Does $sup|A-B| ge |sup A -sup B|$?
$begingroup$
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
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
asked Dec 25 '18 at 15:09
DanielDaniel
268
$endgroup$
add a comment |
$begingroup$
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
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
asked Dec 25 '18 at 15:09
DanielDaniel
268
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
asked Dec 25 '18 at 15:09
DanielDaniel
268
$endgroup$
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
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
asked Dec 25 '18 at 15:09
DanielDaniel
268
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
asked Dec 25 '18 at 15:09
DanielDaniel
268
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
edited Dec 25 '18 at 15:28
mechanodroid
27.5k62447
27.5k62447
asked Dec 25 '18 at 15:09
DanielDaniel
268
asked Dec 25 '18 at 15:09
DanielDaniel
268
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
$endgroup$
add a comment |
$begingroup$
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 $|sup A-sup B|=1$.
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
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.
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
$endgroup$
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.
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
edited Dec 25 '18 at 17:03
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
answered Dec 25 '18 at 15:26
mechanodroidmechanodroid
27.5k62447
27.5k62447
add a comment |
add a comment |
$begingroup$
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 $|sup A-sup B|=1$.
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
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 $|sup A-sup B|=1$.
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
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 $|sup A-sup B|=1$.
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
$endgroup$
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 $|sup A-sup B|=1$.
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
answered Dec 25 '18 at 15:21
Ben WBen W
2,276615
2,276615
add a comment |
add a comment |
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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