# If $X$ is not connected, then $X\subset G_1\cup G_2$

The following proposition is true?

Let $X\subset\mathbb{R}^n$ be a not connected set. Then there are disjoint open sets $G_1,G_2\subset \mathbb{R}^n$ such that $X\subset G_1\cup G_2$ with $X\cap G_1\neq\emptyset$ and $X\cap G_2\neq\emptyset$.

$X$ is connected if $X=A\cup B$ (where $A,B$ are open on $X$ and disjoint) then $A=\emptyset$ or $B=\emptyset$.

Any help would be appreciated.

• It's close to the negation of connectedness for $X$ as a subspace with the extra condition that $G_1$ and $G_2$ are disjoint. But this can easily be taken care of in $\mathbb{R}^n$. – Taylor Jun 17 '15 at 4:00
• @Taylor Let me know how - I haven't been able to fill in the gap in my answer below. I know that you must at least have the space be Hausdorff but I think a stronger separation condition is necessary. – Mario Carneiro Jun 17 '15 at 4:01
• @MarioCarneiro I was thinking that if $G_1$ and $G_2$ are not disjoint, then $M_1=G_1-\overline{G_1\cap G_2}$ and $M_2=G_2-\overline{G_1\cap G_2}$ would work. – Taylor Jun 17 '15 at 4:14
• @Taylor Certainly that will work if anything does; but how do you know that $\overline{G_1\cap G_2}\cap X=\emptyset$? That works if $X$ is closed but in general it still seems problematic. – Mario Carneiro Jun 17 '15 at 4:18
• @MarioCarneiro I don't think that $X$ can intersect $\overline{G_1\cap G_2}$, but I can't think of an explanation. I think this does the trick though: Define $P=\overline{X\cap G_1} \cap \overline{X\cap G_2}$ and let $M_i=G_i-P$ for $i=1,2$. Now $\overline{X\cap M_1}$ and $\overline{X\cap M_2}$ are disjoint so that we can use normality to get the two disjoint open sets as desired. – Taylor Jun 17 '15 at 5:11

I'll start by generalizing to an arbitrary topological space $X$, with a view toward setting $X=\Bbb R^n$:

$(1)$ If $E\subseteq X$ is disconnected, then there exist disjoint open sets $G_1,G_2$ such that $E\subseteq G_1\cup G_2$, $G_1\cap E\ne\emptyset$, and $G_2\cap E\ne\emptyset$.

Treated as a property of topologies, there is a closely related formulation which specifies the disconnection:

$(2)$ If $E=A\cup B$ is a disconnection of $E\subseteq X$, then there exist disjoint open sets $G_1,G_2$ such that $A\subseteq G_1$ and $B\subseteq G_2$.

Obviously $(2)\to(1)$, and if $A$ and $B$ are connected, then $(1)$ and $(2)$ are equivalent (because the disconnection is uniquely determined). It is not clear to me if $(1)\to(2)$ in general, but the most natural approach to proving $(1)$ in fact attempts to prove $(2)$ since the alternative would require finding some other disconnection in $E$ (i.e. $E$ is actually in three or more "pieces").

By definition, since $E$ is not connected, there exist $A,B\subseteq E$ such that $A,B$ are open in $E$, nonempty and disjoint, and $A\cup B=E$. By the definition of openness in $E$, there exist open $G_1,G_2$ such that $A=G_1\cap E$ and $B=G_2\cap E$. Then clearly $G_1\cap E\ne\emptyset$ and $G_2\cap E\ne\emptyset$, and $G_1\cup G_2\supseteq A\cup B=E$. But we don't know that $G_1$ and $G_2$ are disjoint; instead we only get that $G_1\cap G_2\cap E=\emptyset$.

For $X$ a metric space, $(2)$ and hence $(1)$ are true (which answers your question in the positive). Let $x\in G_1$ when $d(x,A)<d(x,B)$ and $x\in G_2$ when $d(x,A)>d(x,B)$. These are both open sets, disjoint, and since there is an open set containing $A$ and disjoint from $B$, this implies that $d(x,B)>0$ when $x\in A$, so $A\subseteq G_1$ and similarly $B\subseteq G_2$.

For general topologies $X$, without weakening the disjointness condition to $G_1\cap G_2\subseteq X\setminus E$, $(1)$ is false. Consider $X=\{a,b,c\}$ with open sets $\emptyset,\{a\},\{a,b\},\{a,c\},\{a,b,c\}$, and let $E=\{b,c\}$. Then the sets $\{b\},\{c\}$ form a separation of $E$, so $E$ is disconnected. But no such $G_1,G_2$ exist; the obvious candidate is $G_1=\{a,b\},G_2=\{a,c\}$ but these are not disjoint.