3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.