Can you write $\mathbb R^2$ as a disjoint union of two totally disconnected sets?

6$\begingroup$ Gerald Edgar answered this here: mathoverflow.net/a/44320/2926 $\endgroup$– Todd Trimble ♦Feb 25, 2015 at 12:05

6$\begingroup$ Although (regarding my prior comment) @Włodzimierz commented below Gerald's answer, raising a possible objection, to which there hasn't been a response. Not wishing to put him on the spot, I hope Włodzimierz (or Gerald or someone else) has some spare time to answer this. $\endgroup$– Todd Trimble ♦Feb 25, 2015 at 14:15

4$\begingroup$ Meta: meta.mathoverflow.net/a/2154/2926 $\endgroup$– Todd Trimble ♦Feb 25, 2015 at 17:30

4$\begingroup$ According to the comments on Gerald Edgar's answer, it is possible to write $\mathbb{R}^2$ as a union of three totally disconnected sets. Is there an easy way to construct such sets? $\endgroup$– Eric WofseyFeb 26, 2015 at 22:37

3$\begingroup$ @EricWofsey  yes, into 3 sets of dimension $0$ (better than just totally disconnected), it's quite pleasing: $\ \mathbb R_{k\,\ 2k}\ $ is the set of all points which have exactly $\ k\ $ coordinates rational, and $\ 2k\ $ irrational $\ (\ k=0\ 1\ 2).\ $ Now you can decompose $\ \mathbb R^n\ $ into $\ n+1\ $ zerodimensional sets for each natural $\ n.$ $\endgroup$– Włodzimierz HolsztyńskiMar 2, 2015 at 11:15
4 Answers
THEOREM There do not exist two disjoint sets $\ A\ B\subseteq\mathbb R^2\ $ which are totally disconected, and which cover the plane: $\ A\cup B=\mathbb R^2$.
PROOF Let $\ A\ B\ $ form a cover of $\ \mathbb R^2\ $ while they are totally disconnected (a proof by contradiction). Thus both are dense hence they have more than one point. Then, since A is disconnected, $\ A=E_0\cup F_0\ $ where the two summands are nonempty, closed with respect to $\ A,\ $ and $\ E_0\cap F_0=\emptyset.\ $ There exist $\ E\ F\ $ closed in $\ \mathbb R^2\ $ such that $ E\cap A = E_0\ $ and $\ F\cap A = F_0.\ $ Then
$$X_0:= E\cap F$$
is a closed subset of $\ \mathbb R^2\ $ which is disjoint with $\ A:\ X_0\cap A=\emptyset.\ $ Consider a continuos function $\ f : E\cup F\rightarrow [1;1]\ $ such that
 $\ f^{1}(0)\ =\ X_0$
 $\ f^{1}(\,[1;0)\,)\ =\ E\setminus X_0$
 $\ f^{1}(\, (0;1]\, )\,\ =\ F\setminus X_0$
and let $\ \phi:\mathbb R^2\rightarrow\mathbb R\ $be a continuous extension of $\ f.\ $ We see that $\ X:=\phi^{1}(0)\ $ is a closed subset of $\ \mathbb R^2\ $ which disconnects sets $\ E_0\ F_0,\ $ thus $\ X\ $ is not totally disconnected. Since $\ X\subseteq B\ $ it follows that $\ B\ $ is not totally disconnected. END of Proof
Sets $\ A\ B\ $ in the theorem are arbitrary (not closed, nor nothing :).
It doesn't matter what $\ X\ $ disconnectsonly that disconnects whatever (and that $\ X\ $ is disjoint from $\ A$).
I've answered the question exactly. Essentially the same proof allows for a more general formulation.

3$\begingroup$ I think you want $X_0$, not $X$ in the second and third bullet points. Do you want the codomain of $\phi$ to be $\mathbb{R}^2$, or just $\mathbb{R}$? I'm probably missing something obvious, but where are you actually using the fact that we're working with $\mathbb{R}^2$, since the proof must break down somewhere if we replace $\mathbb{R}^2$ with $\mathbb{R}$? $\endgroup$– Todd Trimble ♦Feb 26, 2015 at 12:01

3$\begingroup$ @ToddTrimble: I believe the place is “... which disconnects sets $E_0\:F_0$, thus $X$ is not totally disconnected”, which summons the property of $\mathbb R^2$ referred to in Kristal Cantwell’s answer. $\endgroup$ Feb 26, 2015 at 12:24

3$\begingroup$ @WłodzimierzHolsztyński Well, before Emil enlightened me by referring to Kristal's answer (now unfortunately deleted), the line "which disconnects sets $E_0, F_0$, thus $X$ is not totally disconnected" was a mystery to me. Do you have a way of justifying that inference that is different from Kristal's (which invokes Alexander duality)? It would be a good idea to explain in your answer just what special feature of $\mathbb{R}^2$ you are using. $\endgroup$– Todd Trimble ♦Feb 27, 2015 at 0:34

1

8$\begingroup$ "I could justify the mentioned nature of a closed subset of $\mathbb{R}^2$, which separates the $\mathbb{R}^2$ it's not trivial but the great Alexander duality is an overkill." Since you say it's not trivial, would you add that detail so that people like me can understand your answer? Thanks. $\endgroup$– Todd Trimble ♦Feb 27, 2015 at 5:19
Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. Taking $X=\mathbb{R}^2$ answers the question.
[This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^1(X)=0$ and $X\setminus\{p\}$ is connected for each $p$].
I will use proof by contradiction. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$.
So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory.
Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))F(\gamma(t_{k1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers.
Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.
The question complement of a totally disconnected closed set in the plane has an answer by Thurston in which it is shown that the complement of a totally disconnected set of any manifold of dimension greater than or equal to 2 is connected using the Alexander duality theorem.

10$\begingroup$ That answer only applies to closed sets, in which case the question asked here is trivial since no nonempty open set is totally disconnected. $\endgroup$ Feb 25, 2015 at 20:53

$\begingroup$ @EricWofsey  for a contrast, your own (transcendental) example in the other (path) thread was super ! $\endgroup$ Feb 26, 2015 at 21:53

1$\begingroup$ But actually, as noted in the question there, the result (any disconnected subset of the plane can be disconnected by a closed connected set) does not require any assumption of closedness / compactness. (See also the answer I just gave there.) So this does in fact answer the question. $\endgroup$ Apr 27, 2016 at 13:30
I like my (previous) answer in its simple form. Thus I will expand it here (it should not be a waste, I hope), and first of all I will fulfill Todd's request, see Thm 1 below: we want to show that a closed subset of $\ \mathbb R^2,\ $ which separates $\ \mathbb R^2,\ $ is not totally disconnected. The proof is smoother in the compact case, but in general it needs just a minor extra consideration.
A direct dimensionflavor approach is possible. However I feel like applying the elegant Borsuk's theorem based on his separation criterion. Eilenberg and Steenrod cover this material beautifully in their Chapter 11 (it's like a perfectly elementary appendix) to their "Foundations of Algebraic Topology".
BORSUK's THEOREM Let $\ X\ $ be a closed subset of $\ S^n.\ $ Then $\ S^n\setminus X\ $ is disconnected $\ \Leftrightarrow\ $ there exists a continuous map $\ f:X\rightarrow S^{n1}\ $ which is not homotopic to a constant.
On this occasion I also have selected this direction of presentation because Bill Thurston in a respective post wrote about the Alexander duality in a way which harmonized with the above. I read that Bill's post only after Kristal's answer (which didn't really answer) since it referenced to the said Bill's post; the very first comment of this whole thread, just under the QUESTION of the threat, written by me before Kristal's answer, read:
It is not possible. Where does this problem come from?
Now, in the compact context the being totally disconnected is equivalent to being $0$dimensional. This means for the totally disconnected (i.e. $0$dimensional) closed set $\ X\subseteq S^n\ $ every continuous map $\ f:X\rightarrow S^{n1}\ $ is homotopically trivial for $\ n\ge 2.\ $ Thus for a compact $\ X\subseteq\mathbb R^2\ $ which disconnects $\ R^2\ $ the requested theorem already holds, i.e. $\ X\ $ is not totally disconnected (add $\infty$ to the unbounded component of $\ \mathbb R^2\setminus X).\ $ Thus we have to include in the proof only the (nuisance :) noncompact case (while the theorem is formulated for both):
THEOREM 1 Let $\ X\ $ be a closed subset of $\ \mathbb R^2\ $ such that $\ \mathbb R^2\setminus X\ $ is not connected. Then $\ X\ $ is not totally disconnected.
PROOF The case of compact $\ X\ $ was covered above. Now let $\ X\ $ be closed and not compact. Then $\ Y := X\cup\{\infty\}\ $ is compact. Of course $\ Y\ $ disconnects $\ \{\infty\}\cup \mathbb R^2 = S^2.\ $ Thus by the Borsuk's theorem and the remarks which followed, $\ \dim(Y)\ge 1.\ $ However $\ X\ $ is $\sigma$compact, $\ X=\bigcup_{n=1\ 2\ \ldots}X_n\ $ where each $\ X_n\ $ is compact. Thus $\ \dim(\{\infty\}\cup\bigcup_{n=1\ 2\ \ldots}X_n) = \dim Y \ge 1,\ $ and $\ \dim(\{\infty\}) = 0.\ $ Thus $\ \exists_n \dim(X_n)\ge 1.\ $ Such $\ X_n\ $ is not totally disconnected, hence neither is $\ X$. END of Proof
Todd's request's fulfilled; now back to the generalization, while George Lowther has already provided a still more general result (nice!)
The proof in my first answer used only the assumption that set $\ A\ $ was disconnected, and not moreit didn't mention that $\ A\ $ was totally disconnected. Thus (as promised in my first answer) I really proved:
THEOREM 2 If $\ \mathbb R^2 = A\cup B\ $, and $\ A\ $ is not connected, then $\ B\ $ is not totally disconnected.
In other words:
THEOREM 2' If $\ \mathbb R^2 = A\cup B\ $, and $\ B\ $ is totally disconnected then $\ A\ $ is connected.
Actually, all this holds for $\ \mathbb R^n\ $ for every $\ n\ge\ 2\ $ (not just for $\ n=2).\ $ This follows from the Borsuk's theorem for each $\ n\ge 2\ $ just as it does for its special case of $\ n=2$.

$\begingroup$ It's nice to use Borsuk's theorem. Of course, you can also use Janiczewski's theorem, which is perhaps more elementary (see also my new answer to the question referenced by Kristal). $\endgroup$ Apr 27, 2016 at 13:35

$\begingroup$ Dear @LasseRempeGillen, it's Janiszewski. Yes, I'll try to find your answer to the q. referenced by Kristal. $\endgroup$ Apr 27, 2016 at 19:52