# 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.

• This seems to be the Baire category Theorem en.wikipedia.org/wiki/Baire_category_theorem Commented Oct 24, 2021 at 18:02
• I wasn't able to understand that. can you help me with a hint? Commented Oct 24, 2021 at 18:39
• In this case we cannot use it, because in the hypothesis of your problem the space is not complete Commented Oct 24, 2021 at 19:00
• You probably want to solve the dual problem with closures. That is, if $F$ is open and both $F$ and $A$ are dense, then $F \cap A$ is also dense. Commented Oct 24, 2021 at 20:22
• I think I could solve the dual problem if the metric space were complete, here in this case the assumption is weaker. Commented Oct 24, 2021 at 20:50

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.

• I actually thought a lot about that but what if the open set contains elements from both F and A which isn't in A $\cap$ F? Commented Oct 24, 2021 at 19:04
• correct me, in case I'm thinking wrong. But I believe that this situation cannot happen, because if the open contained a point of $F$, as $F$ is closed, that point would be the limit point of a sequence of $F$, then you could build a neighborhood of this limit point totally contained in $F$, the which gives a contradiction again, since the interior of $F$ is empty. Commented Oct 24, 2021 at 20:10
• I found a better proof and added it above, I think it's clearer now! Commented Oct 24, 2021 at 20:41
• I can't see why the last line is a contradiction. also in the third line "if $x \in F$ we are done". can you elaborate? Commented Oct 24, 2021 at 22:33
• the contradiction in the last line is because you have a non-empty open inside $F$. In the third line it's because if $x$ is in $F$ then the inclusion is already valid, so I analyze the case of $x$ outside of $F$. Commented Oct 24, 2021 at 22:55

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 $$(A \cup F)^\circ = \varnothing.$$