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We have to find a condition $B$ that make the following equivalence true:

$$B \implies {\rm int}({\rm cl}(A)) = A$$

A “valid” condition that seems to satisfy this equivalence is that there shouldn’t be a discontinuity point like for $[0, 1) \cup (1,3]$ where $1$ is the discontinuity point. But how do we translate that mathematically?

Of course $(E, d)$ is a metric space and $A$ is a subset of $E$.

The example was given on $\mathbb{R}$ with the usual distance $|x - y|$.

Christian E. Ramirez
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Charbel El Bateh
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  • The condition $B$ is that we have $A\subset int(cl(A))$. Then we also have equality. See also other posts, e.g., this one. Of course, $A$ need not be a subset of a metric space in general. – Dietrich Burde Mar 08 '23 at 20:35
  • Your condition $B$ will include "$A$ is an open set". To elaborate on Dietrich Burde's comment, for any open set $A$ we would have $A = \operatorname{Int}(A)\subset \operatorname{Int}(\overline{A})$. Your question would be for an open set $A$, when is $\operatorname{Int}(\overline{A})\subset A$ true? – William Sun Mar 08 '23 at 20:40
  • Also see this wikipage. – William Sun Mar 08 '23 at 20:48
  • @WilliamSun I am not really following. The condition will include that A is an open set since it is equal to an interior which is open. But where does Int(A) subset of int(compliment(A)) come in? Excuse me for my markdown knowledge. – Charbel El Bateh Mar 08 '23 at 20:49
  • @CharbelElBateh The overline of $A$ stands for the closure of $A$, not complement. – William Sun Mar 08 '23 at 20:52
  • @WilliamSun thank you for the Wikipedia Page reference it is exactly what I am looking for. Your help was really appreciated. – Charbel El Bateh Mar 08 '23 at 21:01

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