$A$ and $B$ nonempty subsets of $mathbb{R}$, if $sup(A)=sup(B)$ then $forall a in A, exists b in B$ such that...












3












$begingroup$



Let $A$ and $B$ nonempty subsets of $mathbb{R}$, such that $sup(A)=sup(B)$ and $sup(A) not in A$



Prove: $forall a in A, exists b in B$ such that $a<b$




I started out with the fact that:
$$ a< sup (A) = sup (B)$$
And of course we know that
$$b leq sup(B)= sup(A)$$
So now we know that $exists b$ that sort of "must be in the middle"
$$ a <b leq sup(B) $$
I don't know how to make this concrete, also is this element simply the supremum?






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What is your definition of the supremum, and what properties have you proven for it? For example, if your definition of $mathrm{sup}(X)$ is that it is a number $s$ such that (i) $s$ is an upper bound for $X$; and (ii) if $t$ is any upper bound for $x$, then $sleq t$; then from $altmathrm{sup}(B)$ you can conclude that $a$ is not an upper bound for $B$, and therefore...
    $endgroup$
    – Arturo Magidin
    Dec 2 '18 at 0:00
















3












$begingroup$



Let $A$ and $B$ nonempty subsets of $mathbb{R}$, such that $sup(A)=sup(B)$ and $sup(A) not in A$



Prove: $forall a in A, exists b in B$ such that $a<b$




I started out with the fact that:
$$ a< sup (A) = sup (B)$$
And of course we know that
$$b leq sup(B)= sup(A)$$
So now we know that $exists b$ that sort of "must be in the middle"
$$ a <b leq sup(B) $$
I don't know how to make this concrete, also is this element simply the supremum?






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What is your definition of the supremum, and what properties have you proven for it? For example, if your definition of $mathrm{sup}(X)$ is that it is a number $s$ such that (i) $s$ is an upper bound for $X$; and (ii) if $t$ is any upper bound for $x$, then $sleq t$; then from $altmathrm{sup}(B)$ you can conclude that $a$ is not an upper bound for $B$, and therefore...
    $endgroup$
    – Arturo Magidin
    Dec 2 '18 at 0:00














3












3








3





$begingroup$



Let $A$ and $B$ nonempty subsets of $mathbb{R}$, such that $sup(A)=sup(B)$ and $sup(A) not in A$



Prove: $forall a in A, exists b in B$ such that $a<b$




I started out with the fact that:
$$ a< sup (A) = sup (B)$$
And of course we know that
$$b leq sup(B)= sup(A)$$
So now we know that $exists b$ that sort of "must be in the middle"
$$ a <b leq sup(B) $$
I don't know how to make this concrete, also is this element simply the supremum?






real-analysis







share|cite|improve this question









$endgroup$





Let $A$ and $B$ nonempty subsets of $mathbb{R}$, such that $sup(A)=sup(B)$ and $sup(A) not in A$



Prove: $forall a in A, exists b in B$ such that $a<b$




I started out with the fact that:
$$ a< sup (A) = sup (B)$$
And of course we know that
$$b leq sup(B)= sup(A)$$
So now we know that $exists b$ that sort of "must be in the middle"
$$ a <b leq sup(B) $$
I don't know how to make this concrete, also is this element simply the supremum?






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 23:54









Wesley StrikWesley Strik

1,667423




1,667423








  • 1




    $begingroup$
    What is your definition of the supremum, and what properties have you proven for it? For example, if your definition of $mathrm{sup}(X)$ is that it is a number $s$ such that (i) $s$ is an upper bound for $X$; and (ii) if $t$ is any upper bound for $x$, then $sleq t$; then from $altmathrm{sup}(B)$ you can conclude that $a$ is not an upper bound for $B$, and therefore...
    $endgroup$
    – Arturo Magidin
    Dec 2 '18 at 0:00














  • 1




    $begingroup$
    What is your definition of the supremum, and what properties have you proven for it? For example, if your definition of $mathrm{sup}(X)$ is that it is a number $s$ such that (i) $s$ is an upper bound for $X$; and (ii) if $t$ is any upper bound for $x$, then $sleq t$; then from $altmathrm{sup}(B)$ you can conclude that $a$ is not an upper bound for $B$, and therefore...
    $endgroup$
    – Arturo Magidin
    Dec 2 '18 at 0:00








1




1




$begingroup$
What is your definition of the supremum, and what properties have you proven for it? For example, if your definition of $mathrm{sup}(X)$ is that it is a number $s$ such that (i) $s$ is an upper bound for $X$; and (ii) if $t$ is any upper bound for $x$, then $sleq t$; then from $altmathrm{sup}(B)$ you can conclude that $a$ is not an upper bound for $B$, and therefore...
$endgroup$
– Arturo Magidin
Dec 2 '18 at 0:00




$begingroup$
What is your definition of the supremum, and what properties have you proven for it? For example, if your definition of $mathrm{sup}(X)$ is that it is a number $s$ such that (i) $s$ is an upper bound for $X$; and (ii) if $t$ is any upper bound for $x$, then $sleq t$; then from $altmathrm{sup}(B)$ you can conclude that $a$ is not an upper bound for $B$, and therefore...
$endgroup$
– Arturo Magidin
Dec 2 '18 at 0:00










4 Answers
4






active

oldest

votes


















2












$begingroup$

The following argument utilizes the second characterization of the supremum found here:



Suppose $a in A$. Since $sup(A) notin A$, we know that $sup(A)-a>0$. So we select $varepsilon=sup(A)-a$. Now we have that there exists $b in B$ such that
begin{aligned} b&>sup(B)-varepsilon \& = sup(B)-left(sup(A)-aright) \& = sup(B)-sup(B)+a \& =a.
end{aligned}






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is a good tool, yes we had to prove this as an earlier exercise in the book. It's a little bit less used than " the minimum of all upper bounds" property, but very useful indeed. Thank you.
    $endgroup$
    – Wesley Strik
    Dec 2 '18 at 10:27



















3












$begingroup$

By contradiction. Assume $exists a in A$ such that $forall b in B$, $bleq a$. Then $sup(B) leq a < sup(A)$, which is a contradiction.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    We know that $ ale sup A=sup B$, and because $sup A notin A$, $a< sup A=sup B$. In particular, because $sup B$ is the least upper bound of $B$, and $a$ is strictly less than it, there exists $bin B$ with $b>a$. If this were not the case, then $a$ would be a smaller lower bound for $B$, which is contradictory.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      oh. I get it now!
      $endgroup$
      – Wesley Strik
      Dec 2 '18 at 13:32



















    2












    $begingroup$

    Note that $a<sup(A)=sup(B)$, so $a$ is not an upper bound of $B$, because $sup(B)$ is the least upper bound of $B$. Therefore $a<b$, for some $bin B$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      The following argument utilizes the second characterization of the supremum found here:



      Suppose $a in A$. Since $sup(A) notin A$, we know that $sup(A)-a>0$. So we select $varepsilon=sup(A)-a$. Now we have that there exists $b in B$ such that
      begin{aligned} b&>sup(B)-varepsilon \& = sup(B)-left(sup(A)-aright) \& = sup(B)-sup(B)+a \& =a.
      end{aligned}






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This is a good tool, yes we had to prove this as an earlier exercise in the book. It's a little bit less used than " the minimum of all upper bounds" property, but very useful indeed. Thank you.
        $endgroup$
        – Wesley Strik
        Dec 2 '18 at 10:27
















      2












      $begingroup$

      The following argument utilizes the second characterization of the supremum found here:



      Suppose $a in A$. Since $sup(A) notin A$, we know that $sup(A)-a>0$. So we select $varepsilon=sup(A)-a$. Now we have that there exists $b in B$ such that
      begin{aligned} b&>sup(B)-varepsilon \& = sup(B)-left(sup(A)-aright) \& = sup(B)-sup(B)+a \& =a.
      end{aligned}






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This is a good tool, yes we had to prove this as an earlier exercise in the book. It's a little bit less used than " the minimum of all upper bounds" property, but very useful indeed. Thank you.
        $endgroup$
        – Wesley Strik
        Dec 2 '18 at 10:27














      2












      2








      2





      $begingroup$

      The following argument utilizes the second characterization of the supremum found here:



      Suppose $a in A$. Since $sup(A) notin A$, we know that $sup(A)-a>0$. So we select $varepsilon=sup(A)-a$. Now we have that there exists $b in B$ such that
      begin{aligned} b&>sup(B)-varepsilon \& = sup(B)-left(sup(A)-aright) \& = sup(B)-sup(B)+a \& =a.
      end{aligned}






      share|cite|improve this answer









      $endgroup$



      The following argument utilizes the second characterization of the supremum found here:



      Suppose $a in A$. Since $sup(A) notin A$, we know that $sup(A)-a>0$. So we select $varepsilon=sup(A)-a$. Now we have that there exists $b in B$ such that
      begin{aligned} b&>sup(B)-varepsilon \& = sup(B)-left(sup(A)-aright) \& = sup(B)-sup(B)+a \& =a.
      end{aligned}







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 2 '18 at 0:22









      Matt A PeltoMatt A Pelto

      2,502620




      2,502620








      • 1




        $begingroup$
        This is a good tool, yes we had to prove this as an earlier exercise in the book. It's a little bit less used than " the minimum of all upper bounds" property, but very useful indeed. Thank you.
        $endgroup$
        – Wesley Strik
        Dec 2 '18 at 10:27














      • 1




        $begingroup$
        This is a good tool, yes we had to prove this as an earlier exercise in the book. It's a little bit less used than " the minimum of all upper bounds" property, but very useful indeed. Thank you.
        $endgroup$
        – Wesley Strik
        Dec 2 '18 at 10:27








      1




      1




      $begingroup$
      This is a good tool, yes we had to prove this as an earlier exercise in the book. It's a little bit less used than " the minimum of all upper bounds" property, but very useful indeed. Thank you.
      $endgroup$
      – Wesley Strik
      Dec 2 '18 at 10:27




      $begingroup$
      This is a good tool, yes we had to prove this as an earlier exercise in the book. It's a little bit less used than " the minimum of all upper bounds" property, but very useful indeed. Thank you.
      $endgroup$
      – Wesley Strik
      Dec 2 '18 at 10:27











      3












      $begingroup$

      By contradiction. Assume $exists a in A$ such that $forall b in B$, $bleq a$. Then $sup(B) leq a < sup(A)$, which is a contradiction.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        By contradiction. Assume $exists a in A$ such that $forall b in B$, $bleq a$. Then $sup(B) leq a < sup(A)$, which is a contradiction.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          By contradiction. Assume $exists a in A$ such that $forall b in B$, $bleq a$. Then $sup(B) leq a < sup(A)$, which is a contradiction.






          share|cite|improve this answer









          $endgroup$



          By contradiction. Assume $exists a in A$ such that $forall b in B$, $bleq a$. Then $sup(B) leq a < sup(A)$, which is a contradiction.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 0:04









          mlerma54mlerma54

          1,177148




          1,177148























              3












              $begingroup$

              We know that $ ale sup A=sup B$, and because $sup A notin A$, $a< sup A=sup B$. In particular, because $sup B$ is the least upper bound of $B$, and $a$ is strictly less than it, there exists $bin B$ with $b>a$. If this were not the case, then $a$ would be a smaller lower bound for $B$, which is contradictory.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                oh. I get it now!
                $endgroup$
                – Wesley Strik
                Dec 2 '18 at 13:32
















              3












              $begingroup$

              We know that $ ale sup A=sup B$, and because $sup A notin A$, $a< sup A=sup B$. In particular, because $sup B$ is the least upper bound of $B$, and $a$ is strictly less than it, there exists $bin B$ with $b>a$. If this were not the case, then $a$ would be a smaller lower bound for $B$, which is contradictory.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                oh. I get it now!
                $endgroup$
                – Wesley Strik
                Dec 2 '18 at 13:32














              3












              3








              3





              $begingroup$

              We know that $ ale sup A=sup B$, and because $sup A notin A$, $a< sup A=sup B$. In particular, because $sup B$ is the least upper bound of $B$, and $a$ is strictly less than it, there exists $bin B$ with $b>a$. If this were not the case, then $a$ would be a smaller lower bound for $B$, which is contradictory.






              share|cite|improve this answer









              $endgroup$



              We know that $ ale sup A=sup B$, and because $sup A notin A$, $a< sup A=sup B$. In particular, because $sup B$ is the least upper bound of $B$, and $a$ is strictly less than it, there exists $bin B$ with $b>a$. If this were not the case, then $a$ would be a smaller lower bound for $B$, which is contradictory.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 2 '18 at 0:05









              Antonios-Alexandros RobotisAntonios-Alexandros Robotis

              9,90241640




              9,90241640












              • $begingroup$
                oh. I get it now!
                $endgroup$
                – Wesley Strik
                Dec 2 '18 at 13:32


















              • $begingroup$
                oh. I get it now!
                $endgroup$
                – Wesley Strik
                Dec 2 '18 at 13:32
















              $begingroup$
              oh. I get it now!
              $endgroup$
              – Wesley Strik
              Dec 2 '18 at 13:32




              $begingroup$
              oh. I get it now!
              $endgroup$
              – Wesley Strik
              Dec 2 '18 at 13:32











              2












              $begingroup$

              Note that $a<sup(A)=sup(B)$, so $a$ is not an upper bound of $B$, because $sup(B)$ is the least upper bound of $B$. Therefore $a<b$, for some $bin B$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note that $a<sup(A)=sup(B)$, so $a$ is not an upper bound of $B$, because $sup(B)$ is the least upper bound of $B$. Therefore $a<b$, for some $bin B$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note that $a<sup(A)=sup(B)$, so $a$ is not an upper bound of $B$, because $sup(B)$ is the least upper bound of $B$. Therefore $a<b$, for some $bin B$.






                  share|cite|improve this answer









                  $endgroup$



                  Note that $a<sup(A)=sup(B)$, so $a$ is not an upper bound of $B$, because $sup(B)$ is the least upper bound of $B$. Therefore $a<b$, for some $bin B$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 0:30









                  egregegreg

                  180k1485202




                  180k1485202






























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