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Is subtracting inequalities allowed? $sup(A+B)= sup(A)+sup(B)$ - proof verification
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Given sets $A$ and $B$ define $A+B = a+b:a in A and b in B $. If these sets are nonempty and bounded above show $ sup(A+B)= sup(A)+sup(B)$.
My Attempt:
Let $sup A= alpha$ and $sup B= beta$. This means that for all $a in A$, $a le alpha$ and for all $b in B$, $b le beta$. Thus adding the inequalities it follows $a+b le alpha + beta$ and hence $alpha + beta$ is an upperbound for $A+B$.
Let $gamma $ be an upper bound for $A+B$. Thus $a+b= gamma$. It follows that $a + beta le gamma$. Thus (1) $beta le gamma -a.$ Similarly $alpha + b le gamma .$ Thus (2) $alpha le gamma -b$.Adding (1) and (2) we get $$alpha + betale 2gamma -a -b$$
$Rightarrow$ $$alpha + beta + a +b le 2gamma$$
Is the following valid: I took $a+b le gamma $ since $gamma$ is an upper bound for $A+B$ and subtracted this from the inequlaity above to get $alpha + betale gamma$. And thus proving the theorem. Only problem is that if take $-gamma le -a-b$ (equivalent to $a+b le gamma $) and add it to the inequality above we get $$alpha + beta + a +b -gamma le 2gamma -a -b$$ which I was told by my tutor is wrong. If this wrong why so?? If so, why isn't subtracting inequalities allowed?
real-analysis proof-verification inequality supremum-and-infimum upper-lower-bounds
edited Jul 21 at 15:33
Jneven
510218
asked Jul 21 at 14:34
Red
1,747733
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up vote
2
down vote
favorite
Given sets $A$ and $B$ define $A+B = a+b:a in A and b in B $. If these sets are nonempty and bounded above show $ sup(A+B)= sup(A)+sup(B)$.
My Attempt:
Let $sup A= alpha$ and $sup B= beta$. This means that for all $a in A$, $a le alpha$ and for all $b in B$, $b le beta$. Thus adding the inequalities it follows $a+b le alpha + beta$ and hence $alpha + beta$ is an upperbound for $A+B$.
Let $gamma $ be an upper bound for $A+B$. Thus $a+b= gamma$. It follows that $a + beta le gamma$. Thus (1) $beta le gamma -a.$ Similarly $alpha + b le gamma .$ Thus (2) $alpha le gamma -b$.Adding (1) and (2) we get $$alpha + betale 2gamma -a -b$$
$Rightarrow$ $$alpha + beta + a +b le 2gamma$$
Is the following valid: I took $a+b le gamma $ since $gamma$ is an upper bound for $A+B$ and subtracted this from the inequlaity above to get $alpha + betale gamma$. And thus proving the theorem. Only problem is that if take $-gamma le -a-b$ (equivalent to $a+b le gamma $) and add it to the inequality above we get $$alpha + beta + a +b -gamma le 2gamma -a -b$$ which I was told by my tutor is wrong. If this wrong why so?? If so, why isn't subtracting inequalities allowed?
real-analysis proof-verification inequality supremum-and-infimum upper-lower-bounds
edited Jul 21 at 15:33
Jneven
510218
asked Jul 21 at 14:34
Red
1,747733
1
$10 > 0$, $20> 0$ but is $10-20 > 0 - 0$?
– NickD
Jul 21 at 14:49
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up vote
2
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up vote
2
down vote
favorite
Given sets $A$ and $B$ define $A+B = a+b:a in A and b in B $. If these sets are nonempty and bounded above show $ sup(A+B)= sup(A)+sup(B)$.
My Attempt:
Let $sup A= alpha$ and $sup B= beta$. This means that for all $a in A$, $a le alpha$ and for all $b in B$, $b le beta$. Thus adding the inequalities it follows $a+b le alpha + beta$ and hence $alpha + beta$ is an upperbound for $A+B$.
Let $gamma $ be an upper bound for $A+B$. Thus $a+b= gamma$. It follows that $a + beta le gamma$. Thus (1) $beta le gamma -a.$ Similarly $alpha + b le gamma .$ Thus (2) $alpha le gamma -b$.Adding (1) and (2) we get $$alpha + betale 2gamma -a -b$$
$Rightarrow$ $$alpha + beta + a +b le 2gamma$$
Is the following valid: I took $a+b le gamma $ since $gamma$ is an upper bound for $A+B$ and subtracted this from the inequlaity above to get $alpha + betale gamma$. And thus proving the theorem. Only problem is that if take $-gamma le -a-b$ (equivalent to $a+b le gamma $) and add it to the inequality above we get $$alpha + beta + a +b -gamma le 2gamma -a -b$$ which I was told by my tutor is wrong. If this wrong why so?? If so, why isn't subtracting inequalities allowed?
real-analysis proof-verification inequality supremum-and-infimum upper-lower-bounds
edited Jul 21 at 15:33
Jneven
510218
asked Jul 21 at 14:34
Red
1,747733
Given sets $A$ and $B$ define $A+B = a+b:a in A and b in B $. If these sets are nonempty and bounded above show $ sup(A+B)= sup(A)+sup(B)$.
My Attempt:
Let $sup A= alpha$ and $sup B= beta$. This means that for all $a in A$, $a le alpha$ and for all $b in B$, $b le beta$. Thus adding the inequalities it follows $a+b le alpha + beta$ and hence $alpha + beta$ is an upperbound for $A+B$.
Let $gamma $ be an upper bound for $A+B$. Thus $a+b= gamma$. It follows that $a + beta le gamma$. Thus (1) $beta le gamma -a.$ Similarly $alpha + b le gamma .$ Thus (2) $alpha le gamma -b$.Adding (1) and (2) we get $$alpha + betale 2gamma -a -b$$
$Rightarrow$ $$alpha + beta + a +b le 2gamma$$
Is the following valid: I took $a+b le gamma $ since $gamma$ is an upper bound for $A+B$ and subtracted this from the inequlaity above to get $alpha + betale gamma$. And thus proving the theorem. Only problem is that if take $-gamma le -a-b$ (equivalent to $a+b le gamma $) and add it to the inequality above we get $$alpha + beta + a +b -gamma le 2gamma -a -b$$ which I was told by my tutor is wrong. If this wrong why so?? If so, why isn't subtracting inequalities allowed?
real-analysis proof-verification inequality supremum-and-infimum upper-lower-bounds
edited Jul 21 at 15:33
Jneven
510218
asked Jul 21 at 14:34
Red
1,747733
edited Jul 21 at 15:33
Jneven
510218
edited Jul 21 at 15:33
Jneven
510218
edited Jul 21 at 15:33
Jneven
510218
510218
asked Jul 21 at 14:34
Red
1,747733
asked Jul 21 at 14:34
Red
1,747733
asked Jul 21 at 14:34
Red
1,747733
1,747733
1
$10 > 0$, $20> 0$ but is $10-20 > 0 - 0$?
– NickD
Jul 21 at 14:49
add a comment |Â
1
$10 > 0$, $20> 0$ but is $10-20 > 0 - 0$?
– NickD
Jul 21 at 14:49
1
1
$10 > 0$, $20> 0$ but is $10-20 > 0 - 0$?
– NickD
Jul 21 at 14:49
$10 > 0$, $20> 0$ but is $10-20 > 0 - 0$?
– NickD
Jul 21 at 14:49
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
We have e.g. $3<4$ and $1<2$, but subtracting would give $3-1 < 4-2 implies 2<2$, which is wrong. In general, note that if $p<q$, we can write this as $p = q - delta$ for some positive $delta$, and similarly $r<s implies r = s - epsilon$ for some positive $epsilon$. Subtracting would give $p - r < q - s iff q - delta - s + epsilon < q - s iff epsilon < delta$, which may or may not be true. In contrast, adding inequalities is allowed because $p + r < q + s iff q - delta + s - epsilon < q + s iff delta + epsilon > 0$, which is true as both $delta$ and $epsilon$ are positive.
In your last paragraph, your inequality is correct but totally useless for the proof, so it's "wrong" in that sense. Similarly if we take an example like $3<5$ and $1<2$, where we want to prove $3 - 1 < 5 - 2$, i.e. $2 < 3$, we can add $3<5$ and $-2<-1$ to get $1<4$, but this is a weaker inequality than the one we want to prove.
Considering your whole proof attempt, your first step of showing that $sup(A) + sup(B)$ (i.e. $alpha + beta$ in your notation) is an upper bound for $A + B$ is useful. Here's a hint for what to do next: this upper bound condition can be rewritten as $A + B leq sup(A) + sup(B)$, so whatever $sup(A+B)$ is, it must be less than or equal to this upper bound on $A + B$, i.e. we have $sup(A + B) leq sup(A) + sup(B)$. Now if we can also show that $sup(A+B) geq sup(A) + sup(B)$, the combination of the two inequalities will give $sup(A+B) = sup(A) + sup(B)$, as required. In order to show $sup(A+B) geq sup(A) + sup(B)$, you should first prove that $sup(A+B)$ is an upper bound for $A + sup(B)$, so $sup(A + sup(B)) leq sup(A+B)$, or writing it the other way round, $sup(A+B) geq sup(A + sup(B))$. Finally, prove that $sup(A + sup(B)) equiv sup(A+B)$ in order to reach the desired result.
answered Jul 21 at 14:59
Prasiortle
463
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up vote
1
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It is wrong. It is not true that an upper bound $gamma$ of $A+B$ can be written as $a+b$ with $ain A$ and $bin B$. Suppose, for instance, that $A=B=0$ and that $gamma=10$.
However, you proved correctly that $sup A+sup B$ is an upper bound of $A+B$. Therefore, $sup(A+B)leqslantsup A+sup B$. What would happen if we had $sup(A+B)<sup A+sup B$? In that case, take $varepsilon=sup A+sup B-sup(A+B)$, take $ain A$ such that $a>sup(A)-fracvarepsilon2$ and take $bin B$ such that $b>sup(B)-fracvarepsilon2$. What does this tell you about $a+b$?
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
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up vote
0
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The example by NickD illustrates the problem with the present proof.
A quick fix: suppose that $gammageq a+b$ for all $ain A$ and $bin B$. We want to show that $gammageqalpha+beta$.
So suppose otherwise that $gamma=alpha+beta-e$ for some $e>0$. Now $alpha-e/2$ is strictly smaller than $alpha$ so there is some $a^*in A$ such that $a^*>alpha-e/2$. Similarly, there is some $b^*in B$ such that $b^*>beta-e/2$. But then:
$$
a^*+b^*in A+Bquadtextandquad a^*+b^*>(alpha-e/2)+(beta-e/2)=gamma.
$$
This is your desired contradiction.
answered Jul 21 at 14:56
yurnero
6,8401824
The example illustrated one error in the proof. It wasn't the first error. This answer quietly avoids the earlier errors, however.
– David K
Jul 21 at 15:27
add a comment |Â
up vote
0
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Subtracting inequalities is wrong, for example if you subtract $-5<2$ from $4<10$ you get $9<8$
For your proof of $$ sup(A+B)= sup(A)+sup(B)$$ where you have already shown that $$ sup(A+B)le sup(A)+sup(B)$$
you need to show that for any number $gamma $ less than $ sup(A)+sup(B)$ you can find an element in A+B which is
greater than $gamma$
Note that $$gamma= sup(A)+sup(B)-epsilon = (sup(A) - epsilon /2) + (sup(B) - epsilon /2)$$
Now you can easily find elements $ain A$ and $bin B$ such that $a+b >gamma$
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have e.g. $3<4$ and $1<2$, but subtracting would give $3-1 < 4-2 implies 2<2$, which is wrong. In general, note that if $p<q$, we can write this as $p = q - delta$ for some positive $delta$, and similarly $r<s implies r = s - epsilon$ for some positive $epsilon$. Subtracting would give $p - r < q - s iff q - delta - s + epsilon < q - s iff epsilon < delta$, which may or may not be true. In contrast, adding inequalities is allowed because $p + r < q + s iff q - delta + s - epsilon < q + s iff delta + epsilon > 0$, which is true as both $delta$ and $epsilon$ are positive.
In your last paragraph, your inequality is correct but totally useless for the proof, so it's "wrong" in that sense. Similarly if we take an example like $3<5$ and $1<2$, where we want to prove $3 - 1 < 5 - 2$, i.e. $2 < 3$, we can add $3<5$ and $-2<-1$ to get $1<4$, but this is a weaker inequality than the one we want to prove.
Considering your whole proof attempt, your first step of showing that $sup(A) + sup(B)$ (i.e. $alpha + beta$ in your notation) is an upper bound for $A + B$ is useful. Here's a hint for what to do next: this upper bound condition can be rewritten as $A + B leq sup(A) + sup(B)$, so whatever $sup(A+B)$ is, it must be less than or equal to this upper bound on $A + B$, i.e. we have $sup(A + B) leq sup(A) + sup(B)$. Now if we can also show that $sup(A+B) geq sup(A) + sup(B)$, the combination of the two inequalities will give $sup(A+B) = sup(A) + sup(B)$, as required. In order to show $sup(A+B) geq sup(A) + sup(B)$, you should first prove that $sup(A+B)$ is an upper bound for $A + sup(B)$, so $sup(A + sup(B)) leq sup(A+B)$, or writing it the other way round, $sup(A+B) geq sup(A + sup(B))$. Finally, prove that $sup(A + sup(B)) equiv sup(A+B)$ in order to reach the desired result.
answered Jul 21 at 14:59
Prasiortle
463
add a comment |Â
up vote
1
down vote
accepted
We have e.g. $3<4$ and $1<2$, but subtracting would give $3-1 < 4-2 implies 2<2$, which is wrong. In general, note that if $p<q$, we can write this as $p = q - delta$ for some positive $delta$, and similarly $r<s implies r = s - epsilon$ for some positive $epsilon$. Subtracting would give $p - r < q - s iff q - delta - s + epsilon < q - s iff epsilon < delta$, which may or may not be true. In contrast, adding inequalities is allowed because $p + r < q + s iff q - delta + s - epsilon < q + s iff delta + epsilon > 0$, which is true as both $delta$ and $epsilon$ are positive.
In your last paragraph, your inequality is correct but totally useless for the proof, so it's "wrong" in that sense. Similarly if we take an example like $3<5$ and $1<2$, where we want to prove $3 - 1 < 5 - 2$, i.e. $2 < 3$, we can add $3<5$ and $-2<-1$ to get $1<4$, but this is a weaker inequality than the one we want to prove.
Considering your whole proof attempt, your first step of showing that $sup(A) + sup(B)$ (i.e. $alpha + beta$ in your notation) is an upper bound for $A + B$ is useful. Here's a hint for what to do next: this upper bound condition can be rewritten as $A + B leq sup(A) + sup(B)$, so whatever $sup(A+B)$ is, it must be less than or equal to this upper bound on $A + B$, i.e. we have $sup(A + B) leq sup(A) + sup(B)$. Now if we can also show that $sup(A+B) geq sup(A) + sup(B)$, the combination of the two inequalities will give $sup(A+B) = sup(A) + sup(B)$, as required. In order to show $sup(A+B) geq sup(A) + sup(B)$, you should first prove that $sup(A+B)$ is an upper bound for $A + sup(B)$, so $sup(A + sup(B)) leq sup(A+B)$, or writing it the other way round, $sup(A+B) geq sup(A + sup(B))$. Finally, prove that $sup(A + sup(B)) equiv sup(A+B)$ in order to reach the desired result.
answered Jul 21 at 14:59
Prasiortle
463
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have e.g. $3<4$ and $1<2$, but subtracting would give $3-1 < 4-2 implies 2<2$, which is wrong. In general, note that if $p<q$, we can write this as $p = q - delta$ for some positive $delta$, and similarly $r<s implies r = s - epsilon$ for some positive $epsilon$. Subtracting would give $p - r < q - s iff q - delta - s + epsilon < q - s iff epsilon < delta$, which may or may not be true. In contrast, adding inequalities is allowed because $p + r < q + s iff q - delta + s - epsilon < q + s iff delta + epsilon > 0$, which is true as both $delta$ and $epsilon$ are positive.
In your last paragraph, your inequality is correct but totally useless for the proof, so it's "wrong" in that sense. Similarly if we take an example like $3<5$ and $1<2$, where we want to prove $3 - 1 < 5 - 2$, i.e. $2 < 3$, we can add $3<5$ and $-2<-1$ to get $1<4$, but this is a weaker inequality than the one we want to prove.
Considering your whole proof attempt, your first step of showing that $sup(A) + sup(B)$ (i.e. $alpha + beta$ in your notation) is an upper bound for $A + B$ is useful. Here's a hint for what to do next: this upper bound condition can be rewritten as $A + B leq sup(A) + sup(B)$, so whatever $sup(A+B)$ is, it must be less than or equal to this upper bound on $A + B$, i.e. we have $sup(A + B) leq sup(A) + sup(B)$. Now if we can also show that $sup(A+B) geq sup(A) + sup(B)$, the combination of the two inequalities will give $sup(A+B) = sup(A) + sup(B)$, as required. In order to show $sup(A+B) geq sup(A) + sup(B)$, you should first prove that $sup(A+B)$ is an upper bound for $A + sup(B)$, so $sup(A + sup(B)) leq sup(A+B)$, or writing it the other way round, $sup(A+B) geq sup(A + sup(B))$. Finally, prove that $sup(A + sup(B)) equiv sup(A+B)$ in order to reach the desired result.
answered Jul 21 at 14:59
Prasiortle
463
We have e.g. $3<4$ and $1<2$, but subtracting would give $3-1 < 4-2 implies 2<2$, which is wrong. In general, note that if $p<q$, we can write this as $p = q - delta$ for some positive $delta$, and similarly $r<s implies r = s - epsilon$ for some positive $epsilon$. Subtracting would give $p - r < q - s iff q - delta - s + epsilon < q - s iff epsilon < delta$, which may or may not be true. In contrast, adding inequalities is allowed because $p + r < q + s iff q - delta + s - epsilon < q + s iff delta + epsilon > 0$, which is true as both $delta$ and $epsilon$ are positive.
In your last paragraph, your inequality is correct but totally useless for the proof, so it's "wrong" in that sense. Similarly if we take an example like $3<5$ and $1<2$, where we want to prove $3 - 1 < 5 - 2$, i.e. $2 < 3$, we can add $3<5$ and $-2<-1$ to get $1<4$, but this is a weaker inequality than the one we want to prove.
Considering your whole proof attempt, your first step of showing that $sup(A) + sup(B)$ (i.e. $alpha + beta$ in your notation) is an upper bound for $A + B$ is useful. Here's a hint for what to do next: this upper bound condition can be rewritten as $A + B leq sup(A) + sup(B)$, so whatever $sup(A+B)$ is, it must be less than or equal to this upper bound on $A + B$, i.e. we have $sup(A + B) leq sup(A) + sup(B)$. Now if we can also show that $sup(A+B) geq sup(A) + sup(B)$, the combination of the two inequalities will give $sup(A+B) = sup(A) + sup(B)$, as required. In order to show $sup(A+B) geq sup(A) + sup(B)$, you should first prove that $sup(A+B)$ is an upper bound for $A + sup(B)$, so $sup(A + sup(B)) leq sup(A+B)$, or writing it the other way round, $sup(A+B) geq sup(A + sup(B))$. Finally, prove that $sup(A + sup(B)) equiv sup(A+B)$ in order to reach the desired result.
answered Jul 21 at 14:59
Prasiortle
463
answered Jul 21 at 14:59
Prasiortle
463
answered Jul 21 at 14:59
Prasiortle
463
answered Jul 21 at 14:59
Prasiortle
463
463
add a comment |Â
add a comment |Â
up vote
1
down vote
It is wrong. It is not true that an upper bound $gamma$ of $A+B$ can be written as $a+b$ with $ain A$ and $bin B$. Suppose, for instance, that $A=B=0$ and that $gamma=10$.
However, you proved correctly that $sup A+sup B$ is an upper bound of $A+B$. Therefore, $sup(A+B)leqslantsup A+sup B$. What would happen if we had $sup(A+B)<sup A+sup B$? In that case, take $varepsilon=sup A+sup B-sup(A+B)$, take $ain A$ such that $a>sup(A)-fracvarepsilon2$ and take $bin B$ such that $b>sup(B)-fracvarepsilon2$. What does this tell you about $a+b$?
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
It is wrong. It is not true that an upper bound $gamma$ of $A+B$ can be written as $a+b$ with $ain A$ and $bin B$. Suppose, for instance, that $A=B=0$ and that $gamma=10$.
However, you proved correctly that $sup A+sup B$ is an upper bound of $A+B$. Therefore, $sup(A+B)leqslantsup A+sup B$. What would happen if we had $sup(A+B)<sup A+sup B$? In that case, take $varepsilon=sup A+sup B-sup(A+B)$, take $ain A$ such that $a>sup(A)-fracvarepsilon2$ and take $bin B$ such that $b>sup(B)-fracvarepsilon2$. What does this tell you about $a+b$?
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is wrong. It is not true that an upper bound $gamma$ of $A+B$ can be written as $a+b$ with $ain A$ and $bin B$. Suppose, for instance, that $A=B=0$ and that $gamma=10$.
However, you proved correctly that $sup A+sup B$ is an upper bound of $A+B$. Therefore, $sup(A+B)leqslantsup A+sup B$. What would happen if we had $sup(A+B)<sup A+sup B$? In that case, take $varepsilon=sup A+sup B-sup(A+B)$, take $ain A$ such that $a>sup(A)-fracvarepsilon2$ and take $bin B$ such that $b>sup(B)-fracvarepsilon2$. What does this tell you about $a+b$?
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
It is wrong. It is not true that an upper bound $gamma$ of $A+B$ can be written as $a+b$ with $ain A$ and $bin B$. Suppose, for instance, that $A=B=0$ and that $gamma=10$.
However, you proved correctly that $sup A+sup B$ is an upper bound of $A+B$. Therefore, $sup(A+B)leqslantsup A+sup B$. What would happen if we had $sup(A+B)<sup A+sup B$? In that case, take $varepsilon=sup A+sup B-sup(A+B)$, take $ain A$ such that $a>sup(A)-fracvarepsilon2$ and take $bin B$ such that $b>sup(B)-fracvarepsilon2$. What does this tell you about $a+b$?
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
answered Jul 21 at 14:54
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
0
down vote
The example by NickD illustrates the problem with the present proof.
A quick fix: suppose that $gammageq a+b$ for all $ain A$ and $bin B$. We want to show that $gammageqalpha+beta$.
So suppose otherwise that $gamma=alpha+beta-e$ for some $e>0$. Now $alpha-e/2$ is strictly smaller than $alpha$ so there is some $a^*in A$ such that $a^*>alpha-e/2$. Similarly, there is some $b^*in B$ such that $b^*>beta-e/2$. But then:
$$
a^*+b^*in A+Bquadtextandquad a^*+b^*>(alpha-e/2)+(beta-e/2)=gamma.
$$
This is your desired contradiction.
answered Jul 21 at 14:56
yurnero
6,8401824
The example illustrated one error in the proof. It wasn't the first error. This answer quietly avoids the earlier errors, however.
– David K
Jul 21 at 15:27
add a comment |Â
up vote
0
down vote
The example by NickD illustrates the problem with the present proof.
A quick fix: suppose that $gammageq a+b$ for all $ain A$ and $bin B$. We want to show that $gammageqalpha+beta$.
So suppose otherwise that $gamma=alpha+beta-e$ for some $e>0$. Now $alpha-e/2$ is strictly smaller than $alpha$ so there is some $a^*in A$ such that $a^*>alpha-e/2$. Similarly, there is some $b^*in B$ such that $b^*>beta-e/2$. But then:
$$
a^*+b^*in A+Bquadtextandquad a^*+b^*>(alpha-e/2)+(beta-e/2)=gamma.
$$
This is your desired contradiction.
answered Jul 21 at 14:56
yurnero
6,8401824
The example illustrated one error in the proof. It wasn't the first error. This answer quietly avoids the earlier errors, however.
– David K
Jul 21 at 15:27
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The example by NickD illustrates the problem with the present proof.
A quick fix: suppose that $gammageq a+b$ for all $ain A$ and $bin B$. We want to show that $gammageqalpha+beta$.
So suppose otherwise that $gamma=alpha+beta-e$ for some $e>0$. Now $alpha-e/2$ is strictly smaller than $alpha$ so there is some $a^*in A$ such that $a^*>alpha-e/2$. Similarly, there is some $b^*in B$ such that $b^*>beta-e/2$. But then:
$$
a^*+b^*in A+Bquadtextandquad a^*+b^*>(alpha-e/2)+(beta-e/2)=gamma.
$$
This is your desired contradiction.
answered Jul 21 at 14:56
yurnero
6,8401824
The example by NickD illustrates the problem with the present proof.
A quick fix: suppose that $gammageq a+b$ for all $ain A$ and $bin B$. We want to show that $gammageqalpha+beta$.
So suppose otherwise that $gamma=alpha+beta-e$ for some $e>0$. Now $alpha-e/2$ is strictly smaller than $alpha$ so there is some $a^*in A$ such that $a^*>alpha-e/2$. Similarly, there is some $b^*in B$ such that $b^*>beta-e/2$. But then:
$$
a^*+b^*in A+Bquadtextandquad a^*+b^*>(alpha-e/2)+(beta-e/2)=gamma.
$$
This is your desired contradiction.
answered Jul 21 at 14:56
yurnero
6,8401824
answered Jul 21 at 14:56
yurnero
6,8401824
answered Jul 21 at 14:56
yurnero
6,8401824
answered Jul 21 at 14:56
yurnero
6,8401824
6,8401824
The example illustrated one error in the proof. It wasn't the first error. This answer quietly avoids the earlier errors, however.
– David K
Jul 21 at 15:27
add a comment |Â
The example illustrated one error in the proof. It wasn't the first error. This answer quietly avoids the earlier errors, however.
– David K
Jul 21 at 15:27
The example illustrated one error in the proof. It wasn't the first error. This answer quietly avoids the earlier errors, however.
– David K
Jul 21 at 15:27
The example illustrated one error in the proof. It wasn't the first error. This answer quietly avoids the earlier errors, however.
– David K
Jul 21 at 15:27
add a comment |Â
up vote
0
down vote
Subtracting inequalities is wrong, for example if you subtract $-5<2$ from $4<10$ you get $9<8$
For your proof of $$ sup(A+B)= sup(A)+sup(B)$$ where you have already shown that $$ sup(A+B)le sup(A)+sup(B)$$
you need to show that for any number $gamma $ less than $ sup(A)+sup(B)$ you can find an element in A+B which is
greater than $gamma$
Note that $$gamma= sup(A)+sup(B)-epsilon = (sup(A) - epsilon /2) + (sup(B) - epsilon /2)$$
Now you can easily find elements $ain A$ and $bin B$ such that $a+b >gamma$
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
Subtracting inequalities is wrong, for example if you subtract $-5<2$ from $4<10$ you get $9<8$
For your proof of $$ sup(A+B)= sup(A)+sup(B)$$ where you have already shown that $$ sup(A+B)le sup(A)+sup(B)$$
you need to show that for any number $gamma $ less than $ sup(A)+sup(B)$ you can find an element in A+B which is
greater than $gamma$
Note that $$gamma= sup(A)+sup(B)-epsilon = (sup(A) - epsilon /2) + (sup(B) - epsilon /2)$$
Now you can easily find elements $ain A$ and $bin B$ such that $a+b >gamma$
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Subtracting inequalities is wrong, for example if you subtract $-5<2$ from $4<10$ you get $9<8$
For your proof of $$ sup(A+B)= sup(A)+sup(B)$$ where you have already shown that $$ sup(A+B)le sup(A)+sup(B)$$
you need to show that for any number $gamma $ less than $ sup(A)+sup(B)$ you can find an element in A+B which is
greater than $gamma$
Note that $$gamma= sup(A)+sup(B)-epsilon = (sup(A) - epsilon /2) + (sup(B) - epsilon /2)$$
Now you can easily find elements $ain A$ and $bin B$ such that $a+b >gamma$
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
Subtracting inequalities is wrong, for example if you subtract $-5<2$ from $4<10$ you get $9<8$
For your proof of $$ sup(A+B)= sup(A)+sup(B)$$ where you have already shown that $$ sup(A+B)le sup(A)+sup(B)$$
you need to show that for any number $gamma $ less than $ sup(A)+sup(B)$ you can find an element in A+B which is
greater than $gamma$
Note that $$gamma= sup(A)+sup(B)-epsilon = (sup(A) - epsilon /2) + (sup(B) - epsilon /2)$$
Now you can easily find elements $ain A$ and $bin B$ such that $a+b >gamma$
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 15:16
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
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1
$10 > 0$, $20> 0$ but is $10-20 > 0 - 0$?
– NickD
Jul 21 at 14:49