A perfect set without any rationals - Perfect minus Rationals?

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This has already been asked several times and well answered here and here so I am asking if my approach is correct or not. My question is as follows




Suppose $E$ is a set in $[0, 1]$ whose decimal expansion consists only of $4, 7$. It can be shown that $E$ is perfect set. Is $E setminus Bbb Q$ perfect?




I cannot find any isolated point but I am confused about $Esetminus Bbb Q$ being closed.




real-analysis




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  • Let $(X,d)$ be a complete metric space. (1). If $X$ is non-empty and has no isolated points then $X$ has a subspace $D$ which is homeomorphic to the Cantor set. So $D$ is a perfect set. (2). If $Y$ is a subspace of $X$ then the subspace topology on $Y$ can be generated by a complete metric $e$ on $Y$ iff $Y$ is a $G_delta$ subset of $X.$.... By (2), $Bbb R$ $Bbb Q$ is completely metrizable. So by (1), $Bbb R$ $Bbb Q$ has a perfect subset.
    – DanielWainfleet
    Jul 17 at 17:40














up vote
1
down vote

favorite












This has already been asked several times and well answered here and here so I am asking if my approach is correct or not. My question is as follows




Suppose $E$ is a set in $[0, 1]$ whose decimal expansion consists only of $4, 7$. It can be shown that $E$ is perfect set. Is $E setminus Bbb Q$ perfect?




I cannot find any isolated point but I am confused about $Esetminus Bbb Q$ being closed.




real-analysis




share|cite|improve this question





















  • Let $(X,d)$ be a complete metric space. (1). If $X$ is non-empty and has no isolated points then $X$ has a subspace $D$ which is homeomorphic to the Cantor set. So $D$ is a perfect set. (2). If $Y$ is a subspace of $X$ then the subspace topology on $Y$ can be generated by a complete metric $e$ on $Y$ iff $Y$ is a $G_delta$ subset of $X.$.... By (2), $Bbb R$ $Bbb Q$ is completely metrizable. So by (1), $Bbb R$ $Bbb Q$ has a perfect subset.
    – DanielWainfleet
    Jul 17 at 17:40












up vote
1
down vote

favorite









up vote
1
down vote

favorite











This has already been asked several times and well answered here and here so I am asking if my approach is correct or not. My question is as follows




Suppose $E$ is a set in $[0, 1]$ whose decimal expansion consists only of $4, 7$. It can be shown that $E$ is perfect set. Is $E setminus Bbb Q$ perfect?




I cannot find any isolated point but I am confused about $Esetminus Bbb Q$ being closed.




real-analysis




share|cite|improve this question













This has already been asked several times and well answered here and here so I am asking if my approach is correct or not. My question is as follows




Suppose $E$ is a set in $[0, 1]$ whose decimal expansion consists only of $4, 7$. It can be shown that $E$ is perfect set. Is $E setminus Bbb Q$ perfect?




I cannot find any isolated point but I am confused about $Esetminus Bbb Q$ being closed.





real-analysis





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share|cite|improve this question




share|cite|improve this question








edited Jul 17 at 14:38









Mason

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1,2321224









asked Jul 17 at 14:26









Hash Nuke

525




525











  • Let $(X,d)$ be a complete metric space. (1). If $X$ is non-empty and has no isolated points then $X$ has a subspace $D$ which is homeomorphic to the Cantor set. So $D$ is a perfect set. (2). If $Y$ is a subspace of $X$ then the subspace topology on $Y$ can be generated by a complete metric $e$ on $Y$ iff $Y$ is a $G_delta$ subset of $X.$.... By (2), $Bbb R$ $Bbb Q$ is completely metrizable. So by (1), $Bbb R$ $Bbb Q$ has a perfect subset.
    – DanielWainfleet
    Jul 17 at 17:40
















  • Let $(X,d)$ be a complete metric space. (1). If $X$ is non-empty and has no isolated points then $X$ has a subspace $D$ which is homeomorphic to the Cantor set. So $D$ is a perfect set. (2). If $Y$ is a subspace of $X$ then the subspace topology on $Y$ can be generated by a complete metric $e$ on $Y$ iff $Y$ is a $G_delta$ subset of $X.$.... By (2), $Bbb R$ $Bbb Q$ is completely metrizable. So by (1), $Bbb R$ $Bbb Q$ has a perfect subset.
    – DanielWainfleet
    Jul 17 at 17:40















Let $(X,d)$ be a complete metric space. (1). If $X$ is non-empty and has no isolated points then $X$ has a subspace $D$ which is homeomorphic to the Cantor set. So $D$ is a perfect set. (2). If $Y$ is a subspace of $X$ then the subspace topology on $Y$ can be generated by a complete metric $e$ on $Y$ iff $Y$ is a $G_delta$ subset of $X.$.... By (2), $Bbb R$ $Bbb Q$ is completely metrizable. So by (1), $Bbb R$ $Bbb Q$ has a perfect subset.
– DanielWainfleet
Jul 17 at 17:40




Let $(X,d)$ be a complete metric space. (1). If $X$ is non-empty and has no isolated points then $X$ has a subspace $D$ which is homeomorphic to the Cantor set. So $D$ is a perfect set. (2). If $Y$ is a subspace of $X$ then the subspace topology on $Y$ can be generated by a complete metric $e$ on $Y$ iff $Y$ is a $G_delta$ subset of $X.$.... By (2), $Bbb R$ $Bbb Q$ is completely metrizable. So by (1), $Bbb R$ $Bbb Q$ has a perfect subset.
– DanielWainfleet
Jul 17 at 17:40










1 Answer
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No, $EsetminusmathbbQ$ is not perfect. For example, consider the sequence:



  • $0.47447444744447444447...$

  • $0.447447444744447444447...$

  • $0.4447447444744447444447...$

  • $0.44447447444744447444447...$

  • $0.444447447444744447444447...$

  • $0.4444447447444744447444447...$


  • ...


Each number in the sequence is in $EsetminusmathbbQ$, but the limit of the sequence is $0.44444....$ which is rational. So $EsetminusmathbbQ$ is not closed.






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  • I see ... thank you :)
    – Hash Nuke
    Jul 17 at 14:38










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










No, $EsetminusmathbbQ$ is not perfect. For example, consider the sequence:



  • $0.47447444744447444447...$

  • $0.447447444744447444447...$

  • $0.4447447444744447444447...$

  • $0.44447447444744447444447...$

  • $0.444447447444744447444447...$

  • $0.4444447447444744447444447...$


  • ...


Each number in the sequence is in $EsetminusmathbbQ$, but the limit of the sequence is $0.44444....$ which is rational. So $EsetminusmathbbQ$ is not closed.






share|cite|improve this answer





















  • I see ... thank you :)
    – Hash Nuke
    Jul 17 at 14:38














up vote
4
down vote



accepted










No, $EsetminusmathbbQ$ is not perfect. For example, consider the sequence:



  • $0.47447444744447444447...$

  • $0.447447444744447444447...$

  • $0.4447447444744447444447...$

  • $0.44447447444744447444447...$

  • $0.444447447444744447444447...$

  • $0.4444447447444744447444447...$


  • ...


Each number in the sequence is in $EsetminusmathbbQ$, but the limit of the sequence is $0.44444....$ which is rational. So $EsetminusmathbbQ$ is not closed.






share|cite|improve this answer





















  • I see ... thank you :)
    – Hash Nuke
    Jul 17 at 14:38












up vote
4
down vote



accepted







up vote
4
down vote



accepted






No, $EsetminusmathbbQ$ is not perfect. For example, consider the sequence:



  • $0.47447444744447444447...$

  • $0.447447444744447444447...$

  • $0.4447447444744447444447...$

  • $0.44447447444744447444447...$

  • $0.444447447444744447444447...$

  • $0.4444447447444744447444447...$


  • ...


Each number in the sequence is in $EsetminusmathbbQ$, but the limit of the sequence is $0.44444....$ which is rational. So $EsetminusmathbbQ$ is not closed.






share|cite|improve this answer













No, $EsetminusmathbbQ$ is not perfect. For example, consider the sequence:



  • $0.47447444744447444447...$

  • $0.447447444744447444447...$

  • $0.4447447444744447444447...$

  • $0.44447447444744447444447...$

  • $0.444447447444744447444447...$

  • $0.4444447447444744447444447...$


  • ...


Each number in the sequence is in $EsetminusmathbbQ$, but the limit of the sequence is $0.44444....$ which is rational. So $EsetminusmathbbQ$ is not closed.







share|cite|improve this answer













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answered Jul 17 at 14:35









Noah Schweber

111k9140263




111k9140263











  • I see ... thank you :)
    – Hash Nuke
    Jul 17 at 14:38
















  • I see ... thank you :)
    – Hash Nuke
    Jul 17 at 14:38















I see ... thank you :)
– Hash Nuke
Jul 17 at 14:38




I see ... thank you :)
– Hash Nuke
Jul 17 at 14:38












 

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