# Show that if A is a open set or a closed set, then $int(\partial A) = \emptyset$. Also find if the converse is true.

Let $$\partial A$$ be a set of boundary points of $$A\subset\mathbb{R}^n$$. Show that if A is a open set or a closed set, then $$\text{int}(\partial A)=\emptyset$$. Also find if the converse is true.

My answer) Suppose A is a open set. Then $$A \cap \partial A = \emptyset$$. (This was mentioned in previous problem, and I already proved it.)

If $$x \in \text{int}(\partial A)$$, then there exists $$\epsilon>0$$ s.t. $$N(x, \epsilon) \subset \partial A$$. However, this means $$N(x, \epsilon) \cap A = \emptyset$$, and $$x$$ cannot be a boundary point. Therefore it is contradiction.

Now, suppose A is a closed set. Then, $$\partial A \subset A$$. (this is also mentioned in previous problem).

If $$x \in \text{int}(\partial A)$$, then there exists $$\epsilon>0$$ s.t. $$N(x, \epsilon) \subset \partial A \subset A$$. This means $$N(x, \epsilon) \cap A^\complement = \emptyset$$, and $$x$$ cannot be a boundary point. This is contradiction, too. Thus, $$\text{int}(\partial A) = \emptyset$$.

This is what I've got so far. The problem is I can't prove the converse. My hunch is that it is related with $$A \subset \text{int}(A) \cup \partial A$$, but I can't progress the idea.

Your work is correct. Now consider $\mathbb R$ with the usual topology and $A = [0,1)$. We have $\partial A = \{0,1\}$; obviously $int (\partial A) = \varnothing$.