Proving interior of a set is empty Given that $F$ is a closed set and $A$ an arbitrary subset in a metric space that $A^\circ = F^\circ = \varnothing$. Prove that $(F \cup A)^\circ = 
\varnothing$
I'm trying to solve this and I've just started with Analysis. I would appreciate it if anyone could give me a hint.
 A: Under the given assumption of closed $F$ and $A$ any, holds $(F \cup A)^\circ \subset F \cup (A)^{\circ}$. In fact, let $x \in (F \cup A)^\circ$, so there is some open set $W$ containing $x$ and $W \subset (F \cup A)^\circ$. If $x\in F$ we are done. Suppose that $x \notin F$. Then $F$ closed, implies that its complement in the given metric space is open. Taking the intersection of $W$ with the complement of $F$, we have an open containing $x$, contained in $A$ (is contained in $(F \cup A)\cap(F^{c}) \subset A$), and which does not intersect $F$, so $x$ is in $A^{\circ}$. In this way the inclusion is proved.
Maybe you could use a contradiction argument, if the interior of the union $F \cup A$ is non-empty, then how $A^{\circ} = \emptyset$, the inclusion above becomes $(F \cup A)^\circ \subset F$. Contradiction about the fact that the interior of $F$ is empty.
A: Proof by Contradiction :
Suppose $\mathcal x \in (A \cup F)^\circ$. Then there exists r such that $N_r(\mathcal x) \subseteq (A \cup F)$.
We know that there exists $\mathcal a \in N_r(\mathcal x), a \notin F$.Otherwise $N_r(\mathcal x) \subseteq F$, Which is in Contradiction with $F^\circ = \varnothing$.
Since F is closed $\exists \ell$ $N_\ell(\mathcal x) \, \cap F = \varnothing$. And we know that any  open ball is an open set. hence $N_r(\mathcal x)$ is open, meaning $\exists \alpha$ $\ N_\alpha(\mathcal x) \ \subseteq N_r(\mathcal x).$
Now by only choosing $\beta < \alpha ,\ell$ we can conclude that  $N_\beta(\mathcal x) \ \subseteq A$. And that is in Contradiction with $F^\circ = \varnothing$.Therefore $(A \cup F)^\circ = \varnothing.$
