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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.

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    $\begingroup$ This seems to be the Baire category Theorem en.wikipedia.org/wiki/Baire_category_theorem $\endgroup$
    – Ilovemath
    Commented Oct 24, 2021 at 18:02
  • $\begingroup$ I wasn't able to understand that. can you help me with a hint? $\endgroup$
    – mehrdad
    Commented Oct 24, 2021 at 18:39
  • $\begingroup$ In this case we cannot use it, because in the hypothesis of your problem the space is not complete $\endgroup$
    – Ilovemath
    Commented Oct 24, 2021 at 19:00
  • $\begingroup$ 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. $\endgroup$
    – Mark Saving
    Commented Oct 24, 2021 at 20:22
  • $\begingroup$ I think I could solve the dual problem if the metric space were complete, here in this case the assumption is weaker. $\endgroup$
    – Ilovemath
    Commented Oct 24, 2021 at 20:50

2 Answers 2

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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.

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    $\begingroup$ 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? $\endgroup$
    – mehrdad
    Commented Oct 24, 2021 at 19:04
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    $\begingroup$ 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. $\endgroup$
    – Ilovemath
    Commented Oct 24, 2021 at 20:10
  • $\begingroup$ I found a better proof and added it above, I think it's clearer now! $\endgroup$
    – Ilovemath
    Commented Oct 24, 2021 at 20:41
  • $\begingroup$ 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? $\endgroup$
    – mehrdad
    Commented Oct 24, 2021 at 22:33
  • $\begingroup$ 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 $F$. $\endgroup$
    – Ilovemath
    Commented Oct 24, 2021 at 22:55
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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 $(A \cup F)^\circ = \varnothing.$

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