Every finite $sigma$-algebra is of the form…?
$begingroup$
Let $mathcal{F}$ be a finite $sigma$-algebra.
The problem asks to show there exists a partition $G = { G_1,dots,G_n }$ of $Omega$ such that for all $A in mathcal{F}$, $A$ is the union of all or some $G_i$:
$$A = bigcup_{i in I} G_i$$
The existence of a partition is immediate from the definition of a $sigma$-algebra, but I'm not sure how to use the fact $mathcal{F}$ is finite to construct the generating set $G$. Specifically, to show that every member of $mathcal{F}$ can be generated from a single partition using only union.
Can someone give me a hint?
Not a homework problem, I am working through a textbook for self-study.
probability self-learning
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
Let $mathcal{F}$ be a finite $sigma$-algebra.
The problem asks to show there exists a partition $G = { G_1,dots,G_n }$ of $Omega$ such that for all $A in mathcal{F}$, $A$ is the union of all or some $G_i$:
$$A = bigcup_{i in I} G_i$$
The existence of a partition is immediate from the definition of a $sigma$-algebra, but I'm not sure how to use the fact $mathcal{F}$ is finite to construct the generating set $G$. Specifically, to show that every member of $mathcal{F}$ can be generated from a single partition using only union.
Can someone give me a hint?
Not a homework problem, I am working through a textbook for self-study.
probability self-learning
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
Let $mathcal{F}$ be a finite $sigma$-algebra.
The problem asks to show there exists a partition $G = { G_1,dots,G_n }$ of $Omega$ such that for all $A in mathcal{F}$, $A$ is the union of all or some $G_i$:
$$A = bigcup_{i in I} G_i$$
The existence of a partition is immediate from the definition of a $sigma$-algebra, but I'm not sure how to use the fact $mathcal{F}$ is finite to construct the generating set $G$. Specifically, to show that every member of $mathcal{F}$ can be generated from a single partition using only union.
Can someone give me a hint?
Not a homework problem, I am working through a textbook for self-study.
probability self-learning
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
$endgroup$
Let $mathcal{F}$ be a finite $sigma$-algebra.
The problem asks to show there exists a partition $G = { G_1,dots,G_n }$ of $Omega$ such that for all $A in mathcal{F}$, $A$ is the union of all or some $G_i$:
$$A = bigcup_{i in I} G_i$$
The existence of a partition is immediate from the definition of a $sigma$-algebra, but I'm not sure how to use the fact $mathcal{F}$ is finite to construct the generating set $G$. Specifically, to show that every member of $mathcal{F}$ can be generated from a single partition using only union.
Can someone give me a hint?
Not a homework problem, I am working through a textbook for self-study.
probability self-learning
probability self-learning
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
edited Dec 7 '18 at 15:37
zhoraster
15.7k21752
15.7k21752
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
asked Dec 7 '18 at 9:49
XiaomiXiaomi
1,057115
1,057115
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $mathcal F ={B_1,B_2,cdots ,B_n}$. Consider all sets of the type $A_1cap A_2cap cdots cap A_n$ where $A_i$ is either $B_i$ or $B_i^{c}$ for each $i$. You get $2^{n}$ such intersection but many of them be just empty. It is now routine to check that the nonempty sets in this collection form a partition of $Omega$ and that each $B_i$ is a union of some sets in this partition. For example, you can write $B_1$ as a union of such sets by fixing $A_1$ to be $B_1$ and varying $A_2,A_3,cdots,A_n$.
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
$begingroup$
Thanks! That is a lot more intuitive
$endgroup$
– Xiaomi
Dec 7 '18 at 9:55
add a comment |
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1 Answer
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1 Answer
1
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oldest
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$begingroup$
Let $mathcal F ={B_1,B_2,cdots ,B_n}$. Consider all sets of the type $A_1cap A_2cap cdots cap A_n$ where $A_i$ is either $B_i$ or $B_i^{c}$ for each $i$. You get $2^{n}$ such intersection but many of them be just empty. It is now routine to check that the nonempty sets in this collection form a partition of $Omega$ and that each $B_i$ is a union of some sets in this partition. For example, you can write $B_1$ as a union of such sets by fixing $A_1$ to be $B_1$ and varying $A_2,A_3,cdots,A_n$.
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
$begingroup$
Thanks! That is a lot more intuitive
$endgroup$
– Xiaomi
Dec 7 '18 at 9:55
add a comment |
$begingroup$
Let $mathcal F ={B_1,B_2,cdots ,B_n}$. Consider all sets of the type $A_1cap A_2cap cdots cap A_n$ where $A_i$ is either $B_i$ or $B_i^{c}$ for each $i$. You get $2^{n}$ such intersection but many of them be just empty. It is now routine to check that the nonempty sets in this collection form a partition of $Omega$ and that each $B_i$ is a union of some sets in this partition. For example, you can write $B_1$ as a union of such sets by fixing $A_1$ to be $B_1$ and varying $A_2,A_3,cdots,A_n$.
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
$begingroup$
Thanks! That is a lot more intuitive
$endgroup$
– Xiaomi
Dec 7 '18 at 9:55
add a comment |
$begingroup$
Let $mathcal F ={B_1,B_2,cdots ,B_n}$. Consider all sets of the type $A_1cap A_2cap cdots cap A_n$ where $A_i$ is either $B_i$ or $B_i^{c}$ for each $i$. You get $2^{n}$ such intersection but many of them be just empty. It is now routine to check that the nonempty sets in this collection form a partition of $Omega$ and that each $B_i$ is a union of some sets in this partition. For example, you can write $B_1$ as a union of such sets by fixing $A_1$ to be $B_1$ and varying $A_2,A_3,cdots,A_n$.
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
Let $mathcal F ={B_1,B_2,cdots ,B_n}$. Consider all sets of the type $A_1cap A_2cap cdots cap A_n$ where $A_i$ is either $B_i$ or $B_i^{c}$ for each $i$. You get $2^{n}$ such intersection but many of them be just empty. It is now routine to check that the nonempty sets in this collection form a partition of $Omega$ and that each $B_i$ is a union of some sets in this partition. For example, you can write $B_1$ as a union of such sets by fixing $A_1$ to be $B_1$ and varying $A_2,A_3,cdots,A_n$.
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
edited Dec 7 '18 at 15:38
zhoraster
15.7k21752
15.7k21752
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
answered Dec 7 '18 at 9:54
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
58.2k42161
$begingroup$
Thanks! That is a lot more intuitive
$endgroup$
– Xiaomi
Dec 7 '18 at 9:55
add a comment |
$begingroup$
Thanks! That is a lot more intuitive
$endgroup$
– Xiaomi
Dec 7 '18 at 9:55
$begingroup$
Thanks! That is a lot more intuitive
$endgroup$
– Xiaomi
Dec 7 '18 at 9:55
$begingroup$
Thanks! That is a lot more intuitive
$endgroup$
– Xiaomi
Dec 7 '18 at 9:55
add a comment |
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