Can the closure of isolated point set be uncountable












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Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.



Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?






real-analysis general-topology







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  • $begingroup$
    What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
    $endgroup$
    – bof
    Jan 6 at 9:52












  • $begingroup$
    Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
    $endgroup$
    – MJD
    Jan 6 at 10:56








  • 1




    $begingroup$
    In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
    $endgroup$
    – bof
    Jan 6 at 11:35












  • $begingroup$
    @bof Nice example!
    $endgroup$
    – Paul
    Jan 6 at 11:44
















1












$begingroup$


Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.



Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?






real-analysis general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
    $endgroup$
    – bof
    Jan 6 at 9:52












  • $begingroup$
    Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
    $endgroup$
    – MJD
    Jan 6 at 10:56








  • 1




    $begingroup$
    In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
    $endgroup$
    – bof
    Jan 6 at 11:35












  • $begingroup$
    @bof Nice example!
    $endgroup$
    – Paul
    Jan 6 at 11:44














1












1








1





$begingroup$


Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.



Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?






real-analysis general-topology







share|cite|improve this question











$endgroup$




Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.



Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?






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 6 at 9:35







Paul

















asked Jan 6 at 9:12









PaulPaul

194111




194111












  • $begingroup$
    What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
    $endgroup$
    – bof
    Jan 6 at 9:52












  • $begingroup$
    Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
    $endgroup$
    – MJD
    Jan 6 at 10:56








  • 1




    $begingroup$
    In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
    $endgroup$
    – bof
    Jan 6 at 11:35












  • $begingroup$
    @bof Nice example!
    $endgroup$
    – Paul
    Jan 6 at 11:44


















  • $begingroup$
    What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
    $endgroup$
    – bof
    Jan 6 at 9:52












  • $begingroup$
    Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
    $endgroup$
    – MJD
    Jan 6 at 10:56








  • 1




    $begingroup$
    In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
    $endgroup$
    – bof
    Jan 6 at 11:35












  • $begingroup$
    @bof Nice example!
    $endgroup$
    – Paul
    Jan 6 at 11:44
















$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52






$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52














$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56






$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56






1




1




$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35






$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35














$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44




$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44










1 Answer
1






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$begingroup$

Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah... When I submited the question, I realized it is trivial...
    $endgroup$
    – Paul
    Jan 6 at 9:19












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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

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3












$begingroup$

Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah... When I submited the question, I realized it is trivial...
    $endgroup$
    – Paul
    Jan 6 at 9:19
















3












$begingroup$

Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah... When I submited the question, I realized it is trivial...
    $endgroup$
    – Paul
    Jan 6 at 9:19














3












3








3





$begingroup$

Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.






share|cite|improve this answer









$endgroup$



Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 9:14









twnlytwnly

1,2211214




1,2211214












  • $begingroup$
    Yeah... When I submited the question, I realized it is trivial...
    $endgroup$
    – Paul
    Jan 6 at 9:19


















  • $begingroup$
    Yeah... When I submited the question, I realized it is trivial...
    $endgroup$
    – Paul
    Jan 6 at 9:19
















$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19




$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19


















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