Partitions of $mathbb{R}^d$
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Let $mathbb{R}^d=Pi_{i=1}^d mathbb{R},$ does there exist a covering ${E_n:nin mathbb{Z}^d}$ of subsets $E_nsubseteq mathbb{R}^d$ that satisfies the following: $1)$ $ E_n$'s are pairwise disjoint, $2) $ for any $x in E_n, yin E_m, $ $x+yin E_{n+m},$ and $3)$ $sup_{nin mathbb{Z}^d}|E_n|<infty?$ Note that I'm using $|M|$ to denote the Lebesgue measure of any measurable set $Msubseteq mathbb{R}^d.$
The collections given by ${n+[0,1)^{d} : nin mathbb{Z}^d }$ easily comes to mind, but it doesn't seem to satisfy condition $2$ with some easy counterexample like $(1+0.5)+(2+0.5)=3+1neq 3+q_3$ where $0leq q_3<1$ in the case $d=1$.
real-analysis
asked Dec 7 '18 at 9:18
KuromeKurome
378114
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add a comment |
$begingroup$
Let $mathbb{R}^d=Pi_{i=1}^d mathbb{R},$ does there exist a covering ${E_n:nin mathbb{Z}^d}$ of subsets $E_nsubseteq mathbb{R}^d$ that satisfies the following: $1)$ $ E_n$'s are pairwise disjoint, $2) $ for any $x in E_n, yin E_m, $ $x+yin E_{n+m},$ and $3)$ $sup_{nin mathbb{Z}^d}|E_n|<infty?$ Note that I'm using $|M|$ to denote the Lebesgue measure of any measurable set $Msubseteq mathbb{R}^d.$
The collections given by ${n+[0,1)^{d} : nin mathbb{Z}^d }$ easily comes to mind, but it doesn't seem to satisfy condition $2$ with some easy counterexample like $(1+0.5)+(2+0.5)=3+1neq 3+q_3$ where $0leq q_3<1$ in the case $d=1$.
real-analysis
asked Dec 7 '18 at 9:18
KuromeKurome
378114
$endgroup$
1
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the first equation looks nonsense to me
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– mathworker21
Dec 7 '18 at 9:26
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Does the empty set cover $mathbb{R}^d?$
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– Kurome
Dec 7 '18 at 9:27
1
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I don't think so if the sets should be measurable. Say focus on the real line. Some $E_n$ must have positive measure. Then $E_{2n}$ contains $E_n+E_n$, so by Steinhaus theorem (sum version math.stackexchange.com/questions/86209/…) it also contains an interval. But then $E_{4n}, E_{6n}$ and so on must contain longer and longer intervals. I haven't thought the details through though.
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– Michal Adamaszek
Dec 7 '18 at 9:40
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Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals.
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– Kurome
Dec 7 '18 at 10:00
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No but if $U$ contains an interval of length $a$ and $V$ contains an interval of length $b$ then surely $U+V$ contains an interval of length $a+b$.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 10:03
add a comment |
$begingroup$
Let $mathbb{R}^d=Pi_{i=1}^d mathbb{R},$ does there exist a covering ${E_n:nin mathbb{Z}^d}$ of subsets $E_nsubseteq mathbb{R}^d$ that satisfies the following: $1)$ $ E_n$'s are pairwise disjoint, $2) $ for any $x in E_n, yin E_m, $ $x+yin E_{n+m},$ and $3)$ $sup_{nin mathbb{Z}^d}|E_n|<infty?$ Note that I'm using $|M|$ to denote the Lebesgue measure of any measurable set $Msubseteq mathbb{R}^d.$
The collections given by ${n+[0,1)^{d} : nin mathbb{Z}^d }$ easily comes to mind, but it doesn't seem to satisfy condition $2$ with some easy counterexample like $(1+0.5)+(2+0.5)=3+1neq 3+q_3$ where $0leq q_3<1$ in the case $d=1$.
real-analysis
asked Dec 7 '18 at 9:18
KuromeKurome
378114
$endgroup$
Let $mathbb{R}^d=Pi_{i=1}^d mathbb{R},$ does there exist a covering ${E_n:nin mathbb{Z}^d}$ of subsets $E_nsubseteq mathbb{R}^d$ that satisfies the following: $1)$ $ E_n$'s are pairwise disjoint, $2) $ for any $x in E_n, yin E_m, $ $x+yin E_{n+m},$ and $3)$ $sup_{nin mathbb{Z}^d}|E_n|<infty?$ Note that I'm using $|M|$ to denote the Lebesgue measure of any measurable set $Msubseteq mathbb{R}^d.$
The collections given by ${n+[0,1)^{d} : nin mathbb{Z}^d }$ easily comes to mind, but it doesn't seem to satisfy condition $2$ with some easy counterexample like $(1+0.5)+(2+0.5)=3+1neq 3+q_3$ where $0leq q_3<1$ in the case $d=1$.
real-analysis
real-analysis
asked Dec 7 '18 at 9:18
KuromeKurome
378114
asked Dec 7 '18 at 9:18
KuromeKurome
378114
edited Dec 7 '18 at 10:14
Kurome
asked Dec 7 '18 at 9:18
KuromeKurome
378114
asked Dec 7 '18 at 9:18
KuromeKurome
378114
asked Dec 7 '18 at 9:18
KuromeKurome
378114
378114
1
$begingroup$
the first equation looks nonsense to me
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– mathworker21
Dec 7 '18 at 9:26
$begingroup$
Does the empty set cover $mathbb{R}^d?$
$endgroup$
– Kurome
Dec 7 '18 at 9:27
1
$begingroup$
I don't think so if the sets should be measurable. Say focus on the real line. Some $E_n$ must have positive measure. Then $E_{2n}$ contains $E_n+E_n$, so by Steinhaus theorem (sum version math.stackexchange.com/questions/86209/…) it also contains an interval. But then $E_{4n}, E_{6n}$ and so on must contain longer and longer intervals. I haven't thought the details through though.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 9:40
$begingroup$
Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals.
$endgroup$
– Kurome
Dec 7 '18 at 10:00
$begingroup$
No but if $U$ contains an interval of length $a$ and $V$ contains an interval of length $b$ then surely $U+V$ contains an interval of length $a+b$.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 10:03
add a comment |
1
$begingroup$
the first equation looks nonsense to me
$endgroup$
– mathworker21
Dec 7 '18 at 9:26
$begingroup$
Does the empty set cover $mathbb{R}^d?$
$endgroup$
– Kurome
Dec 7 '18 at 9:27
1
$begingroup$
I don't think so if the sets should be measurable. Say focus on the real line. Some $E_n$ must have positive measure. Then $E_{2n}$ contains $E_n+E_n$, so by Steinhaus theorem (sum version math.stackexchange.com/questions/86209/…) it also contains an interval. But then $E_{4n}, E_{6n}$ and so on must contain longer and longer intervals. I haven't thought the details through though.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 9:40
$begingroup$
Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals.
$endgroup$
– Kurome
Dec 7 '18 at 10:00
$begingroup$
No but if $U$ contains an interval of length $a$ and $V$ contains an interval of length $b$ then surely $U+V$ contains an interval of length $a+b$.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 10:03
1
1
$begingroup$
the first equation looks nonsense to me
$endgroup$
– mathworker21
Dec 7 '18 at 9:26
$begingroup$
the first equation looks nonsense to me
$endgroup$
– mathworker21
Dec 7 '18 at 9:26
$begingroup$
Does the empty set cover $mathbb{R}^d?$
$endgroup$
– Kurome
Dec 7 '18 at 9:27
$begingroup$
Does the empty set cover $mathbb{R}^d?$
$endgroup$
– Kurome
Dec 7 '18 at 9:27
1
1
$begingroup$
I don't think so if the sets should be measurable. Say focus on the real line. Some $E_n$ must have positive measure. Then $E_{2n}$ contains $E_n+E_n$, so by Steinhaus theorem (sum version math.stackexchange.com/questions/86209/…) it also contains an interval. But then $E_{4n}, E_{6n}$ and so on must contain longer and longer intervals. I haven't thought the details through though.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 9:40
$begingroup$
I don't think so if the sets should be measurable. Say focus on the real line. Some $E_n$ must have positive measure. Then $E_{2n}$ contains $E_n+E_n$, so by Steinhaus theorem (sum version math.stackexchange.com/questions/86209/…) it also contains an interval. But then $E_{4n}, E_{6n}$ and so on must contain longer and longer intervals. I haven't thought the details through though.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 9:40
$begingroup$
Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals.
$endgroup$
– Kurome
Dec 7 '18 at 10:00
$begingroup$
Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals.
$endgroup$
– Kurome
Dec 7 '18 at 10:00
$begingroup$
No but if $U$ contains an interval of length $a$ and $V$ contains an interval of length $b$ then surely $U+V$ contains an interval of length $a+b$.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 10:03
$begingroup$
No but if $U$ contains an interval of length $a$ and $V$ contains an interval of length $b$ then surely $U+V$ contains an interval of length $a+b$.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 10:03
add a comment |
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1
$begingroup$
the first equation looks nonsense to me
$endgroup$
– mathworker21
Dec 7 '18 at 9:26
$begingroup$
Does the empty set cover $mathbb{R}^d?$
$endgroup$
– Kurome
Dec 7 '18 at 9:27
1
$begingroup$
I don't think so if the sets should be measurable. Say focus on the real line. Some $E_n$ must have positive measure. Then $E_{2n}$ contains $E_n+E_n$, so by Steinhaus theorem (sum version math.stackexchange.com/questions/86209/…) it also contains an interval. But then $E_{4n}, E_{6n}$ and so on must contain longer and longer intervals. I haven't thought the details through though.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 9:40
$begingroup$
Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals.
$endgroup$
– Kurome
Dec 7 '18 at 10:00
$begingroup$
No but if $U$ contains an interval of length $a$ and $V$ contains an interval of length $b$ then surely $U+V$ contains an interval of length $a+b$.
$endgroup$
– Michal Adamaszek
Dec 7 '18 at 10:03