Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$.












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Given : $−S = {−s : s in S}.$



Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$. And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$.






supremum-and-infimum







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$endgroup$








  • 2




    $begingroup$
    This is just an application of $x le y iff -x ge -y$. Where did you get stuck?
    $endgroup$
    – Theo Bendit
    Oct 10 '18 at 4:09
















-1












$begingroup$


Given : $−S = {−s : s in S}.$



Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$. And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$.






supremum-and-infimum







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This is just an application of $x le y iff -x ge -y$. Where did you get stuck?
    $endgroup$
    – Theo Bendit
    Oct 10 '18 at 4:09














-1












-1








-1





$begingroup$


Given : $−S = {−s : s in S}.$



Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$. And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$.






supremum-and-infimum







share|cite|improve this question











$endgroup$




Given : $−S = {−s : s in S}.$



Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$. And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$.






supremum-and-infimum




supremum-and-infimum






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edited Dec 23 '18 at 22:05









Eevee Trainer

5,8421936




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asked Oct 10 '18 at 4:05









AriesAries

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  • 2




    $begingroup$
    This is just an application of $x le y iff -x ge -y$. Where did you get stuck?
    $endgroup$
    – Theo Bendit
    Oct 10 '18 at 4:09














  • 2




    $begingroup$
    This is just an application of $x le y iff -x ge -y$. Where did you get stuck?
    $endgroup$
    – Theo Bendit
    Oct 10 '18 at 4:09








2




2




$begingroup$
This is just an application of $x le y iff -x ge -y$. Where did you get stuck?
$endgroup$
– Theo Bendit
Oct 10 '18 at 4:09




$begingroup$
This is just an application of $x le y iff -x ge -y$. Where did you get stuck?
$endgroup$
– Theo Bendit
Oct 10 '18 at 4:09










1 Answer
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2












$begingroup$

Since these proofs are pretty similar to each other, I'll complete one and leave the rest incomplete, with the first proof serving as a hint.





Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$.




To expound upon what was noted by Theo Bendit in the comments, we essentially use the definition of upper/lower bounds and the fact $x leq y Leftrightarrow -x geq -y$ to complete the proof.




Proof ($Rightarrow$):



Suppose $m$ is a lower bound for $S$. Then, for all $s in S$, by definition of lower bound, $m leq s$. Then, as a result, $-m geq -s$ for all $s in S$.



Recall the set $-S$ is defined by $-S = { -s | s in S}$.



Since $-m$ meets that condition - that it is greater than or equal to all elements of $-S$ - then $-m$ is by definition an upper bound for the set $-S$.




Proof ($Leftarrow$):



The converse -- that $-m$ being an upper bound for $-S$ implies $m$ is a lower bound for $S$ -- follows very similar logic to the previous, and thus is an exercise left to the reader. :)





And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$




This one mirrors the previous in that it basically exploits $x leq y Leftrightarrow -x geq -y$ and the definitions of supremum and infimum. So I'll again leave this proof incomplete, just with the hints that:





  • $inf S$ is the element which is the greatest of the lower bounds.


  • $sup S$ is the element which is the least of the upper bounds.

  • To show that an element is either, you not only want to show it is an upper/lower bound by the definition, but also that it is the least/greatest such bound respectively. How you choose to do this is up to you: supposing not, and then finding a contradiction, is how I'd go about it.


Thus, you want to show $inf S$ is not only an upper bound of $-S$ (hint: use the previous proof) and also show it is the least upper bound.






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    $begingroup$

    Since these proofs are pretty similar to each other, I'll complete one and leave the rest incomplete, with the first proof serving as a hint.





    Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$.




    To expound upon what was noted by Theo Bendit in the comments, we essentially use the definition of upper/lower bounds and the fact $x leq y Leftrightarrow -x geq -y$ to complete the proof.




    Proof ($Rightarrow$):



    Suppose $m$ is a lower bound for $S$. Then, for all $s in S$, by definition of lower bound, $m leq s$. Then, as a result, $-m geq -s$ for all $s in S$.



    Recall the set $-S$ is defined by $-S = { -s | s in S}$.



    Since $-m$ meets that condition - that it is greater than or equal to all elements of $-S$ - then $-m$ is by definition an upper bound for the set $-S$.




    Proof ($Leftarrow$):



    The converse -- that $-m$ being an upper bound for $-S$ implies $m$ is a lower bound for $S$ -- follows very similar logic to the previous, and thus is an exercise left to the reader. :)





    And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$




    This one mirrors the previous in that it basically exploits $x leq y Leftrightarrow -x geq -y$ and the definitions of supremum and infimum. So I'll again leave this proof incomplete, just with the hints that:





    • $inf S$ is the element which is the greatest of the lower bounds.


    • $sup S$ is the element which is the least of the upper bounds.

    • To show that an element is either, you not only want to show it is an upper/lower bound by the definition, but also that it is the least/greatest such bound respectively. How you choose to do this is up to you: supposing not, and then finding a contradiction, is how I'd go about it.


    Thus, you want to show $inf S$ is not only an upper bound of $-S$ (hint: use the previous proof) and also show it is the least upper bound.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Since these proofs are pretty similar to each other, I'll complete one and leave the rest incomplete, with the first proof serving as a hint.





      Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$.




      To expound upon what was noted by Theo Bendit in the comments, we essentially use the definition of upper/lower bounds and the fact $x leq y Leftrightarrow -x geq -y$ to complete the proof.




      Proof ($Rightarrow$):



      Suppose $m$ is a lower bound for $S$. Then, for all $s in S$, by definition of lower bound, $m leq s$. Then, as a result, $-m geq -s$ for all $s in S$.



      Recall the set $-S$ is defined by $-S = { -s | s in S}$.



      Since $-m$ meets that condition - that it is greater than or equal to all elements of $-S$ - then $-m$ is by definition an upper bound for the set $-S$.




      Proof ($Leftarrow$):



      The converse -- that $-m$ being an upper bound for $-S$ implies $m$ is a lower bound for $S$ -- follows very similar logic to the previous, and thus is an exercise left to the reader. :)





      And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$




      This one mirrors the previous in that it basically exploits $x leq y Leftrightarrow -x geq -y$ and the definitions of supremum and infimum. So I'll again leave this proof incomplete, just with the hints that:





      • $inf S$ is the element which is the greatest of the lower bounds.


      • $sup S$ is the element which is the least of the upper bounds.

      • To show that an element is either, you not only want to show it is an upper/lower bound by the definition, but also that it is the least/greatest such bound respectively. How you choose to do this is up to you: supposing not, and then finding a contradiction, is how I'd go about it.


      Thus, you want to show $inf S$ is not only an upper bound of $-S$ (hint: use the previous proof) and also show it is the least upper bound.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Since these proofs are pretty similar to each other, I'll complete one and leave the rest incomplete, with the first proof serving as a hint.





        Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$.




        To expound upon what was noted by Theo Bendit in the comments, we essentially use the definition of upper/lower bounds and the fact $x leq y Leftrightarrow -x geq -y$ to complete the proof.




        Proof ($Rightarrow$):



        Suppose $m$ is a lower bound for $S$. Then, for all $s in S$, by definition of lower bound, $m leq s$. Then, as a result, $-m geq -s$ for all $s in S$.



        Recall the set $-S$ is defined by $-S = { -s | s in S}$.



        Since $-m$ meets that condition - that it is greater than or equal to all elements of $-S$ - then $-m$ is by definition an upper bound for the set $-S$.




        Proof ($Leftarrow$):



        The converse -- that $-m$ being an upper bound for $-S$ implies $m$ is a lower bound for $S$ -- follows very similar logic to the previous, and thus is an exercise left to the reader. :)





        And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$




        This one mirrors the previous in that it basically exploits $x leq y Leftrightarrow -x geq -y$ and the definitions of supremum and infimum. So I'll again leave this proof incomplete, just with the hints that:





        • $inf S$ is the element which is the greatest of the lower bounds.


        • $sup S$ is the element which is the least of the upper bounds.

        • To show that an element is either, you not only want to show it is an upper/lower bound by the definition, but also that it is the least/greatest such bound respectively. How you choose to do this is up to you: supposing not, and then finding a contradiction, is how I'd go about it.


        Thus, you want to show $inf S$ is not only an upper bound of $-S$ (hint: use the previous proof) and also show it is the least upper bound.






        share|cite|improve this answer









        $endgroup$



        Since these proofs are pretty similar to each other, I'll complete one and leave the rest incomplete, with the first proof serving as a hint.





        Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$.




        To expound upon what was noted by Theo Bendit in the comments, we essentially use the definition of upper/lower bounds and the fact $x leq y Leftrightarrow -x geq -y$ to complete the proof.




        Proof ($Rightarrow$):



        Suppose $m$ is a lower bound for $S$. Then, for all $s in S$, by definition of lower bound, $m leq s$. Then, as a result, $-m geq -s$ for all $s in S$.



        Recall the set $-S$ is defined by $-S = { -s | s in S}$.



        Since $-m$ meets that condition - that it is greater than or equal to all elements of $-S$ - then $-m$ is by definition an upper bound for the set $-S$.




        Proof ($Leftarrow$):



        The converse -- that $-m$ being an upper bound for $-S$ implies $m$ is a lower bound for $S$ -- follows very similar logic to the previous, and thus is an exercise left to the reader. :)





        And prove that if $S$ is bounded below then its greatest lower bound satisfies $inf S = − sup(−S)$




        This one mirrors the previous in that it basically exploits $x leq y Leftrightarrow -x geq -y$ and the definitions of supremum and infimum. So I'll again leave this proof incomplete, just with the hints that:





        • $inf S$ is the element which is the greatest of the lower bounds.


        • $sup S$ is the element which is the least of the upper bounds.

        • To show that an element is either, you not only want to show it is an upper/lower bound by the definition, but also that it is the least/greatest such bound respectively. How you choose to do this is up to you: supposing not, and then finding a contradiction, is how I'd go about it.


        Thus, you want to show $inf S$ is not only an upper bound of $-S$ (hint: use the previous proof) and also show it is the least upper bound.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 22:22









        Eevee TrainerEevee Trainer

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