Mixing
Prove or disprove that every finite $sigma$-algebra on $Omega$ is generated by a finite partition of $Omega$.
$begingroup$
Prove or disprove that every finite $sigma$-algebra on $Omega$ is generated by a finite partition of $Omega$.
I have a feeling this must be true, but I could not do more than the following:
Being $Sigma$ a finite $sigma$-algebra on $Omega$, I defined a set $S$ as the set of intersections of two arbitrary sets in $Sigma$, but I don't know if there is a partition $P$ which is a subset of $S$ and if there is, if the $sigma$-algebra generated by $P$ is $Sigma$. Any help would be appreciated. Thanks.
And if the fact that every finite $sigma$-algebra can be generated by a finite partition is false, what conditions do the finite $sigma$-algebra need to have in order for it to be true?
measure-theory set-partition
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
$endgroup$
add a comment |
$begingroup$
Prove or disprove that every finite $sigma$-algebra on $Omega$ is generated by a finite partition of $Omega$.
I have a feeling this must be true, but I could not do more than the following:
Being $Sigma$ a finite $sigma$-algebra on $Omega$, I defined a set $S$ as the set of intersections of two arbitrary sets in $Sigma$, but I don't know if there is a partition $P$ which is a subset of $S$ and if there is, if the $sigma$-algebra generated by $P$ is $Sigma$. Any help would be appreciated. Thanks.
And if the fact that every finite $sigma$-algebra can be generated by a finite partition is false, what conditions do the finite $sigma$-algebra need to have in order for it to be true?
measure-theory set-partition
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
$endgroup$
add a comment |
$begingroup$
Prove or disprove that every finite $sigma$-algebra on $Omega$ is generated by a finite partition of $Omega$.
I have a feeling this must be true, but I could not do more than the following:
Being $Sigma$ a finite $sigma$-algebra on $Omega$, I defined a set $S$ as the set of intersections of two arbitrary sets in $Sigma$, but I don't know if there is a partition $P$ which is a subset of $S$ and if there is, if the $sigma$-algebra generated by $P$ is $Sigma$. Any help would be appreciated. Thanks.
And if the fact that every finite $sigma$-algebra can be generated by a finite partition is false, what conditions do the finite $sigma$-algebra need to have in order for it to be true?
measure-theory set-partition
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
$endgroup$
Prove or disprove that every finite $sigma$-algebra on $Omega$ is generated by a finite partition of $Omega$.
I have a feeling this must be true, but I could not do more than the following:
Being $Sigma$ a finite $sigma$-algebra on $Omega$, I defined a set $S$ as the set of intersections of two arbitrary sets in $Sigma$, but I don't know if there is a partition $P$ which is a subset of $S$ and if there is, if the $sigma$-algebra generated by $P$ is $Sigma$. Any help would be appreciated. Thanks.
And if the fact that every finite $sigma$-algebra can be generated by a finite partition is false, what conditions do the finite $sigma$-algebra need to have in order for it to be true?
measure-theory set-partition
measure-theory set-partition
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
asked Jan 7 at 23:27
GarmekainGarmekain
1,345720
1,345720
add a comment |
add a comment |
1 Answer
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Intersection of just two sets from the sigma algebra won't suffice. Let $mathcal A ={A_1,A_2,cdots,A_n}$ be a finite sigma algebra. Verify the following facts: sets of the form $B_1cap B_2capcdots cap B_n$ where $B_i$ is either $A_i$ or $A_i^{c}$ for each $i$ are disjoint. Though many of them may be empty, their union is $Omega$. Also prove that each $A_j$ is a union of those sets of the above form where $B_j=A_j$. Now you can easily verify that the partition we have constructed generates the sigma algebra $mathcal A$.
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
I don't understand the second condition to be verified. What do you mean by saying that each $A_j$ is a union of sets with said form where $B_j=A_j$, Does that mean that each $A_j$ is the union of sets with form $A_1capcdotscap A_n$?
$endgroup$
– Garmekain
Jan 8 at 0:24
1
$begingroup$
@G Consider $A_1$. Now look at all sets of the form $A_1cap B_2cap B_3cap cdotscap B_n$ where $B_2=A_2$ or $B_2=A_2^{c}$,...,$B_n=A_n$ or$B_n=A_n^{c}$. The union of these sets is $A_1$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:40
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Intersection of just two sets from the sigma algebra won't suffice. Let $mathcal A ={A_1,A_2,cdots,A_n}$ be a finite sigma algebra. Verify the following facts: sets of the form $B_1cap B_2capcdots cap B_n$ where $B_i$ is either $A_i$ or $A_i^{c}$ for each $i$ are disjoint. Though many of them may be empty, their union is $Omega$. Also prove that each $A_j$ is a union of those sets of the above form where $B_j=A_j$. Now you can easily verify that the partition we have constructed generates the sigma algebra $mathcal A$.
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
I don't understand the second condition to be verified. What do you mean by saying that each $A_j$ is a union of sets with said form where $B_j=A_j$, Does that mean that each $A_j$ is the union of sets with form $A_1capcdotscap A_n$?
$endgroup$
– Garmekain
Jan 8 at 0:24
1
$begingroup$
@G Consider $A_1$. Now look at all sets of the form $A_1cap B_2cap B_3cap cdotscap B_n$ where $B_2=A_2$ or $B_2=A_2^{c}$,...,$B_n=A_n$ or$B_n=A_n^{c}$. The union of these sets is $A_1$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:40
add a comment |
$begingroup$
Intersection of just two sets from the sigma algebra won't suffice. Let $mathcal A ={A_1,A_2,cdots,A_n}$ be a finite sigma algebra. Verify the following facts: sets of the form $B_1cap B_2capcdots cap B_n$ where $B_i$ is either $A_i$ or $A_i^{c}$ for each $i$ are disjoint. Though many of them may be empty, their union is $Omega$. Also prove that each $A_j$ is a union of those sets of the above form where $B_j=A_j$. Now you can easily verify that the partition we have constructed generates the sigma algebra $mathcal A$.
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
I don't understand the second condition to be verified. What do you mean by saying that each $A_j$ is a union of sets with said form where $B_j=A_j$, Does that mean that each $A_j$ is the union of sets with form $A_1capcdotscap A_n$?
$endgroup$
– Garmekain
Jan 8 at 0:24
1
$begingroup$
@G Consider $A_1$. Now look at all sets of the form $A_1cap B_2cap B_3cap cdotscap B_n$ where $B_2=A_2$ or $B_2=A_2^{c}$,...,$B_n=A_n$ or$B_n=A_n^{c}$. The union of these sets is $A_1$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:40
add a comment |
$begingroup$
Intersection of just two sets from the sigma algebra won't suffice. Let $mathcal A ={A_1,A_2,cdots,A_n}$ be a finite sigma algebra. Verify the following facts: sets of the form $B_1cap B_2capcdots cap B_n$ where $B_i$ is either $A_i$ or $A_i^{c}$ for each $i$ are disjoint. Though many of them may be empty, their union is $Omega$. Also prove that each $A_j$ is a union of those sets of the above form where $B_j=A_j$. Now you can easily verify that the partition we have constructed generates the sigma algebra $mathcal A$.
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
Intersection of just two sets from the sigma algebra won't suffice. Let $mathcal A ={A_1,A_2,cdots,A_n}$ be a finite sigma algebra. Verify the following facts: sets of the form $B_1cap B_2capcdots cap B_n$ where $B_i$ is either $A_i$ or $A_i^{c}$ for each $i$ are disjoint. Though many of them may be empty, their union is $Omega$. Also prove that each $A_j$ is a union of those sets of the above form where $B_j=A_j$. Now you can easily verify that the partition we have constructed generates the sigma algebra $mathcal A$.
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 7 at 23:49
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
56.6k42159
$begingroup$
I don't understand the second condition to be verified. What do you mean by saying that each $A_j$ is a union of sets with said form where $B_j=A_j$, Does that mean that each $A_j$ is the union of sets with form $A_1capcdotscap A_n$?
$endgroup$
– Garmekain
Jan 8 at 0:24
1
$begingroup$
@G Consider $A_1$. Now look at all sets of the form $A_1cap B_2cap B_3cap cdotscap B_n$ where $B_2=A_2$ or $B_2=A_2^{c}$,...,$B_n=A_n$ or$B_n=A_n^{c}$. The union of these sets is $A_1$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:40
add a comment |
$begingroup$
I don't understand the second condition to be verified. What do you mean by saying that each $A_j$ is a union of sets with said form where $B_j=A_j$, Does that mean that each $A_j$ is the union of sets with form $A_1capcdotscap A_n$?
$endgroup$
– Garmekain
Jan 8 at 0:24
1
$begingroup$
@G Consider $A_1$. Now look at all sets of the form $A_1cap B_2cap B_3cap cdotscap B_n$ where $B_2=A_2$ or $B_2=A_2^{c}$,...,$B_n=A_n$ or$B_n=A_n^{c}$. The union of these sets is $A_1$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:40
$begingroup$
I don't understand the second condition to be verified. What do you mean by saying that each $A_j$ is a union of sets with said form where $B_j=A_j$, Does that mean that each $A_j$ is the union of sets with form $A_1capcdotscap A_n$?
$endgroup$
– Garmekain
Jan 8 at 0:24
$begingroup$
I don't understand the second condition to be verified. What do you mean by saying that each $A_j$ is a union of sets with said form where $B_j=A_j$, Does that mean that each $A_j$ is the union of sets with form $A_1capcdotscap A_n$?
$endgroup$
– Garmekain
Jan 8 at 0:24
1
1
$begingroup$
@G Consider $A_1$. Now look at all sets of the form $A_1cap B_2cap B_3cap cdotscap B_n$ where $B_2=A_2$ or $B_2=A_2^{c}$,...,$B_n=A_n$ or$B_n=A_n^{c}$. The union of these sets is $A_1$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:40
$begingroup$
@G Consider $A_1$. Now look at all sets of the form $A_1cap B_2cap B_3cap cdotscap B_n$ where $B_2=A_2$ or $B_2=A_2^{c}$,...,$B_n=A_n$ or$B_n=A_n^{c}$. The union of these sets is $A_1$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:40
add a comment |
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