# Closure of interior of closed set

If $D$ is a closed set, what is the relation in general between the set $D$ and the closure of $\operatorname{Int}D$?

We know that $\operatorname{Int}D\subseteq D$, so $\overline{\operatorname{Int}D}\subseteq \overline{D}$, but since $D$ is closed, we have $\overline{D}=D$, so that $\overline{\operatorname{Int}D}\subseteq D$.

Now, is it true as well that $D\subseteq \overline{\operatorname{Int}D}$? I can't seem to prove it, or give an example of $D$ such that this doesn't hold.

• Aug 26 '13 at 2:43
• It is true if and only if $D$ is the closure of an open set. Aug 26 '13 at 2:50
• For reference, sets where $D = \overline{Int D}$ are called (topologically) regular closed sets: proofwiki.org/wiki/Definition:Regular_Closed_Set Aug 2 '18 at 8:50

Hint: For a counterexample, try to think of a non-empty closed set with empty interior.

In general, the other inclusion doesn't hold. For example, if $D = \{0\}$ is the set containing the single point 0, then its interior is empty. There are a lot of closed sets with this property, like finite sets and Cantor sets.

In general,

$\overline{\operatorname{Int}D}\subseteq D$.

If $X$ is a discrete metric space and $D := X$,

$\overline{\operatorname{Int}D} = D$.

In the case $X = \mathbb{R}^2$,
let $D := \{(0, 0)\}$.

Then,

$\overline{\operatorname{Int}D} \subseteq D$ but $\overline{\operatorname{Int}D} \neq D$.