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I tried to prove that with the set being subset of a space X with metric d, " the interior of the closure of a set equal to the interior of that set". I proved that the interior of,namely, $A$ is included in the interior of the closure of $A$. But I could not prove the reverse, in special because I think that there can be points that are limit points of A and is contained in the interior of the closure, am I wrong?

I am doing this to prove that the closure is equal to the union of the interior points of the closure with the set of all limit points of the set.

Is the aforementioned statement true?

Thank you.

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    $\begingroup$ Think of $\mathbb{Q}\subset \mathbb{R}$. $\endgroup$
    – Daniel Fischer
    Sep 4, 2016 at 12:07
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    $\begingroup$ The claim is true when $A$ is convex $\endgroup$
    – Sasho Nikolov
    Sep 4, 2016 at 14:52
  • $\begingroup$ Thank you for the complementation :) $\endgroup$
    – Yassin Rany
    Sep 4, 2016 at 16:32
  • $\begingroup$ Also useful to ask yourself: What happens to a single point? What happens to the complement of a single point? $\endgroup$
    – Eric Towers
    Sep 4, 2016 at 19:27
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    $\begingroup$ What is the interior of the closure of the complement of a single point (in a metric space)? It's the entire space. But that's not the interior of the complement of a single point. Many (not all) of this kind point-set result can be disposed by considering a trivial nonempty set and its complement, so I suggested that. $\endgroup$
    – Eric Towers
    Sep 5, 2016 at 0:01

2 Answers 2

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The claim isn't true.

The set of rational numbers in the unit interval $[0,1]$ has empty interior, but its closure is the whole interval, so the interior of its closure is the open interval $(0,1)$.

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  • $\begingroup$ Thank you! Just to confirm the ideas, let the whole space be [0,1] with the sme metric that you used, i guess th tis the usual metric, anyway on this metric space, the claim is true, is not? $\endgroup$
    – Yassin Rany
    Sep 4, 2016 at 12:11
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    $\begingroup$ No, the claim is still false in the space $[0,1]$ with the standard metric, for the same reason. $\endgroup$
    – Alon Amit
    Sep 4, 2016 at 12:12
  • $\begingroup$ sorry, i wrote wrong, with the same example i agree that the claim is false, but if the set under analysis, is the set (0,1) ? Thank you for the patience. $\endgroup$
    – Yassin Rany
    Sep 4, 2016 at 12:16
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    $\begingroup$ Another, possibly more intuitive counterexample is $(-1,0)\cup(0,1)$. It's also interesting because here the original set is already open. $\endgroup$
    – celtschk
    Sep 4, 2016 at 13:46
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    $\begingroup$ @LSpice the example remains the rationals in the interval. Maybe I don't understand your question? $\endgroup$
    – Alon Amit
    Sep 4, 2016 at 19:54
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Think about the set $\mathbb R$$-${$0$}

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    $\begingroup$ This is more comment... $\endgroup$
    – nonuser
    May 15, 2022 at 14:28

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