${ x : x,c in mathbb{R} land c>0 land forall j,k in mathbb{Z} ( k ge 0 Rightarrow |x-j2^{-k}| ge c2^{-k}...












2












$begingroup$


We define the subset $Asubset mathbb{R}$ as follows: $xin A$ if and only if there exists $c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds for all $jin mathbb{Z}$ and integers $kgeq 0$. Prove that $A$ is dense




So I tried showing that for any interval $(a,b)in mathbb{R}, (a,b)cap A neq emptyset$. But I am having quite a bit of trouble. Does anyone have a hint on how to proceed?






real-analysis general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Dense in $Bbb R$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 9:21
















2












$begingroup$


We define the subset $Asubset mathbb{R}$ as follows: $xin A$ if and only if there exists $c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds for all $jin mathbb{Z}$ and integers $kgeq 0$. Prove that $A$ is dense




So I tried showing that for any interval $(a,b)in mathbb{R}, (a,b)cap A neq emptyset$. But I am having quite a bit of trouble. Does anyone have a hint on how to proceed?






real-analysis general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Dense in $Bbb R$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 9:21














2












2








2


1



$begingroup$


We define the subset $Asubset mathbb{R}$ as follows: $xin A$ if and only if there exists $c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds for all $jin mathbb{Z}$ and integers $kgeq 0$. Prove that $A$ is dense




So I tried showing that for any interval $(a,b)in mathbb{R}, (a,b)cap A neq emptyset$. But I am having quite a bit of trouble. Does anyone have a hint on how to proceed?






real-analysis general-topology







share|cite|improve this question











$endgroup$




We define the subset $Asubset mathbb{R}$ as follows: $xin A$ if and only if there exists $c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds for all $jin mathbb{Z}$ and integers $kgeq 0$. Prove that $A$ is dense




So I tried showing that for any interval $(a,b)in mathbb{R}, (a,b)cap A neq emptyset$. But I am having quite a bit of trouble. Does anyone have a hint on how to proceed?






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 Dec 3 '18 at 10:24









user21820

38.8k543153




38.8k543153










asked Dec 3 '18 at 9:08









Joe Man AnalysisJoe Man Analysis

33419




33419












  • $begingroup$
    Dense in $Bbb R$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 9:21


















  • $begingroup$
    Dense in $Bbb R$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 9:21
















$begingroup$
Dense in $Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 3 '18 at 9:21




$begingroup$
Dense in $Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 3 '18 at 9:21










2 Answers
2






active

oldest

votes


















7












$begingroup$

If $x $ is a rational number wchich can be represent as $frac{s}{l} $ with $GCD(s,l)=1 $ and such that $2 $ not divide $l$ then $xin A.$ This is because that $$|2^k x -j|geqfrac{1}{|l|}$$ for all $jin mathbb{Z}.$ But set of such $x$ is dense in $mathbb{R}.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what's NWD?....
    $endgroup$
    – mathworker21
    Dec 3 '18 at 9:22










  • $begingroup$
    sorry I mean GCD
    $endgroup$
    – MotylaNogaTomkaMazura
    Dec 3 '18 at 9:23










  • $begingroup$
    greatest common divisor
    $endgroup$
    – MotylaNogaTomkaMazura
    Dec 3 '18 at 9:24






  • 2




    $begingroup$
    Why is this set dense in $mathbb{R}$?
    $endgroup$
    – elrond
    Dec 3 '18 at 10:30






  • 1




    $begingroup$
    @elrond for a fixed prime $p>2$ the set ${k/p^m}$ is dense (it actually is true even if $p$ is not prime). This is because for any $xinmathbb{R}$ and any $m$ there's one of those rationals between $x$ and $x+1/p^m$.
    $endgroup$
    – freakish
    Dec 3 '18 at 10:53





















0












$begingroup$

$x in A Longleftrightarrow$

there exists $c > 0$ with $forall j in Z$, integer $n geq 0$, $c leq |2^nx - j|.$



$x notin A Longleftrightarrow$
$forall c > 0$, there exists $j in Z$, integer $n geq 0$ with $|2^nx - j| < c$
$Longleftrightarrow$

exists $j in Z$, integer $n geq 0$ with $x = j/2^n$
$Longleftrightarrow x$ is a dyadic rational.



Since $R - Q subset R - { x : x text{ dyadic rational } } = A$

and $R - Q$ is dense, $A$ is dense.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    If $x $ is a rational number wchich can be represent as $frac{s}{l} $ with $GCD(s,l)=1 $ and such that $2 $ not divide $l$ then $xin A.$ This is because that $$|2^k x -j|geqfrac{1}{|l|}$$ for all $jin mathbb{Z}.$ But set of such $x$ is dense in $mathbb{R}.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      what's NWD?....
      $endgroup$
      – mathworker21
      Dec 3 '18 at 9:22










    • $begingroup$
      sorry I mean GCD
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:23










    • $begingroup$
      greatest common divisor
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:24






    • 2




      $begingroup$
      Why is this set dense in $mathbb{R}$?
      $endgroup$
      – elrond
      Dec 3 '18 at 10:30






    • 1




      $begingroup$
      @elrond for a fixed prime $p>2$ the set ${k/p^m}$ is dense (it actually is true even if $p$ is not prime). This is because for any $xinmathbb{R}$ and any $m$ there's one of those rationals between $x$ and $x+1/p^m$.
      $endgroup$
      – freakish
      Dec 3 '18 at 10:53


















    7












    $begingroup$

    If $x $ is a rational number wchich can be represent as $frac{s}{l} $ with $GCD(s,l)=1 $ and such that $2 $ not divide $l$ then $xin A.$ This is because that $$|2^k x -j|geqfrac{1}{|l|}$$ for all $jin mathbb{Z}.$ But set of such $x$ is dense in $mathbb{R}.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      what's NWD?....
      $endgroup$
      – mathworker21
      Dec 3 '18 at 9:22










    • $begingroup$
      sorry I mean GCD
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:23










    • $begingroup$
      greatest common divisor
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:24






    • 2




      $begingroup$
      Why is this set dense in $mathbb{R}$?
      $endgroup$
      – elrond
      Dec 3 '18 at 10:30






    • 1




      $begingroup$
      @elrond for a fixed prime $p>2$ the set ${k/p^m}$ is dense (it actually is true even if $p$ is not prime). This is because for any $xinmathbb{R}$ and any $m$ there's one of those rationals between $x$ and $x+1/p^m$.
      $endgroup$
      – freakish
      Dec 3 '18 at 10:53
















    7












    7








    7





    $begingroup$

    If $x $ is a rational number wchich can be represent as $frac{s}{l} $ with $GCD(s,l)=1 $ and such that $2 $ not divide $l$ then $xin A.$ This is because that $$|2^k x -j|geqfrac{1}{|l|}$$ for all $jin mathbb{Z}.$ But set of such $x$ is dense in $mathbb{R}.$






    share|cite|improve this answer











    $endgroup$



    If $x $ is a rational number wchich can be represent as $frac{s}{l} $ with $GCD(s,l)=1 $ and such that $2 $ not divide $l$ then $xin A.$ This is because that $$|2^k x -j|geqfrac{1}{|l|}$$ for all $jin mathbb{Z}.$ But set of such $x$ is dense in $mathbb{R}.$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 3 '18 at 9:22

























    answered Dec 3 '18 at 9:21









    MotylaNogaTomkaMazuraMotylaNogaTomkaMazura

    6,552917




    6,552917












    • $begingroup$
      what's NWD?....
      $endgroup$
      – mathworker21
      Dec 3 '18 at 9:22










    • $begingroup$
      sorry I mean GCD
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:23










    • $begingroup$
      greatest common divisor
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:24






    • 2




      $begingroup$
      Why is this set dense in $mathbb{R}$?
      $endgroup$
      – elrond
      Dec 3 '18 at 10:30






    • 1




      $begingroup$
      @elrond for a fixed prime $p>2$ the set ${k/p^m}$ is dense (it actually is true even if $p$ is not prime). This is because for any $xinmathbb{R}$ and any $m$ there's one of those rationals between $x$ and $x+1/p^m$.
      $endgroup$
      – freakish
      Dec 3 '18 at 10:53




















    • $begingroup$
      what's NWD?....
      $endgroup$
      – mathworker21
      Dec 3 '18 at 9:22










    • $begingroup$
      sorry I mean GCD
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:23










    • $begingroup$
      greatest common divisor
      $endgroup$
      – MotylaNogaTomkaMazura
      Dec 3 '18 at 9:24






    • 2




      $begingroup$
      Why is this set dense in $mathbb{R}$?
      $endgroup$
      – elrond
      Dec 3 '18 at 10:30






    • 1




      $begingroup$
      @elrond for a fixed prime $p>2$ the set ${k/p^m}$ is dense (it actually is true even if $p$ is not prime). This is because for any $xinmathbb{R}$ and any $m$ there's one of those rationals between $x$ and $x+1/p^m$.
      $endgroup$
      – freakish
      Dec 3 '18 at 10:53


















    $begingroup$
    what's NWD?....
    $endgroup$
    – mathworker21
    Dec 3 '18 at 9:22




    $begingroup$
    what's NWD?....
    $endgroup$
    – mathworker21
    Dec 3 '18 at 9:22












    $begingroup$
    sorry I mean GCD
    $endgroup$
    – MotylaNogaTomkaMazura
    Dec 3 '18 at 9:23




    $begingroup$
    sorry I mean GCD
    $endgroup$
    – MotylaNogaTomkaMazura
    Dec 3 '18 at 9:23












    $begingroup$
    greatest common divisor
    $endgroup$
    – MotylaNogaTomkaMazura
    Dec 3 '18 at 9:24




    $begingroup$
    greatest common divisor
    $endgroup$
    – MotylaNogaTomkaMazura
    Dec 3 '18 at 9:24




    2




    2




    $begingroup$
    Why is this set dense in $mathbb{R}$?
    $endgroup$
    – elrond
    Dec 3 '18 at 10:30




    $begingroup$
    Why is this set dense in $mathbb{R}$?
    $endgroup$
    – elrond
    Dec 3 '18 at 10:30




    1




    1




    $begingroup$
    @elrond for a fixed prime $p>2$ the set ${k/p^m}$ is dense (it actually is true even if $p$ is not prime). This is because for any $xinmathbb{R}$ and any $m$ there's one of those rationals between $x$ and $x+1/p^m$.
    $endgroup$
    – freakish
    Dec 3 '18 at 10:53






    $begingroup$
    @elrond for a fixed prime $p>2$ the set ${k/p^m}$ is dense (it actually is true even if $p$ is not prime). This is because for any $xinmathbb{R}$ and any $m$ there's one of those rationals between $x$ and $x+1/p^m$.
    $endgroup$
    – freakish
    Dec 3 '18 at 10:53













    0












    $begingroup$

    $x in A Longleftrightarrow$

    there exists $c > 0$ with $forall j in Z$, integer $n geq 0$, $c leq |2^nx - j|.$



    $x notin A Longleftrightarrow$
    $forall c > 0$, there exists $j in Z$, integer $n geq 0$ with $|2^nx - j| < c$
    $Longleftrightarrow$

    exists $j in Z$, integer $n geq 0$ with $x = j/2^n$
    $Longleftrightarrow x$ is a dyadic rational.



    Since $R - Q subset R - { x : x text{ dyadic rational } } = A$

    and $R - Q$ is dense, $A$ is dense.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $x in A Longleftrightarrow$

      there exists $c > 0$ with $forall j in Z$, integer $n geq 0$, $c leq |2^nx - j|.$



      $x notin A Longleftrightarrow$
      $forall c > 0$, there exists $j in Z$, integer $n geq 0$ with $|2^nx - j| < c$
      $Longleftrightarrow$

      exists $j in Z$, integer $n geq 0$ with $x = j/2^n$
      $Longleftrightarrow x$ is a dyadic rational.



      Since $R - Q subset R - { x : x text{ dyadic rational } } = A$

      and $R - Q$ is dense, $A$ is dense.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $x in A Longleftrightarrow$

        there exists $c > 0$ with $forall j in Z$, integer $n geq 0$, $c leq |2^nx - j|.$



        $x notin A Longleftrightarrow$
        $forall c > 0$, there exists $j in Z$, integer $n geq 0$ with $|2^nx - j| < c$
        $Longleftrightarrow$

        exists $j in Z$, integer $n geq 0$ with $x = j/2^n$
        $Longleftrightarrow x$ is a dyadic rational.



        Since $R - Q subset R - { x : x text{ dyadic rational } } = A$

        and $R - Q$ is dense, $A$ is dense.






        share|cite|improve this answer











        $endgroup$



        $x in A Longleftrightarrow$

        there exists $c > 0$ with $forall j in Z$, integer $n geq 0$, $c leq |2^nx - j|.$



        $x notin A Longleftrightarrow$
        $forall c > 0$, there exists $j in Z$, integer $n geq 0$ with $|2^nx - j| < c$
        $Longleftrightarrow$

        exists $j in Z$, integer $n geq 0$ with $x = j/2^n$
        $Longleftrightarrow x$ is a dyadic rational.



        Since $R - Q subset R - { x : x text{ dyadic rational } } = A$

        and $R - Q$ is dense, $A$ is dense.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 10:29









        Joe Man Analysis

        33419




        33419










        answered Dec 3 '18 at 10:04









        William ElliotWilliam Elliot

        7,5472720




        7,5472720






























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