Closure of interior of closed set

Question

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.

It is true if and only if $D$ is the closure of an open set. — Jonas Meyer, Aug 26 '13 at 02:50
For reference, sets where $D = \overline{Int D}$ are called (topologically) regular closed sets: https://proofwiki.org/wiki/Definition:Regular_Closed_Set — erfink, Aug 02 '18 at 08:50

score 19 · Accepted Answer · answered Aug 26 '13 at 02:40

19

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

answered Aug 26 '13 at 02:40

Jared

31,451

score 12 · Answer 2 · answered Aug 26 '13 at 02:42

12

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.

answered Aug 26 '13 at 02:42

Jake

522
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tchappy ha · Answer 3 · 2021-10-06T10:30:51.907

1

Since $\operatorname{Int}D\subseteq D$, obviously $\overline{\operatorname{Int}D}\subseteq \overline{D}=D$ holds.

edited Oct 06 '21 at 10:30

answered Aug 02 '18 at 06:51

tchappy ha

8,690

Closure of interior of closed set

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