Show that the boundary of $A$ is empty iff $A$ is closed and open. I'm reading Intro to Topology by Mendelson.
The problem at hand is,
Show that $\text{Bdry}(A)=\emptyset$ if and only if $A$ is closed and open.
This was all the problem statement had, but I'm in the chapter covering closure, interior and boundary with respect to topological spaces.
Here is my attempt at the proof,
Suppose that Bdry$(A)=\emptyset$. Then $\bar{A}=A$ $\cup$ Bdry$(A)=A\cup\emptyset=A$. Hence, $\bar{A}=A$, which implies that $A$ is closed. To show that $A$ is also open we will show that $\overline{C(A)}=C(A)$, that is, $C(A)$ is closed and hence $A$ open. We already know that $C(A)\subset\overline{C(A)}$, thus it suffices to only show that $\overline{C(A)}\subset C(A)$. This is the case since we know that Bdry$(A)=\bar{A}\cap\overline{C(A)}=\emptyset$, which implies that $\overline{C(A)}\subset C(\bar{A})=$ Int$(C(A))\subset C(A)$. Thus, $A$ is open. Suppose now that $A$ is both open and closed. Since $A$ is open we know that Int$(A)=A$. Also, since $A$ is closed $\bar{A}=A=$ Int$(A)$. Now we know that Bdry$(A)=\bar{A}\cup\overline{C(A)}=$ Int$(A)$ $\cup$ $C($Int$(A))=\emptyset.$
I used an the identity $\bar{A}=A\cup\text{Bdry}(A)$, which was asked later in the problem set. I'm wondering if I should use this, since I was able to prove it, or attempt the proof assuming I'm unaware of the identity. I have also been trying to write more concise proofs and this one is definitely not one of them. Would removing some of the words help or is there a cleaner approach that can be pointed out?
Thanks for any feedback!
 A: $\emptyset=\partial A=\overline{A}\setminus \mathring{A}$ (the first is the hypothesis, the second a well-known property) together with $\mathring{A} \subset A\subset\overline{A} $ gives you $\mathring{A}=A=\overline{A}$ so that A is clopen (and from the first inequalities the other implication is obvious).
A: We know $\operatorname{Bdry}(A)=\operatorname{cl}(A)\cap \operatorname{cl}(X-A)$. So $\operatorname{Bdry}(A)=\operatorname{cl}(A)\cap (X-\operatorname{int}(A))$. Hence $\operatorname{Bdry}(A)=\operatorname{cl}(A)-\operatorname{int}(A)$. Now use this identity.
A: As you said $\bar{A} = A \cup \operatorname{Bd}(A)$, but there's a more stronger statement that you can and you should prove: $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$.
Now it should be fairly easy to prove that $\operatorname{Bd}(A) = \emptyset \iff \text{A is closed and open}$:
(1)If $\operatorname{Bd}(A)=\emptyset$
Then $\bar{A}=\operatorname{Int}(A)$ and since $\operatorname{Int}(A)\subset A \subset \bar{A}$ we conclude that $\operatorname{Int}(A) = A = \bar{A}$.
$\operatorname{Int}(A) = A$ shows that $A$ is open and $A = \bar{A}$ shows that $A$ is closed.
(2)If $A$ is closed and open.
A is open, then $A=\operatorname{Int}(A)$ and since $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$ we see that $\bar{A} = A \cup \operatorname{Bd}(A)$
But $A$ is also closed!, this means that $\bar{A} = A$ therefore $A = A \cup \operatorname{Bd}(A)$.
Now remember that $\operatorname{Int}(A) \cap \operatorname{Bd}(A) = \emptyset$, then $A \cap \operatorname{Bd}(A) = \emptyset$.
Because of $A = A \cup \operatorname{Bd}(A)$ and $A \cap \operatorname{Bd}(A) = \emptyset$ we conclude $\operatorname{Bd}(A) = \emptyset$
A: (1) If $Bdry(A) = \emptyset$. 
Then, $$\bar A = A \cup Bdry(A) = A \cup \emptyset = A$$ Hence $\bar A = A$, which implies A is closed. 
Since the $Bdry(A) = Bdry(A^c)$ this implies $$\overline {A^c} = A^c \cup Bdry(\overline {A^c}) = A^c \cup Bdry(A) = A ^c \cup \emptyset.$$ Hence $\overline {A^c} = A^c$, which implies $A^c$ is closed, which implies A is open. 
(2) If $A$ is closed 
Then $A = \bar A$. So $$Bdry(A) = \bar A \cap \overline {A^c} = A \cap \overline {A^c} = A \cap (Int(A))^c$$ 
Since $A$ is open, $A$ is a neighborhood of all its points so $A = Int(A)$, 
which implies $A \cap C(Int(A)) = A \cap C(A) = \emptyset$
