# If $A$ is open or closed in $(M,d)$, then $(\partial A)^\circ = \varnothing$

$$(M,d)$$ is a metric space. If $$A\subseteq M$$ is open or closed in $$(M,d)$$, then prove that $$(\partial A)^\circ = \varnothing$$.

We must take two cases: (i) $$A$$ is open (ii) $$A$$ is closed. In either case, we must show that the interior of the boundary of $$A$$ is empty.

My work:

1. Let $$A$$ be open. We want to argue that $$(\partial A)^\circ = \emptyset$$. Suppose there exists $$y\in (\partial A)^\circ$$. Then $$y\in \partial A$$, and there exists some $$\epsilon_y > 0$$ s.t. $$B(y,\epsilon_y)\subset \partial A$$. Since $$y\in\partial A$$, we have $$B(y,\epsilon_y)\cap A\neq \emptyset$$ and $$B(y,\epsilon_y)\cap A^c\neq\emptyset$$. I don't know if this helps but we can also say that $$\partial A\cap A\neq \emptyset$$.

2. Let $$A$$ be closed. Then $$A^c$$ is open. Hoping that the above part would be complete, we get $$(\partial A^c)^\circ = \emptyset$$. Does this help?

How do I complete my proof attempts - and are there other possibly nicer ways of approaching this? Thanks a lot!

P.S. I came across this related post but I haven't seen many of the results stated by the OP there.

• for the interior of $\partial A$, you need $B(x, \varepsilon)\subseteq \partial A$ instead of $A$, no? Commented Feb 14, 2021 at 17:08
• You're right, I'll fix that! Thank you! Commented Feb 14, 2021 at 17:09

On both cases you state that there exists some $$\epsilon_y\in\mathbb{R}^+$$ such that $$B(y,\epsilon_y)\subset A$$. This is never true, as long as $$y\in\partial A\implies y\not\in Aº$$. So you are having the same proof in both cases because you introduce that fact that is a contradiction always.
You can argue it using that $$B(y,\epsilon_y)\subset\partial A$$.
If $$A$$ is closed, then $$\partial A\subset A$$. This means that $$y\in Aº$$, because $$B(y,\epsilon_y)\subset \partial A\subset A$$. This is a contradiction because $$B(y,\epsilon_y)\cap (M\setminus A)\neq\emptyset$$ .
If $$A$$ is open, then $$B(y,\epsilon_y)\cap A\neq\emptyset$$. Using that $$B(y,\epsilon_y)\subset\partial A$$ and $$A=Aº$$, we get $$\partial A\cap Aº\neq\emptyset$$, which is also a contradiction.
• I'm looking at your edit. Now it's all correct. $\partial A\cap A=\emptyset$ does really help. Also, $(\partial A^c)º=\emptyset$ helps too, because you can observe that $\partial A=\partial A^c$. Commented Feb 14, 2021 at 17:21