# In a connected space two closed sets must be connected by a single branch.

Let $X$ be a connected space which is also locally connected. Let $A$ and $B$ be closed non- empty disjoint subsets of $X$.

Prove that there is a connected component $K$ of$X-(A\cup B)$ such that the closure of $K$ intersectan both $A$ and $B$.

Here is what we have so far:

Since X is locally connected and $X-(A\cup B)$ is open we have that the connected components of $X-(A\cup B)$ are open. Since X is connected they cannot be closed, and therefore the closure of each component intersects $A$ or $B$. I want to proceed by contradicción and show that the union of A and the componentes whose closure intersects A is open. But i have not been able to.

Lemma: Let $Y$ be a locally connected space, $F \subset Y$ closed, and $U\subset Y$ a connected open set such that $U \cap F \neq \varnothing$. Then, for every connected component $V$ of $U \setminus F$ we have

$$F \cap \partial_Y V \supset F \cap \partial_U V \neq \varnothing.$$

Proof: Since $Y$ is locally connected, the connected components of the open set $U\setminus F$ are open. Since $U$ is connected, and $U \cap F \neq \varnothing$, a component $V$ of $U \setminus F$ cannot be closed in $U$, so

$$\partial_U V = (\partial V) \cap U \neq \varnothing.$$

But as a connected component, $V$ is closed in $U\setminus F$, so (since $V$ is also open)

$$\varnothing = \partial_{U\setminus F} V = (\partial V) \cap (U\setminus F),$$

hence

$$\varnothing \neq \partial_U V = (\partial V) \cap U = \bigl((\partial V) \cap (U\setminus F)\bigr) \cup \bigl((\partial V) \cap (U\cap F)\bigr) = (\partial V) \cap (U\cap F).$$

Now apply the lemma in your situation. Let $M$ be the union of $A$ with all connected components of $X\setminus (A\cup B)$ whose closure intersects $A$. Then $M$ is open. It is clear that every point of $M$ that belongs to one of the components of $X\setminus (A\cup B)$ is an interior point of $M$, since these components are open. So it remains to see that $A\subset \overset{\Large\circ}{M}$. Pick an arbitrary $a\in A$, and let $U$ be a connected open neighbourhood of $a$ that doesn't intersect $B$. Then $U \subset M$. For if $C$ is a connected component of $X\setminus (A \cup B)$ intersecting $U$, let $W$ be a connected component of $U \cap C$, and $V$ the connected component of $U\setminus A$ containing $W$. Then $C \cup V$ is connected, and $(C\cup V) \cap (A\cup B) = \varnothing$, so $V\subset C$. By the lemma,

$$\varnothing \neq A \cap \partial V = A \cap \overline{V} \subset A \cap \overline{C},$$

therefore $C \subset M$.

With the same argument, the union $N$ of $B$ and all connected components of $X\setminus (A\cup B)$ whose closure intersects $B$ is open. Since $X = M \cup N$, and neither $M$ nor $N$ is empty, it follows that $M \cap N \neq \varnothing$, and that means there is a connected component of $X\setminus (A\cup B)$ whose closure intersects both, $A$ and $B$.