Prove that $sup S=sup R,$ if $SsubseteqBbb{R}$ is bounded from above and $Rsubseteq S$.












3












$begingroup$


Please, is this correct?



Let $S$ be a subset of the real numbers bounded from above and let $Rsubseteq S$ satisfy the following condition: $forall;xin S,;exists; rin R;text{such that};xleq r$.
I want to prove that begin{align} sup S=sup R.end{align}



PROOF



Let $xin S$ be arbitrary, then $exists; rin R;text{such that};xleq r.$ The number, $r$, is an upper bound for the set $S.$ So, $sup Sleq r,;text{for some} ; rin R.$ By definition of $sup,;;yleqsup R, forall; yin R.$ In particular, $y=r,;rleqsup R.$ Thus,
begin{align}tag{1} sup Sleq rleqsup R.end{align}
Since $Rsubseteq S$, we have
begin{align}tag{2} sup Rleq sup S.end{align}
Thus, begin{align} sup S=sup R.end{align}






real-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your $r$ is not necessarily an upper bound for $S$. $r$ depends on $x$. So, how do you conclude that $r$ is an upper bound for $S$? Choose, for example, $S = (0,1)$ and $R= S cap mathbb{Q}$.
    $endgroup$
    – trancelocation
    Jan 10 at 4:22












  • $begingroup$
    @trancelocation: You are indeed, right!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:24
















3












$begingroup$


Please, is this correct?



Let $S$ be a subset of the real numbers bounded from above and let $Rsubseteq S$ satisfy the following condition: $forall;xin S,;exists; rin R;text{such that};xleq r$.
I want to prove that begin{align} sup S=sup R.end{align}



PROOF



Let $xin S$ be arbitrary, then $exists; rin R;text{such that};xleq r.$ The number, $r$, is an upper bound for the set $S.$ So, $sup Sleq r,;text{for some} ; rin R.$ By definition of $sup,;;yleqsup R, forall; yin R.$ In particular, $y=r,;rleqsup R.$ Thus,
begin{align}tag{1} sup Sleq rleqsup R.end{align}
Since $Rsubseteq S$, we have
begin{align}tag{2} sup Rleq sup S.end{align}
Thus, begin{align} sup S=sup R.end{align}






real-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your $r$ is not necessarily an upper bound for $S$. $r$ depends on $x$. So, how do you conclude that $r$ is an upper bound for $S$? Choose, for example, $S = (0,1)$ and $R= S cap mathbb{Q}$.
    $endgroup$
    – trancelocation
    Jan 10 at 4:22












  • $begingroup$
    @trancelocation: You are indeed, right!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:24














3












3








3


1



$begingroup$


Please, is this correct?



Let $S$ be a subset of the real numbers bounded from above and let $Rsubseteq S$ satisfy the following condition: $forall;xin S,;exists; rin R;text{such that};xleq r$.
I want to prove that begin{align} sup S=sup R.end{align}



PROOF



Let $xin S$ be arbitrary, then $exists; rin R;text{such that};xleq r.$ The number, $r$, is an upper bound for the set $S.$ So, $sup Sleq r,;text{for some} ; rin R.$ By definition of $sup,;;yleqsup R, forall; yin R.$ In particular, $y=r,;rleqsup R.$ Thus,
begin{align}tag{1} sup Sleq rleqsup R.end{align}
Since $Rsubseteq S$, we have
begin{align}tag{2} sup Rleq sup S.end{align}
Thus, begin{align} sup S=sup R.end{align}






real-analysis analysis







share|cite|improve this question











$endgroup$




Please, is this correct?



Let $S$ be a subset of the real numbers bounded from above and let $Rsubseteq S$ satisfy the following condition: $forall;xin S,;exists; rin R;text{such that};xleq r$.
I want to prove that begin{align} sup S=sup R.end{align}



PROOF



Let $xin S$ be arbitrary, then $exists; rin R;text{such that};xleq r.$ The number, $r$, is an upper bound for the set $S.$ So, $sup Sleq r,;text{for some} ; rin R.$ By definition of $sup,;;yleqsup R, forall; yin R.$ In particular, $y=r,;rleqsup R.$ Thus,
begin{align}tag{1} sup Sleq rleqsup R.end{align}
Since $Rsubseteq S$, we have
begin{align}tag{2} sup Rleq sup S.end{align}
Thus, begin{align} sup S=sup R.end{align}






real-analysis analysis




real-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 4:38







Omojola Micheal

















asked Jan 10 at 4:15









Omojola MichealOmojola Micheal

2,049424




2,049424








  • 1




    $begingroup$
    Your $r$ is not necessarily an upper bound for $S$. $r$ depends on $x$. So, how do you conclude that $r$ is an upper bound for $S$? Choose, for example, $S = (0,1)$ and $R= S cap mathbb{Q}$.
    $endgroup$
    – trancelocation
    Jan 10 at 4:22












  • $begingroup$
    @trancelocation: You are indeed, right!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:24














  • 1




    $begingroup$
    Your $r$ is not necessarily an upper bound for $S$. $r$ depends on $x$. So, how do you conclude that $r$ is an upper bound for $S$? Choose, for example, $S = (0,1)$ and $R= S cap mathbb{Q}$.
    $endgroup$
    – trancelocation
    Jan 10 at 4:22












  • $begingroup$
    @trancelocation: You are indeed, right!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:24








1




1




$begingroup$
Your $r$ is not necessarily an upper bound for $S$. $r$ depends on $x$. So, how do you conclude that $r$ is an upper bound for $S$? Choose, for example, $S = (0,1)$ and $R= S cap mathbb{Q}$.
$endgroup$
– trancelocation
Jan 10 at 4:22






$begingroup$
Your $r$ is not necessarily an upper bound for $S$. $r$ depends on $x$. So, how do you conclude that $r$ is an upper bound for $S$? Choose, for example, $S = (0,1)$ and $R= S cap mathbb{Q}$.
$endgroup$
– trancelocation
Jan 10 at 4:22














$begingroup$
@trancelocation: You are indeed, right!
$endgroup$
– Omojola Micheal
Jan 10 at 4:24




$begingroup$
@trancelocation: You are indeed, right!
$endgroup$
– Omojola Micheal
Jan 10 at 4:24










2 Answers
2






active

oldest

votes


















2












$begingroup$

It seems you confuse "$color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$" with "$color{blue}{exists}; rin R; color{blue}{forall};xin S,;text{such that};xleq r$".



The sets $S = (0,1)$ and $R= S cap mathbb{Q}$ satisfy the condition: $color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$.
But your conclusion $sup S leq r$ is not true for any $r in R$.



To show that $sup R = sup S$ you may proceed as follows:




  • Show $forall epsilon > 0 exists x_{epsilon}in R: sup S - epsilon < x_{epsilon}$
    $$Rightarrow forall epsilon >0: sup S - epsilon < sup R Rightarrow sup S leq sup R$$
    That's exactly what was to be shown.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) You claimed that "But your conclusion $sup Sleq r$ is not true for any $rin R$"but I claimed $sup Sleq r$ for some $rin R$" instead.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:42












  • $begingroup$
    @OmojolaMicheal: If it is not true for any $r$ in the given case, then there is also no specific $r$, for which it is suddenly true.
    $endgroup$
    – trancelocation
    Jan 10 at 4:45










  • $begingroup$
    I get you now. Thanks!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:46










  • $begingroup$
    You are welcome. :-)
    $endgroup$
    – trancelocation
    Jan 10 at 4:46



















1












$begingroup$

Suppose $sup R<sup S$. Then $sup R$ is not an upper bound of $S$, thus $exists xin S|sup R<xlesup S$. But this is a contradiction, since $forall rin R, rlesup R<x$, and it implies there does not exist any $rge x$. This means $sup R=sup S$.






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









    2












    $begingroup$

    It seems you confuse "$color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$" with "$color{blue}{exists}; rin R; color{blue}{forall};xin S,;text{such that};xleq r$".



    The sets $S = (0,1)$ and $R= S cap mathbb{Q}$ satisfy the condition: $color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$.
    But your conclusion $sup S leq r$ is not true for any $r in R$.



    To show that $sup R = sup S$ you may proceed as follows:




    • Show $forall epsilon > 0 exists x_{epsilon}in R: sup S - epsilon < x_{epsilon}$
      $$Rightarrow forall epsilon >0: sup S - epsilon < sup R Rightarrow sup S leq sup R$$
      That's exactly what was to be shown.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (+1) You claimed that "But your conclusion $sup Sleq r$ is not true for any $rin R$"but I claimed $sup Sleq r$ for some $rin R$" instead.
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:42












    • $begingroup$
      @OmojolaMicheal: If it is not true for any $r$ in the given case, then there is also no specific $r$, for which it is suddenly true.
      $endgroup$
      – trancelocation
      Jan 10 at 4:45










    • $begingroup$
      I get you now. Thanks!
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:46










    • $begingroup$
      You are welcome. :-)
      $endgroup$
      – trancelocation
      Jan 10 at 4:46
















    2












    $begingroup$

    It seems you confuse "$color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$" with "$color{blue}{exists}; rin R; color{blue}{forall};xin S,;text{such that};xleq r$".



    The sets $S = (0,1)$ and $R= S cap mathbb{Q}$ satisfy the condition: $color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$.
    But your conclusion $sup S leq r$ is not true for any $r in R$.



    To show that $sup R = sup S$ you may proceed as follows:




    • Show $forall epsilon > 0 exists x_{epsilon}in R: sup S - epsilon < x_{epsilon}$
      $$Rightarrow forall epsilon >0: sup S - epsilon < sup R Rightarrow sup S leq sup R$$
      That's exactly what was to be shown.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (+1) You claimed that "But your conclusion $sup Sleq r$ is not true for any $rin R$"but I claimed $sup Sleq r$ for some $rin R$" instead.
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:42












    • $begingroup$
      @OmojolaMicheal: If it is not true for any $r$ in the given case, then there is also no specific $r$, for which it is suddenly true.
      $endgroup$
      – trancelocation
      Jan 10 at 4:45










    • $begingroup$
      I get you now. Thanks!
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:46










    • $begingroup$
      You are welcome. :-)
      $endgroup$
      – trancelocation
      Jan 10 at 4:46














    2












    2








    2





    $begingroup$

    It seems you confuse "$color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$" with "$color{blue}{exists}; rin R; color{blue}{forall};xin S,;text{such that};xleq r$".



    The sets $S = (0,1)$ and $R= S cap mathbb{Q}$ satisfy the condition: $color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$.
    But your conclusion $sup S leq r$ is not true for any $r in R$.



    To show that $sup R = sup S$ you may proceed as follows:




    • Show $forall epsilon > 0 exists x_{epsilon}in R: sup S - epsilon < x_{epsilon}$
      $$Rightarrow forall epsilon >0: sup S - epsilon < sup R Rightarrow sup S leq sup R$$
      That's exactly what was to be shown.






    share|cite|improve this answer











    $endgroup$



    It seems you confuse "$color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$" with "$color{blue}{exists}; rin R; color{blue}{forall};xin S,;text{such that};xleq r$".



    The sets $S = (0,1)$ and $R= S cap mathbb{Q}$ satisfy the condition: $color{blue}{forall};xin S,;color{blue}{exists}; rin R;text{such that};xleq r$.
    But your conclusion $sup S leq r$ is not true for any $r in R$.



    To show that $sup R = sup S$ you may proceed as follows:




    • Show $forall epsilon > 0 exists x_{epsilon}in R: sup S - epsilon < x_{epsilon}$
      $$Rightarrow forall epsilon >0: sup S - epsilon < sup R Rightarrow sup S leq sup R$$
      That's exactly what was to be shown.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 10 at 4:43

























    answered Jan 10 at 4:36









    trancelocationtrancelocation

    13.6k1829




    13.6k1829












    • $begingroup$
      (+1) You claimed that "But your conclusion $sup Sleq r$ is not true for any $rin R$"but I claimed $sup Sleq r$ for some $rin R$" instead.
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:42












    • $begingroup$
      @OmojolaMicheal: If it is not true for any $r$ in the given case, then there is also no specific $r$, for which it is suddenly true.
      $endgroup$
      – trancelocation
      Jan 10 at 4:45










    • $begingroup$
      I get you now. Thanks!
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:46










    • $begingroup$
      You are welcome. :-)
      $endgroup$
      – trancelocation
      Jan 10 at 4:46


















    • $begingroup$
      (+1) You claimed that "But your conclusion $sup Sleq r$ is not true for any $rin R$"but I claimed $sup Sleq r$ for some $rin R$" instead.
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:42












    • $begingroup$
      @OmojolaMicheal: If it is not true for any $r$ in the given case, then there is also no specific $r$, for which it is suddenly true.
      $endgroup$
      – trancelocation
      Jan 10 at 4:45










    • $begingroup$
      I get you now. Thanks!
      $endgroup$
      – Omojola Micheal
      Jan 10 at 4:46










    • $begingroup$
      You are welcome. :-)
      $endgroup$
      – trancelocation
      Jan 10 at 4:46
















    $begingroup$
    (+1) You claimed that "But your conclusion $sup Sleq r$ is not true for any $rin R$"but I claimed $sup Sleq r$ for some $rin R$" instead.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:42






    $begingroup$
    (+1) You claimed that "But your conclusion $sup Sleq r$ is not true for any $rin R$"but I claimed $sup Sleq r$ for some $rin R$" instead.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:42














    $begingroup$
    @OmojolaMicheal: If it is not true for any $r$ in the given case, then there is also no specific $r$, for which it is suddenly true.
    $endgroup$
    – trancelocation
    Jan 10 at 4:45




    $begingroup$
    @OmojolaMicheal: If it is not true for any $r$ in the given case, then there is also no specific $r$, for which it is suddenly true.
    $endgroup$
    – trancelocation
    Jan 10 at 4:45












    $begingroup$
    I get you now. Thanks!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:46




    $begingroup$
    I get you now. Thanks!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 4:46












    $begingroup$
    You are welcome. :-)
    $endgroup$
    – trancelocation
    Jan 10 at 4:46




    $begingroup$
    You are welcome. :-)
    $endgroup$
    – trancelocation
    Jan 10 at 4:46











    1












    $begingroup$

    Suppose $sup R<sup S$. Then $sup R$ is not an upper bound of $S$, thus $exists xin S|sup R<xlesup S$. But this is a contradiction, since $forall rin R, rlesup R<x$, and it implies there does not exist any $rge x$. This means $sup R=sup S$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Suppose $sup R<sup S$. Then $sup R$ is not an upper bound of $S$, thus $exists xin S|sup R<xlesup S$. But this is a contradiction, since $forall rin R, rlesup R<x$, and it implies there does not exist any $rge x$. This means $sup R=sup S$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Suppose $sup R<sup S$. Then $sup R$ is not an upper bound of $S$, thus $exists xin S|sup R<xlesup S$. But this is a contradiction, since $forall rin R, rlesup R<x$, and it implies there does not exist any $rge x$. This means $sup R=sup S$.






        share|cite|improve this answer











        $endgroup$



        Suppose $sup R<sup S$. Then $sup R$ is not an upper bound of $S$, thus $exists xin S|sup R<xlesup S$. But this is a contradiction, since $forall rin R, rlesup R<x$, and it implies there does not exist any $rge x$. This means $sup R=sup S$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 10 at 4:48

























        answered Jan 10 at 4:43









        Shubham JohriShubham Johri

        5,515818




        5,515818






























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            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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