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Prove or disprove that the interior and the boundary of a connected set are connected.


I know that if $C$ is connected then $\overline{C}$ is connected but should be the case here

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3 Answers 3

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For the boundary as poton says open interval is a counter example. For interior consider union of closed disks of radius 1 centered at (-1,0) and (1,0) in the plane.

It is surely connected but its interior i.e. union of open disks is not connected

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for the second part consider- the open intervals.

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    $\begingroup$ what about the second part? somebody tell me please. $\endgroup$
    – ayesha
    May 2, 2013 at 6:21
  • $\begingroup$ This probably involves an infinitesimally tin disc that lies on the boundary. No matter how small you make such a disc, it includes some point on the set and a point on the boundary. $\endgroup$
    – Kaz
    May 2, 2013 at 6:26
  • $\begingroup$ could not understand.will you explain please $\endgroup$
    – ayesha
    May 2, 2013 at 6:33
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For the first part: Let $X=\{a,b,c,d\}$ and the topology on $X$ is $\tau=\{X,\phi,\{a\},\{b\},\{a,b\},\{a,b,d\}\}$. Let $A=\{a,b,c\}$ the set $A$ is connected but the interior of it is $\mathrm{int}(A)=\{a,b\}$ is disconnected.

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