Prove that $A$ is open when $A \cap bd(A) = \emptyset$

Question

Before I saw the solution to the problem below, I was attempting to prove it using De Morgan's laws.

$bd(A)$ is the boundary of $A$, and $\overline A$ is the complement of $A$

My attempt

$\overline{A\cap bd(A)}=X$

and by De Morgan's laws

$\overline{A} \cup \overline {bd(A)}=X$

and by the definition of $bd(A)$ and another application of De Morgan

$\overline{A} \cup \overline {cl(A) \cap \overline{A}}=X$

$\overline{A} \cup \overline {cl(A)} \cup A=X$

where $cl(A)$ is the closure of $A$

I was unsure how to proceed from this point. I was considering isolating $A$ by writing $A=X$\ $\overline A \cup \overline{cl(A)}$, but I wasn't sure if I could do this, and even if I could, I still don't see how I'd get that A is open.

Thanks, and here's the problem along with another solution for reference purposes

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Answer 1

I don't think you should use deMorgan's laws in this case. (At least, i don't see how to.)

It's easy just to note that any $a\in A$ has to have a nbhd that doesn't meet $X\setminus A$ by definition of boundary (as in the solution you included).

But, here's an alternate approach: $\partial A=\partial A^c$. Hence $A\cap\partial A=\emptyset\implies A\cap\partial A^c=\emptyset\implies \partial A^c\subset A^c\implies A^c$ is closed. (This uses the sort of complementary fact that a set is closed iff it contains its boundary...)