If $E$ and $F$ are nonempty subsets of $M$ and if $E \cup F$ is connected, show that $\bar{E} \cap \bar{F} \neq \emptyset$

Question

I tried to prove this by contradiction:

So suppose $\bar{E} \cap \bar{F} = \emptyset$, that means $\bar{E}$ and $\bar{F}$ are disjoint.

But I get stuck here... I have no idea what to do... Any help?

Answer 1

HINT: Let $U=M\setminus\operatorname{cl}E$; $U$ is an open set in $M$, and $U\supseteq\operatorname{cl}F$. Show that $U\cap(E\cup F)$ is a clopen subset of $E\cup F$ that is neither empty nor all of $E\cup F$, and conclude that $E\cup F$ is not connected.

Alternative HINT: Define $f:E\cup F\to[0,1]$ by $f(x)=0$ if $x\in E$ and $f(x)=1$ if $x\in F$, and show that $f$ is continuous.

Answer 2

Suppose $\bar{E} \cap \bar{F} = \emptyset$, so $E \subset M \setminus \bar{F}$ and $F \subset M\setminus \bar{E}$. Let $X = E \cup F$.

But $(X \cap (M \setminus \bar{F})) \cap (X \cap (M \setminus \bar{E})) = X \cap (M \setminus (\bar{F} \cup \bar{E})) = \emptyset$

Then $X = (X \cap (M \setminus \bar{F})) \cup (X \cap (M \setminus \bar{E}))$, disjoint open sets.