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

Question

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

Answer 1

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 non-empty, 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\ $ disconnects--only 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.

Answer 2

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 $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, 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 $X$.

Answer 3

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.

Answer 4

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 dimension-flavor 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^{n-1}\ $ 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^{n-1}\ $ 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 :-) non-compact 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 more--it 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 $\ n=2$.