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Can anyone give me an example of compact set of which the derived set is infinitely countable set??

thks in advance, I have no idea about this .

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  • $\begingroup$ The "derived set" is the set of accumulation points within the given set? If so, just take $[0,1]\subset \mathbb R$. $\endgroup$
    – lulu
    Sep 2, 2015 at 13:45
  • $\begingroup$ @lulu In this case the derived set is equal to $[0,1]$, which is uncountable. $\endgroup$ Sep 2, 2015 at 13:48
  • $\begingroup$ @MartinSleziak. Ah! Dangers of hasty reading. Am I right about the definition? $\endgroup$
    – lulu
    Sep 2, 2015 at 13:48
  • $\begingroup$ @lulu It is up to OP to clarify this, but I think the common meaning is the set of all limit points: en.wikipedia.org/wiki/Derived_set_%28mathematics%29 $\endgroup$ Sep 2, 2015 at 13:50

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For example, put $A:=\{1,\frac{1}{2},\frac{1}{4},...\}$.

Then the set $B:=\{0\}\cup\bigcup_n \left(\frac{1}{2^n} + \frac{1}{2^n} A\right)$ is compact (closed and bounded) and has derived set $B'=\{0\}\cup \{\frac{1}{2},\frac{1}{4},...\}$.

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