# $b(A) \cap A=\emptyset \iff A$ is open

Let $A$ be subset of a topological space $X$. Show that: $b(A) \cap A= \emptyset \iff A$ is open.

I started the proof like this: (<==) Let $A$ is open. Then (by a proposition)

$int(A)=A$

Also $b(A)=(int(A) \cup ext(A))^C$

So $b(A) \cap A$ implies $((int(A)\cup ext(A))^C ) \cap A$

implies $((int(A))^C \cap ext(A)^C) \cap int(A)$ (using proposition)

implies $((int(A))^C \cap int(A) \cap (ext(A)^C))$

implies $(\emptyset \cap (ext(A)^C))$

implies $\emptyset$

This is what I have done so far. Is it a right way to prove the implication? And how should I prove the reverse implication?

Let a belong to A then we have suppose a is not in int. A this implies every neighbourhood of a has a point of X-A then we have a is a limit point of X-A thus a belongs to b(A) and A a contradiction. Thus every point is interior and hence A is.open

$x\in\partial A$ if and only if for every neighborhood $U$ of $x$ we have: $$U\cap A\neq\varnothing\text{ and } U\cap A^c\neq\varnothing$$

If $A$ is open we can take $U=A$, to find easily that this is not true.

So we conclude that $x\notin\partial A$ and proved is now that:

$$A\text{ open}\implies A\cap\partial A=\varnothing$$

If conversely $A\cap\partial A=\varnothing$ and $x\in A$ then $x\notin\partial A$.

That means that some neighborhood $U$ of $x$ must exist with: $$U\cap A=\varnothing\text{ or }U\cap A^c=\varnothing$$ We have $x\in U\cap A$ so conclude that $U\cap A^c=\varnothing$ or equivalently $U\subseteq A$.

This can be proved for every $x\in A$ so proved is now that:$$A\cap\partial A=\varnothing\implies A\text{ open}$$