Rudin 2.10 (b) ExampleRudin Theorem $1.11$Nested Interval Property implies Axiom of CompletenessEquivalent...

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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 $mathbb{Q}$ 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 ?










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 ?










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 ?










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






share|cite|improve this question









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Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




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edited 4 hours ago









Lucas Corrêa

1,6151421




1,6151421






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asked 4 hours ago









Mahendra ReddyMahendra Reddy

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New contributor





Mahendra Reddy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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






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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











$endgroup$





















    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.






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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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            Note that $xnotin E_x$ for every $x$







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            answered 3 hours ago









            Andreé RíosAndreé Ríos

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