Mixing
Covering of Intervals
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For a real interval $I=[x-r,x+r]$ and a number $alpha>0$ we write $alpha I:=[x-alpha r,x+alpha r]$, i.e. $alpha I$ is the interval $I$ blown up by a factor $alpha$ around its center.
For example, the Vitali Covering Lemma says that for any finite collection of intervals $([a_j,b_j])_{1leq jleq n}$ there is an index set $Jsubseteq{1,ldots,n}$ such that $([a_j,b_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[a_j,b_j]subseteqbigcup_{jin J}3[a_j,b_j]
$$
Setting: I have two finite collections of intervals $([a_j,b_j])_{1leq jleq n}$ and $([c_j,d_j])_{1leq jleq n}$ with the following three properties.
$|a_j-c_j|leqdelta$ and $|b_j-d_j|leq delta$ for $1leq jleq n$,
$sum_{j=1}^n|(a_j-c_j)-(b_j-d_j)|leqdelta$,
$left|bigcup_{j=1}^n[a_j,b_j]right|leqvarepsilon$, where $|A|$ means the Lebesgue measure of a set $A$.
Here, $delta$ and $varepsilon$ are two positive "small" numbers.
Problem: I would like to show (or disprove) that the Lebesgue measure of $bigcup_{j=1}^n[c_j,d_j]$ is, up to a constant that does not depend on the intervals or $n$, like that of $bigcup_{j=1}^n[a_j,b_j]$ also small.
My attempt: I could prove that this is true if the $[a_j,b_j]$ are mutually disjoint: Indeed, by Vitali's Covering Lemma there is an index set $Jsubseteq{1,ldots,n}$ such that $([c_j,d_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[c_j,d_j]subseteqbigcup_{jin J}3[c_j,d_j].
$$
This implies with the help of 2. and 3. that
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqleft|bigcup_{jin J}3[c_j,d_j]right|
leqsum_{jin J}Big|3[c_j,d_j]Big|
=3sum_{jin J}|c_j-d_j| \
&leq 3sum_{jin J}|(a_j-b_j)-(c_j-d_j)|+3sum_{jin J}|a_j-b_j| \
&leq 3delta+3left|bigcup_{j=1}^n[a_j,b_j]right| \
&leq 3(delta+varepsilon).
end{align*}
However, I have not used 1. In case that the intervals $[a_j,b_j]$ are not pairwise disjoint, I don't know how to proceed. Any ideas are highly appreciated. Thanks in advance!
EDIT: (Thanks to mathworker21): Vitali's Covering Lemma is not necessary here, we can use the estimate
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqsum_{j=1}^n|c_j-d_j|
end{align*}
instead.
measure-theory elementary-set-theory lebesgue-measure
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
$endgroup$
|
show 1 more comment
$begingroup$
For a real interval $I=[x-r,x+r]$ and a number $alpha>0$ we write $alpha I:=[x-alpha r,x+alpha r]$, i.e. $alpha I$ is the interval $I$ blown up by a factor $alpha$ around its center.
For example, the Vitali Covering Lemma says that for any finite collection of intervals $([a_j,b_j])_{1leq jleq n}$ there is an index set $Jsubseteq{1,ldots,n}$ such that $([a_j,b_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[a_j,b_j]subseteqbigcup_{jin J}3[a_j,b_j]
$$
Setting: I have two finite collections of intervals $([a_j,b_j])_{1leq jleq n}$ and $([c_j,d_j])_{1leq jleq n}$ with the following three properties.
$|a_j-c_j|leqdelta$ and $|b_j-d_j|leq delta$ for $1leq jleq n$,
$sum_{j=1}^n|(a_j-c_j)-(b_j-d_j)|leqdelta$,
$left|bigcup_{j=1}^n[a_j,b_j]right|leqvarepsilon$, where $|A|$ means the Lebesgue measure of a set $A$.
Here, $delta$ and $varepsilon$ are two positive "small" numbers.
Problem: I would like to show (or disprove) that the Lebesgue measure of $bigcup_{j=1}^n[c_j,d_j]$ is, up to a constant that does not depend on the intervals or $n$, like that of $bigcup_{j=1}^n[a_j,b_j]$ also small.
My attempt: I could prove that this is true if the $[a_j,b_j]$ are mutually disjoint: Indeed, by Vitali's Covering Lemma there is an index set $Jsubseteq{1,ldots,n}$ such that $([c_j,d_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[c_j,d_j]subseteqbigcup_{jin J}3[c_j,d_j].
$$
This implies with the help of 2. and 3. that
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqleft|bigcup_{jin J}3[c_j,d_j]right|
leqsum_{jin J}Big|3[c_j,d_j]Big|
=3sum_{jin J}|c_j-d_j| \
&leq 3sum_{jin J}|(a_j-b_j)-(c_j-d_j)|+3sum_{jin J}|a_j-b_j| \
&leq 3delta+3left|bigcup_{j=1}^n[a_j,b_j]right| \
&leq 3(delta+varepsilon).
end{align*}
However, I have not used 1. In case that the intervals $[a_j,b_j]$ are not pairwise disjoint, I don't know how to proceed. Any ideas are highly appreciated. Thanks in advance!
EDIT: (Thanks to mathworker21): Vitali's Covering Lemma is not necessary here, we can use the estimate
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqsum_{j=1}^n|c_j-d_j|
end{align*}
instead.
measure-theory elementary-set-theory lebesgue-measure
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
$endgroup$
$begingroup$
Have you run into any problem with constructing a counterexample?
$endgroup$
– Dap
Jan 6 at 15:44
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Yes, I clearly tried to find counterexamples. Condition 2 implies (triangle inequality) that the sum of the lengths of the intervals differ only by $delta$. Without condition 1 my claim is false: If all $[a_j,b_j]=[0,varepsilon]$, say, then 3 is met ($delta=varepsilon$). Now, take $[c_j,d_j]=[(j-1)varepsilon,jvarepsilon]$ for $j=1,ldots,n$. Then 2 is also met, but $left|bigcup_{j=1}^n[c_j,d_j]right|=|[0,nvarepsilon]|=nvarepsilon$. My overall problem is that I have not a clear idea on how geometrically combine 1 and 2. Therefore, I'm not entirely sure that my claim is true, though.
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– sranthrop
Jan 6 at 16:11
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@sranthrop I'm confused why you used Vitali covering lemma in your proof for when the $[a_j,b_j]$'s are mutually disjoint. Can't you just use $|cup_j [c_j,d_j]| le sum_j d_j-c_j$ and run through the argument?
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– mathworker21
Jan 7 at 19:31
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Thanks for asking and pointing this out. I think you are right. I'm working on this problem for weeks now and I'm so focused on some things. Thank you!
$endgroup$
– sranthrop
Jan 7 at 21:02
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how do you go from line 2 to 3 without disjoint intervals $[a_j,b_j]$?
$endgroup$
– Dunham
Jan 7 at 21:11
|
show 1 more comment
$begingroup$
For a real interval $I=[x-r,x+r]$ and a number $alpha>0$ we write $alpha I:=[x-alpha r,x+alpha r]$, i.e. $alpha I$ is the interval $I$ blown up by a factor $alpha$ around its center.
For example, the Vitali Covering Lemma says that for any finite collection of intervals $([a_j,b_j])_{1leq jleq n}$ there is an index set $Jsubseteq{1,ldots,n}$ such that $([a_j,b_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[a_j,b_j]subseteqbigcup_{jin J}3[a_j,b_j]
$$
Setting: I have two finite collections of intervals $([a_j,b_j])_{1leq jleq n}$ and $([c_j,d_j])_{1leq jleq n}$ with the following three properties.
$|a_j-c_j|leqdelta$ and $|b_j-d_j|leq delta$ for $1leq jleq n$,
$sum_{j=1}^n|(a_j-c_j)-(b_j-d_j)|leqdelta$,
$left|bigcup_{j=1}^n[a_j,b_j]right|leqvarepsilon$, where $|A|$ means the Lebesgue measure of a set $A$.
Here, $delta$ and $varepsilon$ are two positive "small" numbers.
Problem: I would like to show (or disprove) that the Lebesgue measure of $bigcup_{j=1}^n[c_j,d_j]$ is, up to a constant that does not depend on the intervals or $n$, like that of $bigcup_{j=1}^n[a_j,b_j]$ also small.
My attempt: I could prove that this is true if the $[a_j,b_j]$ are mutually disjoint: Indeed, by Vitali's Covering Lemma there is an index set $Jsubseteq{1,ldots,n}$ such that $([c_j,d_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[c_j,d_j]subseteqbigcup_{jin J}3[c_j,d_j].
$$
This implies with the help of 2. and 3. that
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqleft|bigcup_{jin J}3[c_j,d_j]right|
leqsum_{jin J}Big|3[c_j,d_j]Big|
=3sum_{jin J}|c_j-d_j| \
&leq 3sum_{jin J}|(a_j-b_j)-(c_j-d_j)|+3sum_{jin J}|a_j-b_j| \
&leq 3delta+3left|bigcup_{j=1}^n[a_j,b_j]right| \
&leq 3(delta+varepsilon).
end{align*}
However, I have not used 1. In case that the intervals $[a_j,b_j]$ are not pairwise disjoint, I don't know how to proceed. Any ideas are highly appreciated. Thanks in advance!
EDIT: (Thanks to mathworker21): Vitali's Covering Lemma is not necessary here, we can use the estimate
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqsum_{j=1}^n|c_j-d_j|
end{align*}
instead.
measure-theory elementary-set-theory lebesgue-measure
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
$endgroup$
For a real interval $I=[x-r,x+r]$ and a number $alpha>0$ we write $alpha I:=[x-alpha r,x+alpha r]$, i.e. $alpha I$ is the interval $I$ blown up by a factor $alpha$ around its center.
For example, the Vitali Covering Lemma says that for any finite collection of intervals $([a_j,b_j])_{1leq jleq n}$ there is an index set $Jsubseteq{1,ldots,n}$ such that $([a_j,b_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[a_j,b_j]subseteqbigcup_{jin J}3[a_j,b_j]
$$
Setting: I have two finite collections of intervals $([a_j,b_j])_{1leq jleq n}$ and $([c_j,d_j])_{1leq jleq n}$ with the following three properties.
$|a_j-c_j|leqdelta$ and $|b_j-d_j|leq delta$ for $1leq jleq n$,
$sum_{j=1}^n|(a_j-c_j)-(b_j-d_j)|leqdelta$,
$left|bigcup_{j=1}^n[a_j,b_j]right|leqvarepsilon$, where $|A|$ means the Lebesgue measure of a set $A$.
Here, $delta$ and $varepsilon$ are two positive "small" numbers.
Problem: I would like to show (or disprove) that the Lebesgue measure of $bigcup_{j=1}^n[c_j,d_j]$ is, up to a constant that does not depend on the intervals or $n$, like that of $bigcup_{j=1}^n[a_j,b_j]$ also small.
My attempt: I could prove that this is true if the $[a_j,b_j]$ are mutually disjoint: Indeed, by Vitali's Covering Lemma there is an index set $Jsubseteq{1,ldots,n}$ such that $([c_j,d_j])_{jin J}$ is a subcollection of pairwise disjoint intervals with
$$
bigcup_{1leq jleq n}[c_j,d_j]subseteqbigcup_{jin J}3[c_j,d_j].
$$
This implies with the help of 2. and 3. that
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqleft|bigcup_{jin J}3[c_j,d_j]right|
leqsum_{jin J}Big|3[c_j,d_j]Big|
=3sum_{jin J}|c_j-d_j| \
&leq 3sum_{jin J}|(a_j-b_j)-(c_j-d_j)|+3sum_{jin J}|a_j-b_j| \
&leq 3delta+3left|bigcup_{j=1}^n[a_j,b_j]right| \
&leq 3(delta+varepsilon).
end{align*}
However, I have not used 1. In case that the intervals $[a_j,b_j]$ are not pairwise disjoint, I don't know how to proceed. Any ideas are highly appreciated. Thanks in advance!
EDIT: (Thanks to mathworker21): Vitali's Covering Lemma is not necessary here, we can use the estimate
begin{align*}
left|bigcup_{j=1}^n[c_j,d_j]right|
&leqsum_{j=1}^n|c_j-d_j|
end{align*}
instead.
measure-theory elementary-set-theory lebesgue-measure
measure-theory elementary-set-theory lebesgue-measure
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
edited Jan 7 at 21:05
sranthrop
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
asked Jan 3 at 2:08
sranthropsranthrop
7,0241925
7,0241925
$begingroup$
Have you run into any problem with constructing a counterexample?
$endgroup$
– Dap
Jan 6 at 15:44
$begingroup$
Yes, I clearly tried to find counterexamples. Condition 2 implies (triangle inequality) that the sum of the lengths of the intervals differ only by $delta$. Without condition 1 my claim is false: If all $[a_j,b_j]=[0,varepsilon]$, say, then 3 is met ($delta=varepsilon$). Now, take $[c_j,d_j]=[(j-1)varepsilon,jvarepsilon]$ for $j=1,ldots,n$. Then 2 is also met, but $left|bigcup_{j=1}^n[c_j,d_j]right|=|[0,nvarepsilon]|=nvarepsilon$. My overall problem is that I have not a clear idea on how geometrically combine 1 and 2. Therefore, I'm not entirely sure that my claim is true, though.
$endgroup$
– sranthrop
Jan 6 at 16:11
$begingroup$
@sranthrop I'm confused why you used Vitali covering lemma in your proof for when the $[a_j,b_j]$'s are mutually disjoint. Can't you just use $|cup_j [c_j,d_j]| le sum_j d_j-c_j$ and run through the argument?
$endgroup$
– mathworker21
Jan 7 at 19:31
$begingroup$
Thanks for asking and pointing this out. I think you are right. I'm working on this problem for weeks now and I'm so focused on some things. Thank you!
$endgroup$
– sranthrop
Jan 7 at 21:02
$begingroup$
how do you go from line 2 to 3 without disjoint intervals $[a_j,b_j]$?
$endgroup$
– Dunham
Jan 7 at 21:11
|
show 1 more comment
$begingroup$
Have you run into any problem with constructing a counterexample?
$endgroup$
– Dap
Jan 6 at 15:44
$begingroup$
Yes, I clearly tried to find counterexamples. Condition 2 implies (triangle inequality) that the sum of the lengths of the intervals differ only by $delta$. Without condition 1 my claim is false: If all $[a_j,b_j]=[0,varepsilon]$, say, then 3 is met ($delta=varepsilon$). Now, take $[c_j,d_j]=[(j-1)varepsilon,jvarepsilon]$ for $j=1,ldots,n$. Then 2 is also met, but $left|bigcup_{j=1}^n[c_j,d_j]right|=|[0,nvarepsilon]|=nvarepsilon$. My overall problem is that I have not a clear idea on how geometrically combine 1 and 2. Therefore, I'm not entirely sure that my claim is true, though.
$endgroup$
– sranthrop
Jan 6 at 16:11
$begingroup$
@sranthrop I'm confused why you used Vitali covering lemma in your proof for when the $[a_j,b_j]$'s are mutually disjoint. Can't you just use $|cup_j [c_j,d_j]| le sum_j d_j-c_j$ and run through the argument?
$endgroup$
– mathworker21
Jan 7 at 19:31
$begingroup$
Thanks for asking and pointing this out. I think you are right. I'm working on this problem for weeks now and I'm so focused on some things. Thank you!
$endgroup$
– sranthrop
Jan 7 at 21:02
$begingroup$
how do you go from line 2 to 3 without disjoint intervals $[a_j,b_j]$?
$endgroup$
– Dunham
Jan 7 at 21:11
$begingroup$
Have you run into any problem with constructing a counterexample?
$endgroup$
– Dap
Jan 6 at 15:44
$begingroup$
Have you run into any problem with constructing a counterexample?
$endgroup$
– Dap
Jan 6 at 15:44
$begingroup$
Yes, I clearly tried to find counterexamples. Condition 2 implies (triangle inequality) that the sum of the lengths of the intervals differ only by $delta$. Without condition 1 my claim is false: If all $[a_j,b_j]=[0,varepsilon]$, say, then 3 is met ($delta=varepsilon$). Now, take $[c_j,d_j]=[(j-1)varepsilon,jvarepsilon]$ for $j=1,ldots,n$. Then 2 is also met, but $left|bigcup_{j=1}^n[c_j,d_j]right|=|[0,nvarepsilon]|=nvarepsilon$. My overall problem is that I have not a clear idea on how geometrically combine 1 and 2. Therefore, I'm not entirely sure that my claim is true, though.
$endgroup$
– sranthrop
Jan 6 at 16:11
$begingroup$
Yes, I clearly tried to find counterexamples. Condition 2 implies (triangle inequality) that the sum of the lengths of the intervals differ only by $delta$. Without condition 1 my claim is false: If all $[a_j,b_j]=[0,varepsilon]$, say, then 3 is met ($delta=varepsilon$). Now, take $[c_j,d_j]=[(j-1)varepsilon,jvarepsilon]$ for $j=1,ldots,n$. Then 2 is also met, but $left|bigcup_{j=1}^n[c_j,d_j]right|=|[0,nvarepsilon]|=nvarepsilon$. My overall problem is that I have not a clear idea on how geometrically combine 1 and 2. Therefore, I'm not entirely sure that my claim is true, though.
$endgroup$
– sranthrop
Jan 6 at 16:11
$begingroup$
@sranthrop I'm confused why you used Vitali covering lemma in your proof for when the $[a_j,b_j]$'s are mutually disjoint. Can't you just use $|cup_j [c_j,d_j]| le sum_j d_j-c_j$ and run through the argument?
$endgroup$
– mathworker21
Jan 7 at 19:31
$begingroup$
@sranthrop I'm confused why you used Vitali covering lemma in your proof for when the $[a_j,b_j]$'s are mutually disjoint. Can't you just use $|cup_j [c_j,d_j]| le sum_j d_j-c_j$ and run through the argument?
$endgroup$
– mathworker21
Jan 7 at 19:31
$begingroup$
Thanks for asking and pointing this out. I think you are right. I'm working on this problem for weeks now and I'm so focused on some things. Thank you!
$endgroup$
– sranthrop
Jan 7 at 21:02
$begingroup$
Thanks for asking and pointing this out. I think you are right. I'm working on this problem for weeks now and I'm so focused on some things. Thank you!
$endgroup$
– sranthrop
Jan 7 at 21:02
$begingroup$
how do you go from line 2 to 3 without disjoint intervals $[a_j,b_j]$?
$endgroup$
– Dunham
Jan 7 at 21:11
$begingroup$
how do you go from line 2 to 3 without disjoint intervals $[a_j,b_j]$?
$endgroup$
– Dunham
Jan 7 at 21:11
|
show 1 more comment
2 Answers
2
active
oldest
votes
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Here is a disproof. Take any $epsilon,delta > 0$. We construct $([a_j,b_j])_{1 le j le n}, ([c_j,d_j])_{1 le j le n}$ such that:
(1) $|a_j-c_j| le delta$ and $|b_j-d_j| le delta$ for each $j$.
(2) $sum_{j=1}^n |(a_j-c_j)-(b_j-d_j)| = 0$.
(3) $|cup_{j=1}^n [a_j,b_j]| = epsilon$.
(4) $|cup_{j=1}^n [c_j,d_j]| = sqrt{epsilon delta n}$.
Take any $L in mathbb{N}$, and let $R = frac{Ldelta}{epsilon}$ and $n = RL$. (It's not a big deal to assume $R in mathbb{N}$). For $0 le l le L-1$ and $1 le r le R$, let $[a_{lR+r},b_{lR+r}] = [l,l+frac{epsilon}{L}]$ and $[c_{lR+r},d_{lR+r}] = [l-delta+(r-1)frac{epsilon}{L},l-delta+rfrac{epsilon}{L}]$.
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
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+1 - although I did find it a "big deal" to see exactly how $epsilon$ and $delta$ are related (both in this answer, and in the question itself).
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– Calum Gilhooley
Jan 9 at 2:49
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@CalumGilhooley I'm a bit confused. In my answer, $epsilon$ and $delta$ were arbitrary. If you are objecting because $frac{Ldelta}{epsilon}$ might not be an integer, then we can just define $R = lceil frac{Ldelta}{epsilon} rceil$ instead. If I'm missing something, let me know.
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– mathworker21
Jan 9 at 7:56
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Thank you, that helps. I think you mean $R=lfloorfrac{Ldelta}{epsilon}rfloor$, which ensures that $[c_{lR},d_{lR}]$ and $[c_{lR+1},d_{lR+1}]$ don't overlap, for $1=1,2,ldots,L-1$. More simply, we can choose $R$ arbitrarily - as "large" as we like - and define $L=lceilfrac{Repsilon}{delta}rceil$. Then $leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = Repsilon$, which is unbounded by any function of $(epsilon,delta)$.
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– Calum Gilhooley
Jan 9 at 15:41
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@CalumGilhooley ah yes, that's simpler. thanks!
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– mathworker21
Jan 9 at 16:04
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Thank you, guys, very much for all your effort and this nice solution, although my claim which I hoped was true unfortunately turned out to be false.
$endgroup$
– sranthrop
Jan 9 at 20:44
add a comment |
$begingroup$
(Disclaimer: this is more of a pedantic comment than an answer!)
Does it change the meaning of the question essentially if one drops
the stipulation that $delta$ and $epsilon$ are "small" - I
confess to being unable to understand what that statement means in
the present context - and asks instead:
Do there exist a number $h > 0$, and functions $f, g$ (of one and
two range-limited real variables, respectively), such that if
$0 < epsilon < h$, and condition 3 holds, then for all $delta$
such that $0 < delta < f(epsilon)$, and all $n$ and
$a_j, b_j, c_j, d_j$ ($1 leqslant j leqslant n$) such that
conditions 1 and 2 hold, we have
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert leqslant
g(epsilon, delta)$?
A more demanding (but I think less plausible) reformulation of the
question is as follows:
Do there exist numbers $h, h' > 0$, and a function $g$ (of two
range-limited real variables), such that if $0 < epsilon < h$ and
$0 < delta < h'$, then for all $n$ and $a_j, b_j, c_j, d_j$
($1 leqslant j leqslant n$) such that conditions 1 to 3 hold, we
have $left|bigcup_{j=1}^n[c_j,d_j]right| leqslant
g(epsilon, delta)$?
If the second reformulation accurately represents the meaning of the
question, then a special case of mathworker21's argument still gives
the answer "no".
On the other hand, if the first reformulation is satisfactory, then
mathworker21's construction cannot be applied, and the question
remains open.
[Not so! See the addendum below.]
Disposing of the second reformulation first:
Given $h, h' > 0$, take any $epsilon$ such
that $0 < epsilon < min{h, h', 1}$. Let $delta = epsilon$, let
$k$ be a positive integer, and, in mathworker21's construction, let
$L = R = k$, $n = k^2$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$.
Because this measure is not bounded above by any function of
$epsilon$ and $delta$, the required function $g$ cannot exist.
$square$
On the first reformulation (which I think is a more reasonable
interpretation of the question), we may still choose any value of
$epsilon$ that is "small enough", i.e. $epsilon < h$, and try
to derive a contradiction; but now we must allow $delta$ to take
on any "sufficiently small" value, i.e. any value less than some
unspecified $f(epsilon)$. In these circumstances, I don't see how
mathworker21's construction can be carried out.
(I don't know what the answer is in this case, but it seems best to
check my understanding of the question before possibly barking up
the wrong tree!)
Even if the first reformulation is valid, the same general
construction does still apply. This "answer" is therefore
probably best viewed as a long-winded comment in confirmation of
mathworker21's answer.
Given $h > 0$, and a function $f$, take any $epsilon$ such that
$0 < epsilon < min{h, 1}$, and any positive integer $k$ such
that $frac{epsilon}{k} < f(epsilon)$.
Define $delta = frac{epsilon}{k}$, In mathworker21's
construction, let $R = k$, $L = k^2$, $n = k^3$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$, much
as before. Again, the required function $g$ cannot exist.
$square$
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
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Thanks a lot for this very detailed discussion and answer to my question!
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– sranthrop
Jan 9 at 20:45
$begingroup$
I've done something useful? ... Excuse me, I think I'm going to have to lie down for a minute! :)
$endgroup$
– Calum Gilhooley
Jan 9 at 20:51
add a comment |
Your Answer
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$begingroup$
Here is a disproof. Take any $epsilon,delta > 0$. We construct $([a_j,b_j])_{1 le j le n}, ([c_j,d_j])_{1 le j le n}$ such that:
(1) $|a_j-c_j| le delta$ and $|b_j-d_j| le delta$ for each $j$.
(2) $sum_{j=1}^n |(a_j-c_j)-(b_j-d_j)| = 0$.
(3) $|cup_{j=1}^n [a_j,b_j]| = epsilon$.
(4) $|cup_{j=1}^n [c_j,d_j]| = sqrt{epsilon delta n}$.
Take any $L in mathbb{N}$, and let $R = frac{Ldelta}{epsilon}$ and $n = RL$. (It's not a big deal to assume $R in mathbb{N}$). For $0 le l le L-1$ and $1 le r le R$, let $[a_{lR+r},b_{lR+r}] = [l,l+frac{epsilon}{L}]$ and $[c_{lR+r},d_{lR+r}] = [l-delta+(r-1)frac{epsilon}{L},l-delta+rfrac{epsilon}{L}]$.
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
$endgroup$
$begingroup$
+1 - although I did find it a "big deal" to see exactly how $epsilon$ and $delta$ are related (both in this answer, and in the question itself).
$endgroup$
– Calum Gilhooley
Jan 9 at 2:49
$begingroup$
@CalumGilhooley I'm a bit confused. In my answer, $epsilon$ and $delta$ were arbitrary. If you are objecting because $frac{Ldelta}{epsilon}$ might not be an integer, then we can just define $R = lceil frac{Ldelta}{epsilon} rceil$ instead. If I'm missing something, let me know.
$endgroup$
– mathworker21
Jan 9 at 7:56
$begingroup$
Thank you, that helps. I think you mean $R=lfloorfrac{Ldelta}{epsilon}rfloor$, which ensures that $[c_{lR},d_{lR}]$ and $[c_{lR+1},d_{lR+1}]$ don't overlap, for $1=1,2,ldots,L-1$. More simply, we can choose $R$ arbitrarily - as "large" as we like - and define $L=lceilfrac{Repsilon}{delta}rceil$. Then $leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = Repsilon$, which is unbounded by any function of $(epsilon,delta)$.
$endgroup$
– Calum Gilhooley
Jan 9 at 15:41
$begingroup$
@CalumGilhooley ah yes, that's simpler. thanks!
$endgroup$
– mathworker21
Jan 9 at 16:04
$begingroup$
Thank you, guys, very much for all your effort and this nice solution, although my claim which I hoped was true unfortunately turned out to be false.
$endgroup$
– sranthrop
Jan 9 at 20:44
add a comment |
$begingroup$
Here is a disproof. Take any $epsilon,delta > 0$. We construct $([a_j,b_j])_{1 le j le n}, ([c_j,d_j])_{1 le j le n}$ such that:
(1) $|a_j-c_j| le delta$ and $|b_j-d_j| le delta$ for each $j$.
(2) $sum_{j=1}^n |(a_j-c_j)-(b_j-d_j)| = 0$.
(3) $|cup_{j=1}^n [a_j,b_j]| = epsilon$.
(4) $|cup_{j=1}^n [c_j,d_j]| = sqrt{epsilon delta n}$.
Take any $L in mathbb{N}$, and let $R = frac{Ldelta}{epsilon}$ and $n = RL$. (It's not a big deal to assume $R in mathbb{N}$). For $0 le l le L-1$ and $1 le r le R$, let $[a_{lR+r},b_{lR+r}] = [l,l+frac{epsilon}{L}]$ and $[c_{lR+r},d_{lR+r}] = [l-delta+(r-1)frac{epsilon}{L},l-delta+rfrac{epsilon}{L}]$.
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
$endgroup$
$begingroup$
+1 - although I did find it a "big deal" to see exactly how $epsilon$ and $delta$ are related (both in this answer, and in the question itself).
$endgroup$
– Calum Gilhooley
Jan 9 at 2:49
$begingroup$
@CalumGilhooley I'm a bit confused. In my answer, $epsilon$ and $delta$ were arbitrary. If you are objecting because $frac{Ldelta}{epsilon}$ might not be an integer, then we can just define $R = lceil frac{Ldelta}{epsilon} rceil$ instead. If I'm missing something, let me know.
$endgroup$
– mathworker21
Jan 9 at 7:56
$begingroup$
Thank you, that helps. I think you mean $R=lfloorfrac{Ldelta}{epsilon}rfloor$, which ensures that $[c_{lR},d_{lR}]$ and $[c_{lR+1},d_{lR+1}]$ don't overlap, for $1=1,2,ldots,L-1$. More simply, we can choose $R$ arbitrarily - as "large" as we like - and define $L=lceilfrac{Repsilon}{delta}rceil$. Then $leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = Repsilon$, which is unbounded by any function of $(epsilon,delta)$.
$endgroup$
– Calum Gilhooley
Jan 9 at 15:41
$begingroup$
@CalumGilhooley ah yes, that's simpler. thanks!
$endgroup$
– mathworker21
Jan 9 at 16:04
$begingroup$
Thank you, guys, very much for all your effort and this nice solution, although my claim which I hoped was true unfortunately turned out to be false.
$endgroup$
– sranthrop
Jan 9 at 20:44
add a comment |
$begingroup$
Here is a disproof. Take any $epsilon,delta > 0$. We construct $([a_j,b_j])_{1 le j le n}, ([c_j,d_j])_{1 le j le n}$ such that:
(1) $|a_j-c_j| le delta$ and $|b_j-d_j| le delta$ for each $j$.
(2) $sum_{j=1}^n |(a_j-c_j)-(b_j-d_j)| = 0$.
(3) $|cup_{j=1}^n [a_j,b_j]| = epsilon$.
(4) $|cup_{j=1}^n [c_j,d_j]| = sqrt{epsilon delta n}$.
Take any $L in mathbb{N}$, and let $R = frac{Ldelta}{epsilon}$ and $n = RL$. (It's not a big deal to assume $R in mathbb{N}$). For $0 le l le L-1$ and $1 le r le R$, let $[a_{lR+r},b_{lR+r}] = [l,l+frac{epsilon}{L}]$ and $[c_{lR+r},d_{lR+r}] = [l-delta+(r-1)frac{epsilon}{L},l-delta+rfrac{epsilon}{L}]$.
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
$endgroup$
Here is a disproof. Take any $epsilon,delta > 0$. We construct $([a_j,b_j])_{1 le j le n}, ([c_j,d_j])_{1 le j le n}$ such that:
(1) $|a_j-c_j| le delta$ and $|b_j-d_j| le delta$ for each $j$.
(2) $sum_{j=1}^n |(a_j-c_j)-(b_j-d_j)| = 0$.
(3) $|cup_{j=1}^n [a_j,b_j]| = epsilon$.
(4) $|cup_{j=1}^n [c_j,d_j]| = sqrt{epsilon delta n}$.
Take any $L in mathbb{N}$, and let $R = frac{Ldelta}{epsilon}$ and $n = RL$. (It's not a big deal to assume $R in mathbb{N}$). For $0 le l le L-1$ and $1 le r le R$, let $[a_{lR+r},b_{lR+r}] = [l,l+frac{epsilon}{L}]$ and $[c_{lR+r},d_{lR+r}] = [l-delta+(r-1)frac{epsilon}{L},l-delta+rfrac{epsilon}{L}]$.
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
edited Jan 8 at 7:15
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
answered Jan 8 at 6:56
mathworker21mathworker21
8,8971928
8,8971928
$begingroup$
+1 - although I did find it a "big deal" to see exactly how $epsilon$ and $delta$ are related (both in this answer, and in the question itself).
$endgroup$
– Calum Gilhooley
Jan 9 at 2:49
$begingroup$
@CalumGilhooley I'm a bit confused. In my answer, $epsilon$ and $delta$ were arbitrary. If you are objecting because $frac{Ldelta}{epsilon}$ might not be an integer, then we can just define $R = lceil frac{Ldelta}{epsilon} rceil$ instead. If I'm missing something, let me know.
$endgroup$
– mathworker21
Jan 9 at 7:56
$begingroup$
Thank you, that helps. I think you mean $R=lfloorfrac{Ldelta}{epsilon}rfloor$, which ensures that $[c_{lR},d_{lR}]$ and $[c_{lR+1},d_{lR+1}]$ don't overlap, for $1=1,2,ldots,L-1$. More simply, we can choose $R$ arbitrarily - as "large" as we like - and define $L=lceilfrac{Repsilon}{delta}rceil$. Then $leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = Repsilon$, which is unbounded by any function of $(epsilon,delta)$.
$endgroup$
– Calum Gilhooley
Jan 9 at 15:41
$begingroup$
@CalumGilhooley ah yes, that's simpler. thanks!
$endgroup$
– mathworker21
Jan 9 at 16:04
$begingroup$
Thank you, guys, very much for all your effort and this nice solution, although my claim which I hoped was true unfortunately turned out to be false.
$endgroup$
– sranthrop
Jan 9 at 20:44
add a comment |
$begingroup$
+1 - although I did find it a "big deal" to see exactly how $epsilon$ and $delta$ are related (both in this answer, and in the question itself).
$endgroup$
– Calum Gilhooley
Jan 9 at 2:49
$begingroup$
@CalumGilhooley I'm a bit confused. In my answer, $epsilon$ and $delta$ were arbitrary. If you are objecting because $frac{Ldelta}{epsilon}$ might not be an integer, then we can just define $R = lceil frac{Ldelta}{epsilon} rceil$ instead. If I'm missing something, let me know.
$endgroup$
– mathworker21
Jan 9 at 7:56
$begingroup$
Thank you, that helps. I think you mean $R=lfloorfrac{Ldelta}{epsilon}rfloor$, which ensures that $[c_{lR},d_{lR}]$ and $[c_{lR+1},d_{lR+1}]$ don't overlap, for $1=1,2,ldots,L-1$. More simply, we can choose $R$ arbitrarily - as "large" as we like - and define $L=lceilfrac{Repsilon}{delta}rceil$. Then $leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = Repsilon$, which is unbounded by any function of $(epsilon,delta)$.
$endgroup$
– Calum Gilhooley
Jan 9 at 15:41
$begingroup$
@CalumGilhooley ah yes, that's simpler. thanks!
$endgroup$
– mathworker21
Jan 9 at 16:04
$begingroup$
Thank you, guys, very much for all your effort and this nice solution, although my claim which I hoped was true unfortunately turned out to be false.
$endgroup$
– sranthrop
Jan 9 at 20:44
$begingroup$
+1 - although I did find it a "big deal" to see exactly how $epsilon$ and $delta$ are related (both in this answer, and in the question itself).
$endgroup$
– Calum Gilhooley
Jan 9 at 2:49
$begingroup$
+1 - although I did find it a "big deal" to see exactly how $epsilon$ and $delta$ are related (both in this answer, and in the question itself).
$endgroup$
– Calum Gilhooley
Jan 9 at 2:49
$begingroup$
@CalumGilhooley I'm a bit confused. In my answer, $epsilon$ and $delta$ were arbitrary. If you are objecting because $frac{Ldelta}{epsilon}$ might not be an integer, then we can just define $R = lceil frac{Ldelta}{epsilon} rceil$ instead. If I'm missing something, let me know.
$endgroup$
– mathworker21
Jan 9 at 7:56
$begingroup$
@CalumGilhooley I'm a bit confused. In my answer, $epsilon$ and $delta$ were arbitrary. If you are objecting because $frac{Ldelta}{epsilon}$ might not be an integer, then we can just define $R = lceil frac{Ldelta}{epsilon} rceil$ instead. If I'm missing something, let me know.
$endgroup$
– mathworker21
Jan 9 at 7:56
$begingroup$
Thank you, that helps. I think you mean $R=lfloorfrac{Ldelta}{epsilon}rfloor$, which ensures that $[c_{lR},d_{lR}]$ and $[c_{lR+1},d_{lR+1}]$ don't overlap, for $1=1,2,ldots,L-1$. More simply, we can choose $R$ arbitrarily - as "large" as we like - and define $L=lceilfrac{Repsilon}{delta}rceil$. Then $leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = Repsilon$, which is unbounded by any function of $(epsilon,delta)$.
$endgroup$
– Calum Gilhooley
Jan 9 at 15:41
$begingroup$
Thank you, that helps. I think you mean $R=lfloorfrac{Ldelta}{epsilon}rfloor$, which ensures that $[c_{lR},d_{lR}]$ and $[c_{lR+1},d_{lR+1}]$ don't overlap, for $1=1,2,ldots,L-1$. More simply, we can choose $R$ arbitrarily - as "large" as we like - and define $L=lceilfrac{Repsilon}{delta}rceil$. Then $leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = Repsilon$, which is unbounded by any function of $(epsilon,delta)$.
$endgroup$
– Calum Gilhooley
Jan 9 at 15:41
$begingroup$
@CalumGilhooley ah yes, that's simpler. thanks!
$endgroup$
– mathworker21
Jan 9 at 16:04
$begingroup$
@CalumGilhooley ah yes, that's simpler. thanks!
$endgroup$
– mathworker21
Jan 9 at 16:04
$begingroup$
Thank you, guys, very much for all your effort and this nice solution, although my claim which I hoped was true unfortunately turned out to be false.
$endgroup$
– sranthrop
Jan 9 at 20:44
$begingroup$
Thank you, guys, very much for all your effort and this nice solution, although my claim which I hoped was true unfortunately turned out to be false.
$endgroup$
– sranthrop
Jan 9 at 20:44
add a comment |
$begingroup$
(Disclaimer: this is more of a pedantic comment than an answer!)
Does it change the meaning of the question essentially if one drops
the stipulation that $delta$ and $epsilon$ are "small" - I
confess to being unable to understand what that statement means in
the present context - and asks instead:
Do there exist a number $h > 0$, and functions $f, g$ (of one and
two range-limited real variables, respectively), such that if
$0 < epsilon < h$, and condition 3 holds, then for all $delta$
such that $0 < delta < f(epsilon)$, and all $n$ and
$a_j, b_j, c_j, d_j$ ($1 leqslant j leqslant n$) such that
conditions 1 and 2 hold, we have
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert leqslant
g(epsilon, delta)$?
A more demanding (but I think less plausible) reformulation of the
question is as follows:
Do there exist numbers $h, h' > 0$, and a function $g$ (of two
range-limited real variables), such that if $0 < epsilon < h$ and
$0 < delta < h'$, then for all $n$ and $a_j, b_j, c_j, d_j$
($1 leqslant j leqslant n$) such that conditions 1 to 3 hold, we
have $left|bigcup_{j=1}^n[c_j,d_j]right| leqslant
g(epsilon, delta)$?
If the second reformulation accurately represents the meaning of the
question, then a special case of mathworker21's argument still gives
the answer "no".
On the other hand, if the first reformulation is satisfactory, then
mathworker21's construction cannot be applied, and the question
remains open.
[Not so! See the addendum below.]
Disposing of the second reformulation first:
Given $h, h' > 0$, take any $epsilon$ such
that $0 < epsilon < min{h, h', 1}$. Let $delta = epsilon$, let
$k$ be a positive integer, and, in mathworker21's construction, let
$L = R = k$, $n = k^2$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$.
Because this measure is not bounded above by any function of
$epsilon$ and $delta$, the required function $g$ cannot exist.
$square$
On the first reformulation (which I think is a more reasonable
interpretation of the question), we may still choose any value of
$epsilon$ that is "small enough", i.e. $epsilon < h$, and try
to derive a contradiction; but now we must allow $delta$ to take
on any "sufficiently small" value, i.e. any value less than some
unspecified $f(epsilon)$. In these circumstances, I don't see how
mathworker21's construction can be carried out.
(I don't know what the answer is in this case, but it seems best to
check my understanding of the question before possibly barking up
the wrong tree!)
Even if the first reformulation is valid, the same general
construction does still apply. This "answer" is therefore
probably best viewed as a long-winded comment in confirmation of
mathworker21's answer.
Given $h > 0$, and a function $f$, take any $epsilon$ such that
$0 < epsilon < min{h, 1}$, and any positive integer $k$ such
that $frac{epsilon}{k} < f(epsilon)$.
Define $delta = frac{epsilon}{k}$, In mathworker21's
construction, let $R = k$, $L = k^2$, $n = k^3$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$, much
as before. Again, the required function $g$ cannot exist.
$square$
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
$endgroup$
$begingroup$
Thanks a lot for this very detailed discussion and answer to my question!
$endgroup$
– sranthrop
Jan 9 at 20:45
$begingroup$
I've done something useful? ... Excuse me, I think I'm going to have to lie down for a minute! :)
$endgroup$
– Calum Gilhooley
Jan 9 at 20:51
add a comment |
$begingroup$
(Disclaimer: this is more of a pedantic comment than an answer!)
Does it change the meaning of the question essentially if one drops
the stipulation that $delta$ and $epsilon$ are "small" - I
confess to being unable to understand what that statement means in
the present context - and asks instead:
Do there exist a number $h > 0$, and functions $f, g$ (of one and
two range-limited real variables, respectively), such that if
$0 < epsilon < h$, and condition 3 holds, then for all $delta$
such that $0 < delta < f(epsilon)$, and all $n$ and
$a_j, b_j, c_j, d_j$ ($1 leqslant j leqslant n$) such that
conditions 1 and 2 hold, we have
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert leqslant
g(epsilon, delta)$?
A more demanding (but I think less plausible) reformulation of the
question is as follows:
Do there exist numbers $h, h' > 0$, and a function $g$ (of two
range-limited real variables), such that if $0 < epsilon < h$ and
$0 < delta < h'$, then for all $n$ and $a_j, b_j, c_j, d_j$
($1 leqslant j leqslant n$) such that conditions 1 to 3 hold, we
have $left|bigcup_{j=1}^n[c_j,d_j]right| leqslant
g(epsilon, delta)$?
If the second reformulation accurately represents the meaning of the
question, then a special case of mathworker21's argument still gives
the answer "no".
On the other hand, if the first reformulation is satisfactory, then
mathworker21's construction cannot be applied, and the question
remains open.
[Not so! See the addendum below.]
Disposing of the second reformulation first:
Given $h, h' > 0$, take any $epsilon$ such
that $0 < epsilon < min{h, h', 1}$. Let $delta = epsilon$, let
$k$ be a positive integer, and, in mathworker21's construction, let
$L = R = k$, $n = k^2$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$.
Because this measure is not bounded above by any function of
$epsilon$ and $delta$, the required function $g$ cannot exist.
$square$
On the first reformulation (which I think is a more reasonable
interpretation of the question), we may still choose any value of
$epsilon$ that is "small enough", i.e. $epsilon < h$, and try
to derive a contradiction; but now we must allow $delta$ to take
on any "sufficiently small" value, i.e. any value less than some
unspecified $f(epsilon)$. In these circumstances, I don't see how
mathworker21's construction can be carried out.
(I don't know what the answer is in this case, but it seems best to
check my understanding of the question before possibly barking up
the wrong tree!)
Even if the first reformulation is valid, the same general
construction does still apply. This "answer" is therefore
probably best viewed as a long-winded comment in confirmation of
mathworker21's answer.
Given $h > 0$, and a function $f$, take any $epsilon$ such that
$0 < epsilon < min{h, 1}$, and any positive integer $k$ such
that $frac{epsilon}{k} < f(epsilon)$.
Define $delta = frac{epsilon}{k}$, In mathworker21's
construction, let $R = k$, $L = k^2$, $n = k^3$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$, much
as before. Again, the required function $g$ cannot exist.
$square$
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
$endgroup$
$begingroup$
Thanks a lot for this very detailed discussion and answer to my question!
$endgroup$
– sranthrop
Jan 9 at 20:45
$begingroup$
I've done something useful? ... Excuse me, I think I'm going to have to lie down for a minute! :)
$endgroup$
– Calum Gilhooley
Jan 9 at 20:51
add a comment |
$begingroup$
(Disclaimer: this is more of a pedantic comment than an answer!)
Does it change the meaning of the question essentially if one drops
the stipulation that $delta$ and $epsilon$ are "small" - I
confess to being unable to understand what that statement means in
the present context - and asks instead:
Do there exist a number $h > 0$, and functions $f, g$ (of one and
two range-limited real variables, respectively), such that if
$0 < epsilon < h$, and condition 3 holds, then for all $delta$
such that $0 < delta < f(epsilon)$, and all $n$ and
$a_j, b_j, c_j, d_j$ ($1 leqslant j leqslant n$) such that
conditions 1 and 2 hold, we have
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert leqslant
g(epsilon, delta)$?
A more demanding (but I think less plausible) reformulation of the
question is as follows:
Do there exist numbers $h, h' > 0$, and a function $g$ (of two
range-limited real variables), such that if $0 < epsilon < h$ and
$0 < delta < h'$, then for all $n$ and $a_j, b_j, c_j, d_j$
($1 leqslant j leqslant n$) such that conditions 1 to 3 hold, we
have $left|bigcup_{j=1}^n[c_j,d_j]right| leqslant
g(epsilon, delta)$?
If the second reformulation accurately represents the meaning of the
question, then a special case of mathworker21's argument still gives
the answer "no".
On the other hand, if the first reformulation is satisfactory, then
mathworker21's construction cannot be applied, and the question
remains open.
[Not so! See the addendum below.]
Disposing of the second reformulation first:
Given $h, h' > 0$, take any $epsilon$ such
that $0 < epsilon < min{h, h', 1}$. Let $delta = epsilon$, let
$k$ be a positive integer, and, in mathworker21's construction, let
$L = R = k$, $n = k^2$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$.
Because this measure is not bounded above by any function of
$epsilon$ and $delta$, the required function $g$ cannot exist.
$square$
On the first reformulation (which I think is a more reasonable
interpretation of the question), we may still choose any value of
$epsilon$ that is "small enough", i.e. $epsilon < h$, and try
to derive a contradiction; but now we must allow $delta$ to take
on any "sufficiently small" value, i.e. any value less than some
unspecified $f(epsilon)$. In these circumstances, I don't see how
mathworker21's construction can be carried out.
(I don't know what the answer is in this case, but it seems best to
check my understanding of the question before possibly barking up
the wrong tree!)
Even if the first reformulation is valid, the same general
construction does still apply. This "answer" is therefore
probably best viewed as a long-winded comment in confirmation of
mathworker21's answer.
Given $h > 0$, and a function $f$, take any $epsilon$ such that
$0 < epsilon < min{h, 1}$, and any positive integer $k$ such
that $frac{epsilon}{k} < f(epsilon)$.
Define $delta = frac{epsilon}{k}$, In mathworker21's
construction, let $R = k$, $L = k^2$, $n = k^3$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$, much
as before. Again, the required function $g$ cannot exist.
$square$
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
$endgroup$
(Disclaimer: this is more of a pedantic comment than an answer!)
Does it change the meaning of the question essentially if one drops
the stipulation that $delta$ and $epsilon$ are "small" - I
confess to being unable to understand what that statement means in
the present context - and asks instead:
Do there exist a number $h > 0$, and functions $f, g$ (of one and
two range-limited real variables, respectively), such that if
$0 < epsilon < h$, and condition 3 holds, then for all $delta$
such that $0 < delta < f(epsilon)$, and all $n$ and
$a_j, b_j, c_j, d_j$ ($1 leqslant j leqslant n$) such that
conditions 1 and 2 hold, we have
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert leqslant
g(epsilon, delta)$?
A more demanding (but I think less plausible) reformulation of the
question is as follows:
Do there exist numbers $h, h' > 0$, and a function $g$ (of two
range-limited real variables), such that if $0 < epsilon < h$ and
$0 < delta < h'$, then for all $n$ and $a_j, b_j, c_j, d_j$
($1 leqslant j leqslant n$) such that conditions 1 to 3 hold, we
have $left|bigcup_{j=1}^n[c_j,d_j]right| leqslant
g(epsilon, delta)$?
If the second reformulation accurately represents the meaning of the
question, then a special case of mathworker21's argument still gives
the answer "no".
On the other hand, if the first reformulation is satisfactory, then
mathworker21's construction cannot be applied, and the question
remains open.
[Not so! See the addendum below.]
Disposing of the second reformulation first:
Given $h, h' > 0$, take any $epsilon$ such
that $0 < epsilon < min{h, h', 1}$. Let $delta = epsilon$, let
$k$ be a positive integer, and, in mathworker21's construction, let
$L = R = k$, $n = k^2$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$.
Because this measure is not bounded above by any function of
$epsilon$ and $delta$, the required function $g$ cannot exist.
$square$
On the first reformulation (which I think is a more reasonable
interpretation of the question), we may still choose any value of
$epsilon$ that is "small enough", i.e. $epsilon < h$, and try
to derive a contradiction; but now we must allow $delta$ to take
on any "sufficiently small" value, i.e. any value less than some
unspecified $f(epsilon)$. In these circumstances, I don't see how
mathworker21's construction can be carried out.
(I don't know what the answer is in this case, but it seems best to
check my understanding of the question before possibly barking up
the wrong tree!)
Even if the first reformulation is valid, the same general
construction does still apply. This "answer" is therefore
probably best viewed as a long-winded comment in confirmation of
mathworker21's answer.
Given $h > 0$, and a function $f$, take any $epsilon$ such that
$0 < epsilon < min{h, 1}$, and any positive integer $k$ such
that $frac{epsilon}{k} < f(epsilon)$.
Define $delta = frac{epsilon}{k}$, In mathworker21's
construction, let $R = k$, $L = k^2$, $n = k^3$. Then
$leftlvertbigcup_{j=1}^n[c_j,d_j]rightrvert = kepsilon$, much
as before. Again, the required function $g$ cannot exist.
$square$
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
edited Jan 9 at 2:42
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
answered Jan 9 at 1:24
Calum GilhooleyCalum Gilhooley
4,132529
4,132529
$begingroup$
Thanks a lot for this very detailed discussion and answer to my question!
$endgroup$
– sranthrop
Jan 9 at 20:45
$begingroup$
I've done something useful? ... Excuse me, I think I'm going to have to lie down for a minute! :)
$endgroup$
– Calum Gilhooley
Jan 9 at 20:51
add a comment |
$begingroup$
Thanks a lot for this very detailed discussion and answer to my question!
$endgroup$
– sranthrop
Jan 9 at 20:45
$begingroup$
I've done something useful? ... Excuse me, I think I'm going to have to lie down for a minute! :)
$endgroup$
– Calum Gilhooley
Jan 9 at 20:51
$begingroup$
Thanks a lot for this very detailed discussion and answer to my question!
$endgroup$
– sranthrop
Jan 9 at 20:45
$begingroup$
Thanks a lot for this very detailed discussion and answer to my question!
$endgroup$
– sranthrop
Jan 9 at 20:45
$begingroup$
I've done something useful? ... Excuse me, I think I'm going to have to lie down for a minute! :)
$endgroup$
– Calum Gilhooley
Jan 9 at 20:51
$begingroup$
I've done something useful? ... Excuse me, I think I'm going to have to lie down for a minute! :)
$endgroup$
– Calum Gilhooley
Jan 9 at 20:51
add a comment |
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$begingroup$
Have you run into any problem with constructing a counterexample?
$endgroup$
– Dap
Jan 6 at 15:44
$begingroup$
Yes, I clearly tried to find counterexamples. Condition 2 implies (triangle inequality) that the sum of the lengths of the intervals differ only by $delta$. Without condition 1 my claim is false: If all $[a_j,b_j]=[0,varepsilon]$, say, then 3 is met ($delta=varepsilon$). Now, take $[c_j,d_j]=[(j-1)varepsilon,jvarepsilon]$ for $j=1,ldots,n$. Then 2 is also met, but $left|bigcup_{j=1}^n[c_j,d_j]right|=|[0,nvarepsilon]|=nvarepsilon$. My overall problem is that I have not a clear idea on how geometrically combine 1 and 2. Therefore, I'm not entirely sure that my claim is true, though.
$endgroup$
– sranthrop
Jan 6 at 16:11
$begingroup$
@sranthrop I'm confused why you used Vitali covering lemma in your proof for when the $[a_j,b_j]$'s are mutually disjoint. Can't you just use $|cup_j [c_j,d_j]| le sum_j d_j-c_j$ and run through the argument?
$endgroup$
– mathworker21
Jan 7 at 19:31
$begingroup$
Thanks for asking and pointing this out. I think you are right. I'm working on this problem for weeks now and I'm so focused on some things. Thank you!
$endgroup$
– sranthrop
Jan 7 at 21:02
$begingroup$
how do you go from line 2 to 3 without disjoint intervals $[a_j,b_j]$?
$endgroup$
– Dunham
Jan 7 at 21:11