Rudin 2.10 (b) Example












4












$begingroup$


Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_{x}$ be the set of real numbers $y$ such that $0< y< x$. Then





  1. $bigcap_{x in A} E_{x}$ is empty.


The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_{x}$ if $x < y$. Hence $y notin bigcap_{x epsilon A} E_{x}$



I didn't understand the proof & also here is my understanding.



Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.



Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.



Please explain how rudin got this result ?






real-analysis set-theory







share|cite|improve this question









New contributor




Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
    $endgroup$
    – zbrads2
    4 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    4 hours ago
















4












$begingroup$


Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_{x}$ be the set of real numbers $y$ such that $0< y< x$. Then





  1. $bigcap_{x in A} E_{x}$ is empty.


The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_{x}$ if $x < y$. Hence $y notin bigcap_{x epsilon A} E_{x}$



I didn't understand the proof & also here is my understanding.



Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.



Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.



Please explain how rudin got this result ?






real-analysis set-theory







share|cite|improve this question









New contributor




Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
    $endgroup$
    – zbrads2
    4 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    4 hours ago














4












4








4





$begingroup$


Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_{x}$ be the set of real numbers $y$ such that $0< y< x$. Then





  1. $bigcap_{x in A} E_{x}$ is empty.


The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_{x}$ if $x < y$. Hence $y notin bigcap_{x epsilon A} E_{x}$



I didn't understand the proof & also here is my understanding.



Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.



Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.



Please explain how rudin got this result ?






real-analysis set-theory







share|cite|improve this question









New contributor




Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let $A$ be the set of real numbers such that $0 < x leq 1$ . For every $x in A$, let $E_{x}$ be the set of real numbers $y$ such that $0< y< x$. Then





  1. $bigcap_{x in A} E_{x}$ is empty.


The Proof provided in textbook:
We note that for every $y > 0$, $y notin E_{x}$ if $x < y$. Hence $y notin bigcap_{x epsilon A} E_{x}$



I didn't understand the proof & also here is my understanding.



Counter Argument-1: We can always find a real number between every $(0,x)$ where x is $x > 0$. So, It can not be empty.



Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.



Please explain how rudin got this result ?






real-analysis set-theory




real-analysis set-theory






share|cite|improve this question









New contributor




Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Lucas Corrêa

1,6151421




1,6151421






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asked 5 hours ago









Mahendra ReddyMahendra Reddy

212




212




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New contributor





Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
    $endgroup$
    – zbrads2
    4 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    4 hours ago


















  • $begingroup$
    He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
    $endgroup$
    – zbrads2
    4 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    4 hours ago
















$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
4 hours ago




$begingroup$
He is saying that for any $y$, you can find an $x$ in the interval such that $xleq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty.
$endgroup$
– zbrads2
4 hours ago












$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 hours ago




$begingroup$
In addition to Martin's answer, the Nested Interval Property requires closed intervals.
$endgroup$
– Lucas Corrêa
4 hours ago










2 Answers
2






active

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3












$begingroup$

Note that $bigcap_{xin A}E_xsubset(0,infty)$. If $yin bigcap_{xin A}E_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).



Regarding your comment, the nested interval property is about compact sets. These are open and not closed.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Note that $xnotin E_x$ for every $x$






    share|cite|improve this answer








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    Andreé Ríos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $begingroup$

    Note that $bigcap_{xin A}E_xsubset(0,infty)$. If $yin bigcap_{xin A}E_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).



    Regarding your comment, the nested interval property is about compact sets. These are open and not closed.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Note that $bigcap_{xin A}E_xsubset(0,infty)$. If $yin bigcap_{xin A}E_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).



      Regarding your comment, the nested interval property is about compact sets. These are open and not closed.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Note that $bigcap_{xin A}E_xsubset(0,infty)$. If $yin bigcap_{xin A}E_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).



        Regarding your comment, the nested interval property is about compact sets. These are open and not closed.






        share|cite|improve this answer











        $endgroup$



        Note that $bigcap_{xin A}E_xsubset(0,infty)$. If $yin bigcap_{xin A}E_xsubset(0,infty)$, this means that $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).



        Regarding your comment, the nested interval property is about compact sets. These are open and not closed.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 4 hours ago









        Martin ArgeramiMartin Argerami

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            Note that $xnotin E_x$ for every $x$






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            Note that $xnotin E_x$ for every $x$







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            answered 4 hours ago









            Andreé RíosAndreé Ríos

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