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Rudin 2.10 (b) Example


Rudin Theorem $1.11$Nested Interval Property implies Axiom of CompletenessEquivalent definitions of Lebesgue Measurability (Rudin and Royden)Baby Rudin Problem 2.29False proofs claiming that $mathbbQ$ is uncountable.Supremum of closed setsIs Abbott's Proof of the Uncountabilty of Real Numbers too strong?Existence of Nth root, by Rudin soft questionProof of uncountability of irrationals without using completeness of real numbersbaby rudin chapter 2 problem 21(b)- help with alternative method













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
    3 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    3 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
    3 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    3 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 3 hours ago









Lucas Corrêa

1,6151421




1,6151421






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.









asked 4 hours ago









Mahendra ReddyMahendra Reddy

212




212




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.





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
    3 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    3 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
    3 hours ago










  • $begingroup$
    In addition to Martin's answer, the Nested Interval Property requires closed intervals.
    $endgroup$
    – Lucas Corrêa
    3 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
3 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
3 hours ago












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




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










2 Answers
2






active

oldest

votes


















3












$begingroup$

Note that $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, 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








    New contributor




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






    $endgroup$












    • $begingroup$
      Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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      3 hours ago











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    3












    $begingroup$

    Note that $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, 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 $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, 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 $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, 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 $bigcap_xin AE_xsubset(0,infty)$. If $yin bigcap_xin AE_xsubset(0,infty)$, 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 3 hours ago

























        answered 4 hours ago









        Martin ArgeramiMartin Argerami

        130k1184185




        130k1184185





















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            Note that $xnotin E_x$ for every $x$






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              Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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            answered 3 hours ago









            Andreé RíosAndreé Ríos

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