Countable subsets of Euclidean Space.












1












$begingroup$


For an uncountable subset $Esubsetmathbb{R}^n$, I know the following results are true,





  1. $E'$ is uncountable, closed and perfect.

  2. Atmost countably many points of $E$ are not in $E'$


Question: Does this results hold true for countable and finite subsets.



For finite sets it is easy to see because if $E$ is finite, $E'=emptyset$. $E'$ is closed and perfect(vacuously true).



For countable sets??



Edits: $E'$ denote set of all limit points of the set $E$.






real-analysis general-topology







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$endgroup$












  • $begingroup$
    What does $E'$ mean here?
    $endgroup$
    – D. Brogan
    Jan 8 at 17:00










  • $begingroup$
    $E'$ denote set of all limit points of the set $E$.@D.Brogan
    $endgroup$
    – StammeringMathematician
    Jan 8 at 17:05










  • $begingroup$
    Consider $mathbb Z$ and $mathbb Q$ in the space $mathbb R$ of real numbers.
    $endgroup$
    – Dave L. Renfro
    Jan 8 at 17:38
















1












$begingroup$


For an uncountable subset $Esubsetmathbb{R}^n$, I know the following results are true,





  1. $E'$ is uncountable, closed and perfect.

  2. Atmost countably many points of $E$ are not in $E'$


Question: Does this results hold true for countable and finite subsets.



For finite sets it is easy to see because if $E$ is finite, $E'=emptyset$. $E'$ is closed and perfect(vacuously true).



For countable sets??



Edits: $E'$ denote set of all limit points of the set $E$.






real-analysis general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does $E'$ mean here?
    $endgroup$
    – D. Brogan
    Jan 8 at 17:00










  • $begingroup$
    $E'$ denote set of all limit points of the set $E$.@D.Brogan
    $endgroup$
    – StammeringMathematician
    Jan 8 at 17:05










  • $begingroup$
    Consider $mathbb Z$ and $mathbb Q$ in the space $mathbb R$ of real numbers.
    $endgroup$
    – Dave L. Renfro
    Jan 8 at 17:38














1












1








1


1



$begingroup$


For an uncountable subset $Esubsetmathbb{R}^n$, I know the following results are true,





  1. $E'$ is uncountable, closed and perfect.

  2. Atmost countably many points of $E$ are not in $E'$


Question: Does this results hold true for countable and finite subsets.



For finite sets it is easy to see because if $E$ is finite, $E'=emptyset$. $E'$ is closed and perfect(vacuously true).



For countable sets??



Edits: $E'$ denote set of all limit points of the set $E$.






real-analysis general-topology







share|cite|improve this question











$endgroup$




For an uncountable subset $Esubsetmathbb{R}^n$, I know the following results are true,





  1. $E'$ is uncountable, closed and perfect.

  2. Atmost countably many points of $E$ are not in $E'$


Question: Does this results hold true for countable and finite subsets.



For finite sets it is easy to see because if $E$ is finite, $E'=emptyset$. $E'$ is closed and perfect(vacuously true).



For countable sets??



Edits: $E'$ denote set of all limit points of the set $E$.






real-analysis general-topology




real-analysis general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 17:05







StammeringMathematician

















asked Jan 8 at 16:41









StammeringMathematicianStammeringMathematician

2,7061324




2,7061324












  • $begingroup$
    What does $E'$ mean here?
    $endgroup$
    – D. Brogan
    Jan 8 at 17:00










  • $begingroup$
    $E'$ denote set of all limit points of the set $E$.@D.Brogan
    $endgroup$
    – StammeringMathematician
    Jan 8 at 17:05










  • $begingroup$
    Consider $mathbb Z$ and $mathbb Q$ in the space $mathbb R$ of real numbers.
    $endgroup$
    – Dave L. Renfro
    Jan 8 at 17:38


















  • $begingroup$
    What does $E'$ mean here?
    $endgroup$
    – D. Brogan
    Jan 8 at 17:00










  • $begingroup$
    $E'$ denote set of all limit points of the set $E$.@D.Brogan
    $endgroup$
    – StammeringMathematician
    Jan 8 at 17:05










  • $begingroup$
    Consider $mathbb Z$ and $mathbb Q$ in the space $mathbb R$ of real numbers.
    $endgroup$
    – Dave L. Renfro
    Jan 8 at 17:38
















$begingroup$
What does $E'$ mean here?
$endgroup$
– D. Brogan
Jan 8 at 17:00




$begingroup$
What does $E'$ mean here?
$endgroup$
– D. Brogan
Jan 8 at 17:00












$begingroup$
$E'$ denote set of all limit points of the set $E$.@D.Brogan
$endgroup$
– StammeringMathematician
Jan 8 at 17:05




$begingroup$
$E'$ denote set of all limit points of the set $E$.@D.Brogan
$endgroup$
– StammeringMathematician
Jan 8 at 17:05












$begingroup$
Consider $mathbb Z$ and $mathbb Q$ in the space $mathbb R$ of real numbers.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:38




$begingroup$
Consider $mathbb Z$ and $mathbb Q$ in the space $mathbb R$ of real numbers.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:38










1 Answer
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$begingroup$

I believe the claim is false for countable sets. Consider $E=left{frac{1}{n};|; ninmathbb{N}right}.$ Then $E'={0},$ so $E'$ is closed, but not perfect since finite sets in $mathbb R$ have no limit points.



In general, if $X$ is a separable metric space and $Esubset X,$ then $E'$ is always closed (exercise!), but $E'$ is not perfect unless $E$ is uncountable (the proof is the same for that of $mathbb R$) or $E$ is finite, as you have observed.



For the second property, it is true that in separable metric spaces uncountable subsets have uncountably many limit points. It is also true that uncountable sets in these metric spaces contain at least one of their limit points [see this answer for a proof]. Thus, if $Esetminus E'$ were uncountable, it would contain one of its limit points, which is also a limit point of $E,$ and hence a member of $E'.$ So $(Esetminus E')cap E'neq emptyset,$ a contradiction.



Thus, to summarize, all of the conclusions still hold, except that $E'$ is perfect.






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    $begingroup$

    I believe the claim is false for countable sets. Consider $E=left{frac{1}{n};|; ninmathbb{N}right}.$ Then $E'={0},$ so $E'$ is closed, but not perfect since finite sets in $mathbb R$ have no limit points.



    In general, if $X$ is a separable metric space and $Esubset X,$ then $E'$ is always closed (exercise!), but $E'$ is not perfect unless $E$ is uncountable (the proof is the same for that of $mathbb R$) or $E$ is finite, as you have observed.



    For the second property, it is true that in separable metric spaces uncountable subsets have uncountably many limit points. It is also true that uncountable sets in these metric spaces contain at least one of their limit points [see this answer for a proof]. Thus, if $Esetminus E'$ were uncountable, it would contain one of its limit points, which is also a limit point of $E,$ and hence a member of $E'.$ So $(Esetminus E')cap E'neq emptyset,$ a contradiction.



    Thus, to summarize, all of the conclusions still hold, except that $E'$ is perfect.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I believe the claim is false for countable sets. Consider $E=left{frac{1}{n};|; ninmathbb{N}right}.$ Then $E'={0},$ so $E'$ is closed, but not perfect since finite sets in $mathbb R$ have no limit points.



      In general, if $X$ is a separable metric space and $Esubset X,$ then $E'$ is always closed (exercise!), but $E'$ is not perfect unless $E$ is uncountable (the proof is the same for that of $mathbb R$) or $E$ is finite, as you have observed.



      For the second property, it is true that in separable metric spaces uncountable subsets have uncountably many limit points. It is also true that uncountable sets in these metric spaces contain at least one of their limit points [see this answer for a proof]. Thus, if $Esetminus E'$ were uncountable, it would contain one of its limit points, which is also a limit point of $E,$ and hence a member of $E'.$ So $(Esetminus E')cap E'neq emptyset,$ a contradiction.



      Thus, to summarize, all of the conclusions still hold, except that $E'$ is perfect.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I believe the claim is false for countable sets. Consider $E=left{frac{1}{n};|; ninmathbb{N}right}.$ Then $E'={0},$ so $E'$ is closed, but not perfect since finite sets in $mathbb R$ have no limit points.



        In general, if $X$ is a separable metric space and $Esubset X,$ then $E'$ is always closed (exercise!), but $E'$ is not perfect unless $E$ is uncountable (the proof is the same for that of $mathbb R$) or $E$ is finite, as you have observed.



        For the second property, it is true that in separable metric spaces uncountable subsets have uncountably many limit points. It is also true that uncountable sets in these metric spaces contain at least one of their limit points [see this answer for a proof]. Thus, if $Esetminus E'$ were uncountable, it would contain one of its limit points, which is also a limit point of $E,$ and hence a member of $E'.$ So $(Esetminus E')cap E'neq emptyset,$ a contradiction.



        Thus, to summarize, all of the conclusions still hold, except that $E'$ is perfect.






        share|cite|improve this answer









        $endgroup$



        I believe the claim is false for countable sets. Consider $E=left{frac{1}{n};|; ninmathbb{N}right}.$ Then $E'={0},$ so $E'$ is closed, but not perfect since finite sets in $mathbb R$ have no limit points.



        In general, if $X$ is a separable metric space and $Esubset X,$ then $E'$ is always closed (exercise!), but $E'$ is not perfect unless $E$ is uncountable (the proof is the same for that of $mathbb R$) or $E$ is finite, as you have observed.



        For the second property, it is true that in separable metric spaces uncountable subsets have uncountably many limit points. It is also true that uncountable sets in these metric spaces contain at least one of their limit points [see this answer for a proof]. Thus, if $Esetminus E'$ were uncountable, it would contain one of its limit points, which is also a limit point of $E,$ and hence a member of $E'.$ So $(Esetminus E')cap E'neq emptyset,$ a contradiction.



        Thus, to summarize, all of the conclusions still hold, except that $E'$ is perfect.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 19:38









        D. BroganD. Brogan

        763513




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