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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.

Any help please. Thanks in advance.

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  • $\begingroup$ Those sets are called preopen. $\endgroup$
    – Berci
    Commented Feb 11, 2021 at 10:30
  • $\begingroup$ A set which is equal to the interior of its closure is called regular-open. & a set which is the closure of its interior is called regular-closed. $\endgroup$
    – DanielWainfleet
    Commented Feb 12, 2021 at 14:44

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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. $\mathbb{Q}\subseteq\mathbb{R}$.

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  • $\begingroup$ Therefore can we conclude that a non-open set which satisfies this property has to be a dense subset, Or do we have a example of a non-dense subset as well. $\endgroup$
    – m pandey
    Commented Feb 11, 2021 at 15:48
  • $\begingroup$ @mpandey no, it doesn't have to be dense. Consider $\mathbb{Q}\cap(0,1)$ inside $\mathbb{R}$. $\endgroup$
    – freakish
    Commented Feb 11, 2021 at 16:23
  • $\begingroup$ I mean the set has to be dense in the interior of its closure. Am I correct ? $\endgroup$
    – m pandey
    Commented Feb 11, 2021 at 16:31
  • $\begingroup$ @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. $\endgroup$
    – freakish
    Commented Feb 11, 2021 at 16:32
  • $\begingroup$ Yes. But if a set A is contained in the set B , then A may not be dense in B. So if a non open set A is contained in the interior of its closure (say B) then is it true that A is dense in B. $\endgroup$
    – m pandey
    Commented Feb 11, 2021 at 16:39

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