Showing that $sup(Acdot B) = sup(A)cdotsup(B)$ for $A$ and $B$ subsets of non-negative reals












1












$begingroup$



While $A cdot B={x cdot y mid x in A, y in B}$, show that, for $A$, $B subseteq [0,infty)$,
$$sup(A cdot B)= sup(A) cdot sup(B)$$




My demonstration:



First:



$$forall ain A, alesup(A)$$
$$forall bin B, blesup(B)$$



so,
$$begin{align}
acdot blesup(A)cdotsup(B)
&implies
acdot b le sup(A cdot B)lesup(A)cdotsup(B) \
&implies
sup(Acdot B)lesup(A)cdotsup(B) tag{1}
end{align}$$



Second: Using $(1)$, while $bne0$,
$$a le frac{sup(Acdot B)}{b} implies sup(A)le frac{sup(Acdot B)}{b}$$



so
$$begin{align}
blefrac{sup(Acdot B)}{sup(A)}
&implies sup(B)le frac{sup(Acdot B)}{sup(A)} \
&implies sup(A)cdotsup(B)le sup(Acdot B) tag{2}
end{align}$$



Then $(1)$ and $(2)$ imply
$$sup(A cdot B)= sup(A) cdot sup(B) tag{3}$$




Is the demonstration correct? Can the case $b = 0$ be discarded for being trivial?











share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    While $A cdot B={x cdot y mid x in A, y in B}$, show that, for $A$, $B subseteq [0,infty)$,
    $$sup(A cdot B)= sup(A) cdot sup(B)$$




    My demonstration:



    First:



    $$forall ain A, alesup(A)$$
    $$forall bin B, blesup(B)$$



    so,
    $$begin{align}
    acdot blesup(A)cdotsup(B)
    &implies
    acdot b le sup(A cdot B)lesup(A)cdotsup(B) \
    &implies
    sup(Acdot B)lesup(A)cdotsup(B) tag{1}
    end{align}$$



    Second: Using $(1)$, while $bne0$,
    $$a le frac{sup(Acdot B)}{b} implies sup(A)le frac{sup(Acdot B)}{b}$$



    so
    $$begin{align}
    blefrac{sup(Acdot B)}{sup(A)}
    &implies sup(B)le frac{sup(Acdot B)}{sup(A)} \
    &implies sup(A)cdotsup(B)le sup(Acdot B) tag{2}
    end{align}$$



    Then $(1)$ and $(2)$ imply
    $$sup(A cdot B)= sup(A) cdot sup(B) tag{3}$$




    Is the demonstration correct? Can the case $b = 0$ be discarded for being trivial?











    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      While $A cdot B={x cdot y mid x in A, y in B}$, show that, for $A$, $B subseteq [0,infty)$,
      $$sup(A cdot B)= sup(A) cdot sup(B)$$




      My demonstration:



      First:



      $$forall ain A, alesup(A)$$
      $$forall bin B, blesup(B)$$



      so,
      $$begin{align}
      acdot blesup(A)cdotsup(B)
      &implies
      acdot b le sup(A cdot B)lesup(A)cdotsup(B) \
      &implies
      sup(Acdot B)lesup(A)cdotsup(B) tag{1}
      end{align}$$



      Second: Using $(1)$, while $bne0$,
      $$a le frac{sup(Acdot B)}{b} implies sup(A)le frac{sup(Acdot B)}{b}$$



      so
      $$begin{align}
      blefrac{sup(Acdot B)}{sup(A)}
      &implies sup(B)le frac{sup(Acdot B)}{sup(A)} \
      &implies sup(A)cdotsup(B)le sup(Acdot B) tag{2}
      end{align}$$



      Then $(1)$ and $(2)$ imply
      $$sup(A cdot B)= sup(A) cdot sup(B) tag{3}$$




      Is the demonstration correct? Can the case $b = 0$ be discarded for being trivial?











      share|cite|improve this question











      $endgroup$





      While $A cdot B={x cdot y mid x in A, y in B}$, show that, for $A$, $B subseteq [0,infty)$,
      $$sup(A cdot B)= sup(A) cdot sup(B)$$




      My demonstration:



      First:



      $$forall ain A, alesup(A)$$
      $$forall bin B, blesup(B)$$



      so,
      $$begin{align}
      acdot blesup(A)cdotsup(B)
      &implies
      acdot b le sup(A cdot B)lesup(A)cdotsup(B) \
      &implies
      sup(Acdot B)lesup(A)cdotsup(B) tag{1}
      end{align}$$



      Second: Using $(1)$, while $bne0$,
      $$a le frac{sup(Acdot B)}{b} implies sup(A)le frac{sup(Acdot B)}{b}$$



      so
      $$begin{align}
      blefrac{sup(Acdot B)}{sup(A)}
      &implies sup(B)le frac{sup(Acdot B)}{sup(A)} \
      &implies sup(A)cdotsup(B)le sup(Acdot B) tag{2}
      end{align}$$



      Then $(1)$ and $(2)$ imply
      $$sup(A cdot B)= sup(A) cdot sup(B) tag{3}$$




      Is the demonstration correct? Can the case $b = 0$ be discarded for being trivial?








      calculus proof-verification supremum-and-infimum






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      edited Jan 13 at 8:17









      Blue

      49.1k870156




      49.1k870156










      asked Jan 13 at 7:52









      Francisco SalazarFrancisco Salazar

      85




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          $begingroup$

          It is basically correct. But you should have started your proof by saying that the cases in which $A={0}$ or $B={0}$ are trivial and that you will assume that $A,Bneq{0}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks!!I will be more rigorous next time.
            $endgroup$
            – Francisco Salazar
            Jan 13 at 8:07











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          0












          $begingroup$

          It is basically correct. But you should have started your proof by saying that the cases in which $A={0}$ or $B={0}$ are trivial and that you will assume that $A,Bneq{0}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks!!I will be more rigorous next time.
            $endgroup$
            – Francisco Salazar
            Jan 13 at 8:07
















          0












          $begingroup$

          It is basically correct. But you should have started your proof by saying that the cases in which $A={0}$ or $B={0}$ are trivial and that you will assume that $A,Bneq{0}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks!!I will be more rigorous next time.
            $endgroup$
            – Francisco Salazar
            Jan 13 at 8:07














          0












          0








          0





          $begingroup$

          It is basically correct. But you should have started your proof by saying that the cases in which $A={0}$ or $B={0}$ are trivial and that you will assume that $A,Bneq{0}$.






          share|cite|improve this answer









          $endgroup$



          It is basically correct. But you should have started your proof by saying that the cases in which $A={0}$ or $B={0}$ are trivial and that you will assume that $A,Bneq{0}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 8:01









          José Carlos SantosJosé Carlos Santos

          168k23132236




          168k23132236












          • $begingroup$
            thanks!!I will be more rigorous next time.
            $endgroup$
            – Francisco Salazar
            Jan 13 at 8:07


















          • $begingroup$
            thanks!!I will be more rigorous next time.
            $endgroup$
            – Francisco Salazar
            Jan 13 at 8:07
















          $begingroup$
          thanks!!I will be more rigorous next time.
          $endgroup$
          – Francisco Salazar
          Jan 13 at 8:07




          $begingroup$
          thanks!!I will be more rigorous next time.
          $endgroup$
          – Francisco Salazar
          Jan 13 at 8:07


















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