3$ looks like enter image description here

  • $\begingroup$ Yes. I now get it. My idea of crossing dense subsets of $R$ with $R$ doesn't produce connected sets. I'll delete my previous comments. $\endgroup$ – coffeemath Nov 3 '12 at 21:01
  • $\begingroup$ I would still like to know for sure whether my $U_i$ are connected...anyone have an idea? $\endgroup$ – Hew Wolff Nov 7 '12 at 1:04

UPDATED 2012-12-15

For the path-connected case, the original post shows that we can achieve $1$ and $2$. I claim that we can also achieve continuum-many. I'd still like to know about all the cardinalities in between.

Claim. There is a solution with continuum-many path-connected $U_i$.

Proof. The basic idea is this: since the original construction is a one-dimensional foliation almost everywhere, what if we try to make it a foliation?

Start with $S^2$ rather than $\mathbb{R}^2$. Take the northern hemisphere $H_0$, whose boundary is the equator $S^1$. Suppose $f_0: S^1 \rightarrow S^1$ is a reflection. Pick a point $x \in S^1$. Define $p_0$ as the line segment from $x$ to $f_0(x)$, projected straight up to $H_0$. $p_0$ is a path properly embedded in $H_0$, although it's degenerate when $f_0(x) = x$. As $x$ varies, the different paths $p_0$ give a partition (I think it's called a foliation) of $H_0$.

Suppose $f_1$ is also a reflection of $S^1$. Use that similarly to define a path $p_1$ in the southern hemisphere $H_1$ starting at $p_0(x)$. Define $f = f_1 \circ f_0$, which is a rotation of $S^1$, and $p$ as the path composition $p_0 p_1$. Then $p$ is a path in $S^2$ from $x$ to $f(x)$ which meets $S^1$ at $x$, $f_0(x)$, and $f(x)$. Repeat this step to extend $p$ from $[0, 1]$ to $[0, \infty)$ so that $p(i) = f^i(x)$ for all integers $i \ge 0$. Since $f$ is a homeomorphism of $S^1$, we can also run the process backwards to extend $p$ to a path $\mathbb{R} \rightarrow S^2$ with $p(i) = f^i(x)$ for all integers $i$.

Now let $x$ vary over $S^1$, giving us paths $p_x$. If $A_x$ is the orbit of $x$ under the action by iterations of $f$, then the intersection of $p_x$ and $S^1$ is $A_x \cup f_0(A_x)$. For any point $y$ in that intersection, $p_y$ is basically the same as $p_x$: the domain has been translated, but the image is the same. So if we define $U_x = p_x(\mathbb{R})$, the choice of $f_0$ and $f_1$ gives us a partition $\{U_x\}$ of $S^2$ into paths.

Then of course we choose the $f_i$ so that $f$ is a rotation by a nasty angle $\alpha$ with $\alpha / 2 \pi$ irrational. Each $A_x$ is dense in $S^1$, and each $p_x$ is dense in $S^2$. The family $\{U_x\}$ is a partition of $S^2$ into dense path-connected subsets.

The original question was about $\mathbb{R}^2$, so puncture $S^2$ by removing a point. That cuts one of the $U_x$ in half, but each half is still path-connected and (it's pretty clear) dense. It's also pretty clear that the cardinality of $\{U_x\}$ is the continuum. For example, each $U_x$ has measure $0$. QED.

  • $\begingroup$ I'm also curious to know whether we could modify this partition slightly to get path-connected partitions that are finite. $\endgroup$ – Hew Wolff Nov 17 '12 at 16:59
  • $\begingroup$ I don't see how the open arcs going from $a_{i-1},a_{i+1}$ to boundary points on the opposite $\alpha_j$ are path-connected to $a_i$ $\endgroup$ – mercio Dec 15 '12 at 17:56
  • $\begingroup$ Thank you @mercio, the $n = 3$ argument does not work. I will withdraw it and keep thinking. $\endgroup$ – Hew Wolff Dec 16 '12 at 3:19

Here, I write the answer for the path-connected case. As you mentioned such a partition for $n=1, 2$ is possible. But, for any countable cardinal $n\geq 3$ this is not possible. First, remember that a subset of a metric space is said to be of the first category if it is the union of at most countably many nowhere dense sets. Indeed, Mary-Elen Rudin in https://projecteuclid.org/euclid.dmj/1077374790 showed that if there are three disjoint path-connected sets which are dense in $\mathbb{R}^2$, then one of them is of the first category. Now, as $\mathbb{R}^2$ is not of the first category and a countable union of the first category sets is of the first category, the claim is proved.

Here https://msp.org/involve/2014/7-2/p01.xhtml, you can see an interesting application of this result which gives a new proof of the non-existence of bijection continuous map between $\mathbb{R}^2$ and $\mathbb{R}^n$ for $n\neq 2$.


You can always take a dense partition of $\mathbb R$ in $n$ subsets and consider the radii. For instance, if $V_1, \dots, V_n$ is a dense partition of $\mathbb R$, then $$ U_i = \{ (x,y) \in \mathbb R^2 \, | \, \sqrt{x^2 + y^2} \in V_i \} $$ is a dense subset of $\mathbb R^2$. It's not connected, but in the same way you used the real line to make $U_1$ and $U_2$ in your case, you can play around with the lines $y = ix$ for instance, to make sure $U_i$ is connected. It would probably look like this :

$$ U_i = \left( \{ (x,y) \in \mathbb R^2 \, | \, \sqrt{x^2 + y^2} \in V_i, (x > 0 \quad \Rightarrow \quad y \neq jx, j = 1, 2, \dots, i-1, i+1, \dots n, ) \} \cup \{ (x,y) \in \mathbb R^2 \, | \, y = i x , x > 0\}\right) $$ and you can put $0 \in U_1$ to make all your conditions satisfied. It's pretty complicated to write down but it's all pretty simple ; I used the partition of $\mathbb R$ to get my partitions of $\mathbb R^2$ dense, and then "connected them" using the lines going to the origin.

Hope that helps,

  • $\begingroup$ If $n\geq 3$, then I think $U_i$ isn't connected. For example, if $n=3$ and $i=1$, then the pieces of the circles in $U_1$ that are "between" the lines $y=2x$ and $y=3x$ are not connected to the rest of $U_1$. $\endgroup$ – Adam Lowrance Nov 3 '12 at 1:10
  • $\begingroup$ Oh yes, you are so right. This just doesn't work. I'll leave my answer there just for inspiration but I'll admit it doesn't work. $\endgroup$ – Patrick Da Silva Nov 3 '12 at 1:18

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