Prove that if $x\in int(A)$, then $x\notin \partial A$ Let $int(A)$ be the interior of $A$, $\partial A$ the boundary of $A$ and $A$ a subset of a normed vector space $(E,\parallel \cdot \parallel)$ . Prove that if $x\in int(A)$, then $x\notin \partial A$.
In the book "Calculus on Normed Vector Spaces", written by Rodney Coleman, the definition of boundary of a set $A$ is $\partial A=\overline{A}\cap\overline{A^c}$, where $\overline{A}$ is the closure of $A$ and $A^c=E\setminus A$. Please, don't use the following proposition to prove the problem above(you can use it if you prove it, because I also don't know how to prove the following proposition):

$a\in\partial A$ if and only if every neighborhood of $a$ intersects
  both $A$ and $A^c$.

Honestly, I couldn't move forward trying to solve this problem. Only thing I proved and tried to use to solve it is $\overline{A}=\partial A\cup A$.
 A: The complementary space $int(A)^c$ of $int(A)$ is a closed subset wich contains $A^c$, so it contains its closure. This implies that $int(A)\cap \overline{A^c}\cap\overline{A}\subset int(A)\cap \overline{A^c}\subset int(A)\cap int(A)^c$ is empty. 
A: Recall that $a\in \overline{A}$ if every open neighbourhood of $a$ has non-empty intersection with $A$ other than $a.$
Let us prove the proposition that you stated, that is, 


$a\in\partial A$ if and only if every neighborhood of $a$ intersects
    both $A$ and $A^c$.


If $a\in \partial A,$ then any open neighbourhood of $a$ will intersect with $A,$ since $a\in \overline{A}.$
Similarly, any open neighbourhood of $a$ will also intersect with $A^c,$ since $a\in \overline{A^c}.$
Therefore, every open neighbourhood of $a$ will intersect both $A$ and $A^c.$
Conversely, if every open neighbourhood of $a$ intersects $A,$ then $a\in\overline{A}.$
Similarly, if every open neighbourhood of $a$ intersects $A^c,$ then $a\in\overline{A^c}.$
It follows that $a\in \overline{A}\cap \overline{A^c}=\partial A.$
Now, let us prove that if $a\in int(A),$ then $a\not\in\partial A.$
If $a\in int(A),$ then there exists an open neighbourhood $U$ of $a$ such that $U\subseteq A.$
This implies that 
$$U \cap A^c = \emptyset.$$
Therefore, by the proposition above, $a\not\in\partial A.$
A: Observe that,  \begin{align}x\in \partial A\cap \operatorname{int}(A)&\implies x\in \partial A\land x\in \operatorname{int}(A)\\ &\implies x\in \overline{A}\cap \overline{X\setminus A}\land x\in\operatorname{int}(A)\\&\implies x\in \overline{A}\land x\in \overline{X\setminus A}\land x\in\operatorname{int}(A)\\&\implies x\in \overline{X\setminus A}\land x\notin X\setminus\operatorname{int}(A)\tag{1}\\&\implies x\in \overline{X\setminus A}\land x\notin \overline{X\setminus A}\tag{2}\end{align}$(1)\implies (2)$ because, $$\operatorname{int}(A)\subseteq A\implies X\setminus A\subseteq X\setminus \operatorname{int}(A)\implies \overline{X\setminus A}\subseteq X\setminus\operatorname{int}(A)$$and we are done.
A: Assume that $x \in \operatorname{int}(A)$ and $x \in \partial A$.
What the former means is that there is $r > 0$ such that $B(x,r) \subseteq A$, while the latter becomes $x \in \overline A \cap \overline {A^c}$, i.e. $x \in \overline A$ and $x \in \overline {A^c}$.
The fact that $x \in \overline {A^c}$, by definition, tells us that $x$ is in every closed set containing $A^c$. We know that $B(x,r)$ is open, so $B(x,r)^c$ is closed. Also, $B(x,r) \subseteq A$ means that $B(x,r)^c \supseteq A^c$, so in fact we have $x \in B(x,r)^c$.
Therefore, we have arrived at a contradiction by unfolding definitions alone.
