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Let $X$ a connected metric space. Is $int X$ connected?

I have a proposition that says: if $X\subseteq Y\subseteq \overline{X}$ and $X$ is connected then $Y$ is connected. But I'm having some problems with using this result. Because every set is open in itself we would have $X = intX \subseteq\overline{X}$ then $intX$ would be connected. Is this correct? What if $X$ is a connected subset of a (not necessarely connected) metric space?

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    $\begingroup$ What is $int X$ when $X$ is a metric space? $\endgroup$
    – xyzzyz
    Oct 31, 2018 at 21:02
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    $\begingroup$ And what is $\bar{X}$ when $X$ is a metric space? $\endgroup$
    – mfl
    Oct 31, 2018 at 21:08
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    $\begingroup$ Are you sure $X$ is not a connected subset of a metric space? Otherwise, $X$ is closed so it is its own closure and $X$ is open so it is its own interior. $\endgroup$
    – John Douma
    Oct 31, 2018 at 21:10
  • $\begingroup$ @JohnDouma I stated the question as it was proposed to me. I don't really know for sure. $\endgroup$
    – AnalyticHarmony
    Oct 31, 2018 at 23:03
  • $\begingroup$ @xyzzyz please see my comment on Maurizio Moreschi's answer $\endgroup$
    – AnalyticHarmony
    Oct 31, 2018 at 23:13

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The answer is negative. Consider two full closed disks touching in one point in the euclidean plane, with induced topology. If you take the interior, you get the disjoint union of two open disks, which is not connected anymore.

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    $\begingroup$ Not in the induced topology, surely? In the ambient topology, i.e. that of the plane, yes. $\endgroup$
    – Calum Gilhooley
    Oct 31, 2018 at 21:29
  • $\begingroup$ I am not sure I get what you're trying to say, can you rephrase? $\endgroup$
    – Maurizio Moreschi
    Oct 31, 2018 at 21:32
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    $\begingroup$ It's much the same point as was made in several comments on the question. I assume you mean the topology induced on the union of the touching discs by considering it as a subset of the plane (with its usual topology). But any space (whether or not it has been defined as a subset of another space) is open - and is therefore its own interior - in any topology that is defined on it. $\endgroup$
    – Calum Gilhooley
    Oct 31, 2018 at 21:42
  • $\begingroup$ Ah yes, of course! I assumed the interior was taken in some ambient space, otherwise the question would not be very interesting... $\endgroup$
    – Maurizio Moreschi
    Oct 31, 2018 at 21:45
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    $\begingroup$ Yes connectedness is intrinsic, there is no problem in there. The problem is just that you want to take the interior of a set, which is meaningful only if you are working in an ambient space (because in its very topology, every space is open by definition). $\endgroup$
    – Maurizio Moreschi
    Oct 31, 2018 at 23:14

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