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.


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 $D := \{(0, 0)\}$.


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


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