Showing that $sup(Acdot B) = sup(A)cdotsup(B)$ for $A$ and $B$ subsets of non-negative reals
$begingroup$
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?
calculus proof-verification supremum-and-infimum
edited Jan 13 at 8:17
Blue
49.1k870156
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
$endgroup$
add a comment |
$begingroup$
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?
calculus proof-verification supremum-and-infimum
edited Jan 13 at 8:17
Blue
49.1k870156
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
$endgroup$
add a comment |
$begingroup$
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?
calculus proof-verification supremum-and-infimum
edited Jan 13 at 8:17
Blue
49.1k870156
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
$endgroup$
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
calculus proof-verification supremum-and-infimum
edited Jan 13 at 8:17
Blue
49.1k870156
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
edited Jan 13 at 8:17
Blue
49.1k870156
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
edited Jan 13 at 8:17
Blue
49.1k870156
edited Jan 13 at 8:17
Blue
49.1k870156
edited Jan 13 at 8:17
Blue
49.1k870156
49.1k870156
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
asked Jan 13 at 7:52
Francisco SalazarFrancisco Salazar
85
85
add a comment |
add a comment |
1 Answer
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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 $A,Bneq{0}$.
answered Jan 13 at 8:01
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
thanks!!I will be more rigorous next time.
$endgroup$
– Francisco Salazar
Jan 13 at 8:07
add a comment |
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1 Answer
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active
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1 Answer
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active
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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 $A,Bneq{0}$.
answered Jan 13 at 8:01
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
thanks!!I will be more rigorous next time.
$endgroup$
– Francisco Salazar
Jan 13 at 8:07
add a comment |
$begingroup$
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 $A,Bneq{0}$.
answered Jan 13 at 8:01
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
thanks!!I will be more rigorous next time.
$endgroup$
– Francisco Salazar
Jan 13 at 8:07
add a comment |
$begingroup$
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 $A,Bneq{0}$.
answered Jan 13 at 8:01
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
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 $A,Bneq{0}$.
answered Jan 13 at 8:01
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 13 at 8:01
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 13 at 8:01
José Carlos SantosJosé Carlos Santos
168k23132236
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
add a comment |
$begingroup$
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
add a comment |
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