# Relation between a set $A$ and the interior of its closure

Let $$A$$ be a subset of a metric space $$(X , d)$$.

I know that if $$A$$ is open then it is contained in the interior of its closure. But if $$A$$ is not open then is it true that A is not contained in the interior of its closure. There are examples of non open sets which are not contained in the interior of their closure. For example any non empty finite subset of $$\mathbb{R}$$ (with usual metric) satisfies this statement as the interior their closure is empty. However, is this true for any non open subset of a general metric space.

If $$D\subseteq X$$ is dense, then $$\overline{D}=X$$ (by definition) and so $$int(\overline{D})=X$$. Therefore $$D\subseteq int(\overline{D})$$. So all you need is to take some non-open dense subset, e.g. $$\mathbb{Q}\subseteq\mathbb{R}$$.
• @mpandey no, it doesn't have to be dense. Consider $\mathbb{Q}\cap(0,1)$ inside $\mathbb{R}$. Commented Feb 11, 2021 at 16:23
• @mpandey Sure, but that is a circular argument. If it is dense in the interior of its closure then it has to be subset of the interior of the closure to begin with. Also every set $D$ is dense in $\overline{D}$. And thus in $int(\overline{D})$ as well if a subset of. Commented Feb 11, 2021 at 16:32