I'm asked to prove that, with the usual topology for $\mathbb{R}$, if the intersection of the closure of a set with the closure of the complement of the same set equals the empty set, then it must be the case that such set is either the empty set or the real numbers. This is:
If $\overline{X}\cap\overline{X^c} = \emptyset$ then $X=\mathbb{R}$ or $X=\emptyset$.
Assuming $X\neq\emptyset$, I'm struggling with the "complicated" block of the proof, this is that $\mathbb{R}\subset X$. So far my approach has been a proof by contradiction, taking $x\in\mathbb{R}$ and assuming $x\notin X$. Thus $x\in X^c$ and, since $X^c \subset \overline{X^c}$, $x$ is contained in the open set $\overline{X^c}$. Moreover, by hypothesis, if $x \in \overline{X^c}$ then $x \notin \overline{X},\ x\in\overline{X}^{\ c}$.
Additionaly, given that $\overline{X}\cap\overline{X^c} = \emptyset$, I obtain $\overline{X^c} \subset \overline{X}^{\ c}$.
Do you have any thoughts on how to proceed from here towards a contradiction? Or is there a more elegant way to prove this?
Thank you very much.
Best Answer
HINT
I would start with noticing that $\overline{X} = \text{int}(X)\cup\partial X$ and $\overline{X^{c}} = \text{ext}(X)\cup\partial X$.
Consequently, $\overline{X}\cap\overline{X^{c}} = \partial X = \varnothing$.
Now it remains the question: what are the subsets of $\mathbb{R}$ with no boundary points?
Hopefully this helps!