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Is there a subset $A$ of $\Bbb R$ such that $A\cap A^\prime=\emptyset$ and $\overline{A}$ uncountable?

My Attempt:

$$A\cap A^\prime=\emptyset\quad \Rightarrow \quad \overline{A}=A\cup A'=(A\cap A^\prime)\cup A=A$$ so $A$ is closed and $A'\subseteq A$ thus $A'=A\cap A^\prime=\emptyset$.

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  • $\begingroup$ Does $A'$ mean complement of $A$ $\endgroup$ – Abishanka Saha Dec 13 '17 at 12:29
  • $\begingroup$ No, set of all limit points. limit point and we know $\overline{A}=A^\prime\cup A$ $\endgroup$ – C.F.G Dec 13 '17 at 12:31
  • $\begingroup$ Do you know that every irrational number is a limit of some rational sequence? $\endgroup$ – user160738 Dec 13 '17 at 12:36
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    $\begingroup$ @user160738: Do you know that every rational number is also a limit of some rational sequence? $\endgroup$ – Asaf Karagila Dec 13 '17 at 12:42
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    $\begingroup$ @Onil: But any interior point of an interval is also a limit point of the interval. $\endgroup$ – Asaf Karagila Dec 13 '17 at 12:45
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A point in the set $A$ that is not its limit point must be its isolated point. The condition that $A\cap A'=\emptyset$ is equivalent to the statement that the set $A$ consists only of isolated points. Its closure $\overline{A}$ can be uncountable. You can find a counter-intuitive example in the entry isolated point on wikipedia.

https://en.wikipedia.org/wiki/Isolated_point

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