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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.

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

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  • $\begingroup$ Oh, it was false then. Thank you! $\endgroup$
    – user432019
    Apr 23, 2017 at 10:39

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