Does $sup|A-B| ge |sup A -sup B|$?












1












$begingroup$


I have strong intuition that "yes" is the answer, because the difference between any two points between $A$ and $B$ can be only bigger (or equal) to the difference between two specific points. But I couldn't prove it.
Would like to see a prove, or a counter example of course.






supremum-and-infimum







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What are $A$ and $B$?
    $endgroup$
    – gt6989b
    Dec 25 '18 at 15:18






  • 2




    $begingroup$
    If they are sets, which I assume they are, how do you even calculate $left| A - B right|$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 25 '18 at 15:19










  • $begingroup$
    sup|A-B| is well-defined also when A,B are sets, which they can be (or finite groups as the given example below) @AniruddhaDeshmukh
    $endgroup$
    – Daniel
    Dec 25 '18 at 17:50
















1












$begingroup$


I have strong intuition that "yes" is the answer, because the difference between any two points between $A$ and $B$ can be only bigger (or equal) to the difference between two specific points. But I couldn't prove it.
Would like to see a prove, or a counter example of course.






supremum-and-infimum







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What are $A$ and $B$?
    $endgroup$
    – gt6989b
    Dec 25 '18 at 15:18






  • 2




    $begingroup$
    If they are sets, which I assume they are, how do you even calculate $left| A - B right|$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 25 '18 at 15:19










  • $begingroup$
    sup|A-B| is well-defined also when A,B are sets, which they can be (or finite groups as the given example below) @AniruddhaDeshmukh
    $endgroup$
    – Daniel
    Dec 25 '18 at 17:50














1












1








1


0



$begingroup$


I have strong intuition that "yes" is the answer, because the difference between any two points between $A$ and $B$ can be only bigger (or equal) to the difference between two specific points. But I couldn't prove it.
Would like to see a prove, or a counter example of course.






supremum-and-infimum







share|cite|improve this question











$endgroup$




I have strong intuition that "yes" is the answer, because the difference between any two points between $A$ and $B$ can be only bigger (or equal) to the difference between two specific points. But I couldn't prove it.
Would like to see a prove, or a counter example of course.






supremum-and-infimum




supremum-and-infimum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 15:28









mechanodroid

27.5k62447




27.5k62447










asked Dec 25 '18 at 15:09









DanielDaniel

268




268








  • 1




    $begingroup$
    What are $A$ and $B$?
    $endgroup$
    – gt6989b
    Dec 25 '18 at 15:18






  • 2




    $begingroup$
    If they are sets, which I assume they are, how do you even calculate $left| A - B right|$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 25 '18 at 15:19










  • $begingroup$
    sup|A-B| is well-defined also when A,B are sets, which they can be (or finite groups as the given example below) @AniruddhaDeshmukh
    $endgroup$
    – Daniel
    Dec 25 '18 at 17:50














  • 1




    $begingroup$
    What are $A$ and $B$?
    $endgroup$
    – gt6989b
    Dec 25 '18 at 15:18






  • 2




    $begingroup$
    If they are sets, which I assume they are, how do you even calculate $left| A - B right|$?
    $endgroup$
    – Aniruddha Deshmukh
    Dec 25 '18 at 15:19










  • $begingroup$
    sup|A-B| is well-defined also when A,B are sets, which they can be (or finite groups as the given example below) @AniruddhaDeshmukh
    $endgroup$
    – Daniel
    Dec 25 '18 at 17:50








1




1




$begingroup$
What are $A$ and $B$?
$endgroup$
– gt6989b
Dec 25 '18 at 15:18




$begingroup$
What are $A$ and $B$?
$endgroup$
– gt6989b
Dec 25 '18 at 15:18




2




2




$begingroup$
If they are sets, which I assume they are, how do you even calculate $left| A - B right|$?
$endgroup$
– Aniruddha Deshmukh
Dec 25 '18 at 15:19




$begingroup$
If they are sets, which I assume they are, how do you even calculate $left| A - B right|$?
$endgroup$
– Aniruddha Deshmukh
Dec 25 '18 at 15:19












$begingroup$
sup|A-B| is well-defined also when A,B are sets, which they can be (or finite groups as the given example below) @AniruddhaDeshmukh
$endgroup$
– Daniel
Dec 25 '18 at 17:50




$begingroup$
sup|A-B| is well-defined also when A,B are sets, which they can be (or finite groups as the given example below) @AniruddhaDeshmukh
$endgroup$
– Daniel
Dec 25 '18 at 17:50










2 Answers
2






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$begingroup$

The answer is yes, and your intuition is correct.



Let $varepsilon > 0$ and find $a in A$ such that $sup A - a < varepsilon$.



For any $b in B$ we have
$$sup|A-B| ge |a-b| ge a-b > (sup A - varepsilon) - sup B ge (sup A - sup B) - varepsilon$$
since $b le sup B$. Now $varepsilon$ was arbitrary so it follows $sup|A-B| ge sup A - sup B$.



The inequality $sup|A-B| ge sup B - sup A$ follows by symmetry.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Yes. Find $ain A$ and $bin B$ to $epsilon$-approximate $sup A$ and $sup B$. Then $|a-b|$ will approximate $|sup A-sup B|$, where $sup|A-B|geq|a-b|$. Now let $epsilonto 0$.



    The reverse inequality does not hold like you had originally also wanted. Let $A={-1,0}$ and $B={0,1}$ so that $sup|A-B|=2$, whereas $sup A=0$ and $sup B=1$ so that $|sup A-sup B|=1$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      2












      $begingroup$

      The answer is yes, and your intuition is correct.



      Let $varepsilon > 0$ and find $a in A$ such that $sup A - a < varepsilon$.



      For any $b in B$ we have
      $$sup|A-B| ge |a-b| ge a-b > (sup A - varepsilon) - sup B ge (sup A - sup B) - varepsilon$$
      since $b le sup B$. Now $varepsilon$ was arbitrary so it follows $sup|A-B| ge sup A - sup B$.



      The inequality $sup|A-B| ge sup B - sup A$ follows by symmetry.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        The answer is yes, and your intuition is correct.



        Let $varepsilon > 0$ and find $a in A$ such that $sup A - a < varepsilon$.



        For any $b in B$ we have
        $$sup|A-B| ge |a-b| ge a-b > (sup A - varepsilon) - sup B ge (sup A - sup B) - varepsilon$$
        since $b le sup B$. Now $varepsilon$ was arbitrary so it follows $sup|A-B| ge sup A - sup B$.



        The inequality $sup|A-B| ge sup B - sup A$ follows by symmetry.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          The answer is yes, and your intuition is correct.



          Let $varepsilon > 0$ and find $a in A$ such that $sup A - a < varepsilon$.



          For any $b in B$ we have
          $$sup|A-B| ge |a-b| ge a-b > (sup A - varepsilon) - sup B ge (sup A - sup B) - varepsilon$$
          since $b le sup B$. Now $varepsilon$ was arbitrary so it follows $sup|A-B| ge sup A - sup B$.



          The inequality $sup|A-B| ge sup B - sup A$ follows by symmetry.






          share|cite|improve this answer











          $endgroup$



          The answer is yes, and your intuition is correct.



          Let $varepsilon > 0$ and find $a in A$ such that $sup A - a < varepsilon$.



          For any $b in B$ we have
          $$sup|A-B| ge |a-b| ge a-b > (sup A - varepsilon) - sup B ge (sup A - sup B) - varepsilon$$
          since $b le sup B$. Now $varepsilon$ was arbitrary so it follows $sup|A-B| ge sup A - sup B$.



          The inequality $sup|A-B| ge sup B - sup A$ follows by symmetry.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 17:03

























          answered Dec 25 '18 at 15:26









          mechanodroidmechanodroid

          27.5k62447




          27.5k62447























              0












              $begingroup$

              Yes. Find $ain A$ and $bin B$ to $epsilon$-approximate $sup A$ and $sup B$. Then $|a-b|$ will approximate $|sup A-sup B|$, where $sup|A-B|geq|a-b|$. Now let $epsilonto 0$.



              The reverse inequality does not hold like you had originally also wanted. Let $A={-1,0}$ and $B={0,1}$ so that $sup|A-B|=2$, whereas $sup A=0$ and $sup B=1$ so that $|sup A-sup B|=1$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Yes. Find $ain A$ and $bin B$ to $epsilon$-approximate $sup A$ and $sup B$. Then $|a-b|$ will approximate $|sup A-sup B|$, where $sup|A-B|geq|a-b|$. Now let $epsilonto 0$.



                The reverse inequality does not hold like you had originally also wanted. Let $A={-1,0}$ and $B={0,1}$ so that $sup|A-B|=2$, whereas $sup A=0$ and $sup B=1$ so that $|sup A-sup B|=1$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Yes. Find $ain A$ and $bin B$ to $epsilon$-approximate $sup A$ and $sup B$. Then $|a-b|$ will approximate $|sup A-sup B|$, where $sup|A-B|geq|a-b|$. Now let $epsilonto 0$.



                  The reverse inequality does not hold like you had originally also wanted. Let $A={-1,0}$ and $B={0,1}$ so that $sup|A-B|=2$, whereas $sup A=0$ and $sup B=1$ so that $|sup A-sup B|=1$.






                  share|cite|improve this answer









                  $endgroup$



                  Yes. Find $ain A$ and $bin B$ to $epsilon$-approximate $sup A$ and $sup B$. Then $|a-b|$ will approximate $|sup A-sup B|$, where $sup|A-B|geq|a-b|$. Now let $epsilonto 0$.



                  The reverse inequality does not hold like you had originally also wanted. Let $A={-1,0}$ and $B={0,1}$ so that $sup|A-B|=2$, whereas $sup A=0$ and $sup B=1$ so that $|sup A-sup B|=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 15:21









                  Ben WBen W

                  2,276615




                  2,276615






























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