Prove that the set used in archimedean property is closed












-1












$begingroup$


Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.



What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.



Is my approach right ?



Please help!






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
    $endgroup$
    – DanielWainfleet
    Dec 17 '18 at 6:26


















-1












$begingroup$


Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.



What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.



Is my approach right ?



Please help!






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
    $endgroup$
    – DanielWainfleet
    Dec 17 '18 at 6:26
















-1












-1








-1





$begingroup$


Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.



What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.



Is my approach right ?



Please help!






real-analysis







share|cite|improve this question











$endgroup$




Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.



What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.



Is my approach right ?



Please help!






real-analysis




real-analysis






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 4:12









Tianlalu

3,08621038




3,08621038










asked Dec 17 '18 at 4:07









Debarghya MukherjeeDebarghya Mukherjee

352




352












  • $begingroup$
    I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
    $endgroup$
    – DanielWainfleet
    Dec 17 '18 at 6:26




















  • $begingroup$
    I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
    $endgroup$
    – DanielWainfleet
    Dec 17 '18 at 6:26


















$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26






$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26












2 Answers
2






active

oldest

votes


















0












$begingroup$

You have a couple options:



1) Show the compliment is open.
That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.



Clearly the first case is easy. The second is slightly harder.



2) Show the set contains all its limit points.
This might be a bit harder.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So do we need to apply archime dean property
    $endgroup$
    – Debarghya Mukherjee
    Dec 17 '18 at 4:33










  • $begingroup$
    For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
    $endgroup$
    – gdepaul
    Dec 17 '18 at 4:36












  • $begingroup$
    Complement, not compliment. Related to the word "complete".
    $endgroup$
    – DanielWainfleet
    Dec 17 '18 at 6:21



















0












$begingroup$

(gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)



And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.



I don't see this approach working.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    You have a couple options:



    1) Show the compliment is open.
    That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.



    Clearly the first case is easy. The second is slightly harder.



    2) Show the set contains all its limit points.
    This might be a bit harder.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So do we need to apply archime dean property
      $endgroup$
      – Debarghya Mukherjee
      Dec 17 '18 at 4:33










    • $begingroup$
      For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
      $endgroup$
      – gdepaul
      Dec 17 '18 at 4:36












    • $begingroup$
      Complement, not compliment. Related to the word "complete".
      $endgroup$
      – DanielWainfleet
      Dec 17 '18 at 6:21
















    0












    $begingroup$

    You have a couple options:



    1) Show the compliment is open.
    That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.



    Clearly the first case is easy. The second is slightly harder.



    2) Show the set contains all its limit points.
    This might be a bit harder.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So do we need to apply archime dean property
      $endgroup$
      – Debarghya Mukherjee
      Dec 17 '18 at 4:33










    • $begingroup$
      For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
      $endgroup$
      – gdepaul
      Dec 17 '18 at 4:36












    • $begingroup$
      Complement, not compliment. Related to the word "complete".
      $endgroup$
      – DanielWainfleet
      Dec 17 '18 at 6:21














    0












    0








    0





    $begingroup$

    You have a couple options:



    1) Show the compliment is open.
    That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.



    Clearly the first case is easy. The second is slightly harder.



    2) Show the set contains all its limit points.
    This might be a bit harder.






    share|cite|improve this answer









    $endgroup$



    You have a couple options:



    1) Show the compliment is open.
    That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.



    Clearly the first case is easy. The second is slightly harder.



    2) Show the set contains all its limit points.
    This might be a bit harder.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 17 '18 at 4:23









    gdepaulgdepaul

    613




    613












    • $begingroup$
      So do we need to apply archime dean property
      $endgroup$
      – Debarghya Mukherjee
      Dec 17 '18 at 4:33










    • $begingroup$
      For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
      $endgroup$
      – gdepaul
      Dec 17 '18 at 4:36












    • $begingroup$
      Complement, not compliment. Related to the word "complete".
      $endgroup$
      – DanielWainfleet
      Dec 17 '18 at 6:21


















    • $begingroup$
      So do we need to apply archime dean property
      $endgroup$
      – Debarghya Mukherjee
      Dec 17 '18 at 4:33










    • $begingroup$
      For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
      $endgroup$
      – gdepaul
      Dec 17 '18 at 4:36












    • $begingroup$
      Complement, not compliment. Related to the word "complete".
      $endgroup$
      – DanielWainfleet
      Dec 17 '18 at 6:21
















    $begingroup$
    So do we need to apply archime dean property
    $endgroup$
    – Debarghya Mukherjee
    Dec 17 '18 at 4:33




    $begingroup$
    So do we need to apply archime dean property
    $endgroup$
    – Debarghya Mukherjee
    Dec 17 '18 at 4:33












    $begingroup$
    For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
    $endgroup$
    – gdepaul
    Dec 17 '18 at 4:36






    $begingroup$
    For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
    $endgroup$
    – gdepaul
    Dec 17 '18 at 4:36














    $begingroup$
    Complement, not compliment. Related to the word "complete".
    $endgroup$
    – DanielWainfleet
    Dec 17 '18 at 6:21




    $begingroup$
    Complement, not compliment. Related to the word "complete".
    $endgroup$
    – DanielWainfleet
    Dec 17 '18 at 6:21











    0












    $begingroup$

    (gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)



    And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.



    I don't see this approach working.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      (gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)



      And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.



      I don't see this approach working.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        (gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)



        And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.



        I don't see this approach working.






        share|cite|improve this answer









        $endgroup$



        (gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)



        And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.



        I don't see this approach working.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 4:24









        jmerryjmerry

        4,334514




        4,334514






























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