Given two dense disjoint subsests their interior is empty

Question

I need to prove the following statement: Let $(E,d)$ be a metric space and let $A, B$ be two disjoint dense subsets of $E.$ Then $\mathring A=\mathring B=\emptyset.$

By definition, $\overline A=E=\overline B.$ Also, $\overline A= \mathring A \, \cup \partial A.$ Thus, $\emptyset=E^c=(\mathring A \, \cup \partial A)^c=(\mathring A)^c \, \cap (\partial A)^c.$ For the boundary $\partial A$ one could eventually write $\partial A=\overline A \,\cap (A)^c. $ But this does not bring me further. I suspect that I should somehow connect $A$ and $B$ and the fact that they are disjoint. Also $E$ is not only a set but a metric space.

Can somebody help me figure out the solution. Many thanks.

Answer 1

Let $x\in\mathring A$.

Then set $\mathring A$ is open and not empty so that $\mathring A\cap B\neq\varnothing$ because $B$ is dense.

(Further explaining: observe that ${\left(\mathring A\right)}^{\complement}$ is closed, so if $\mathring A\cap B=\varnothing$ or equivalently $B\subseteq{\left(\mathring A\right)}^{\complement}$ then $\overline B\subseteq{\left(\mathring A\right)}^{\complement}$ so that $x\notin \overline B$, contradicting that $B$ is dense.)

Then also $A\cap B\neq\varnothing$ because $\mathring A\subseteq A$ contradiction that the sets $A$ and $B$ are given to be disjoint.

We conclude that $\mathring A$ must be empty and on a similar way we also find that $\mathring B$ must be empty.

Answer 2

If $A^{0}$ is non empty it must intersect $B$ because $B$ is dense. But then $B$ intersects $A$ which is a contradiction. So $A^{0}$ is empty. Similarly $B^{0}$ is empty.