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Does every set have a power set?
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While reading the probability space in Wikipedia, I'd found the usual formulation is a triplet, which is $displaystyle (Omega ,mathcal F,P)$.
Upon my understanding, the middle $mathcal F$ is a power set of $Omega$ which will be allocated with real-valued probabiilty by $P$.
If every set in this nature has power set, there might be no necessity of introduction of $mathcal F$ I guess however, I've never thought of a set which doesn't have its power set.
Is there any set that doesn't have power set? or if not, which means every set has its power set, is there any plausible reason that $mathcal F$ is introduced in probability formulation?
probability elementary-set-theory
edited 8 mins ago
peterh
2,15441631
asked 2 hours ago
Beverlie
1,098318
add a comment |Â
up vote
3
down vote
favorite
While reading the probability space in Wikipedia, I'd found the usual formulation is a triplet, which is $displaystyle (Omega ,mathcal F,P)$.
Upon my understanding, the middle $mathcal F$ is a power set of $Omega$ which will be allocated with real-valued probabiilty by $P$.
If every set in this nature has power set, there might be no necessity of introduction of $mathcal F$ I guess however, I've never thought of a set which doesn't have its power set.
Is there any set that doesn't have power set? or if not, which means every set has its power set, is there any plausible reason that $mathcal F$ is introduced in probability formulation?
probability elementary-set-theory
edited 8 mins ago
peterh
2,15441631
asked 2 hours ago
Beverlie
1,098318
Could you say more specifically where you read that $mathcal F$ is the power set of $Omega$? This is not true in general, and I can't find where the Wikipedia article on probability spaces says that.
– joriki
2 hours ago
@joriki your link is what I exactly intended to refer in OP. as you mentioned, I thought myself intuitively reasonable to regard it as a power set.. but it's not. Thanks for the correction point!
– Beverlie
2 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
While reading the probability space in Wikipedia, I'd found the usual formulation is a triplet, which is $displaystyle (Omega ,mathcal F,P)$.
Upon my understanding, the middle $mathcal F$ is a power set of $Omega$ which will be allocated with real-valued probabiilty by $P$.
If every set in this nature has power set, there might be no necessity of introduction of $mathcal F$ I guess however, I've never thought of a set which doesn't have its power set.
Is there any set that doesn't have power set? or if not, which means every set has its power set, is there any plausible reason that $mathcal F$ is introduced in probability formulation?
probability elementary-set-theory
edited 8 mins ago
peterh
2,15441631
asked 2 hours ago
Beverlie
1,098318
While reading the probability space in Wikipedia, I'd found the usual formulation is a triplet, which is $displaystyle (Omega ,mathcal F,P)$.
Upon my understanding, the middle $mathcal F$ is a power set of $Omega$ which will be allocated with real-valued probabiilty by $P$.
If every set in this nature has power set, there might be no necessity of introduction of $mathcal F$ I guess however, I've never thought of a set which doesn't have its power set.
Is there any set that doesn't have power set? or if not, which means every set has its power set, is there any plausible reason that $mathcal F$ is introduced in probability formulation?
probability elementary-set-theory
probability elementary-set-theory
edited 8 mins ago
peterh
2,15441631
asked 2 hours ago
Beverlie
1,098318
edited 8 mins ago
peterh
2,15441631
asked 2 hours ago
Beverlie
1,098318
edited 8 mins ago
peterh
2,15441631
edited 8 mins ago
peterh
2,15441631
edited 8 mins ago
peterh
2,15441631
2,15441631
asked 2 hours ago
Beverlie
1,098318
asked 2 hours ago
Beverlie
1,098318
asked 2 hours ago
Beverlie
1,098318
1,098318
Could you say more specifically where you read that $mathcal F$ is the power set of $Omega$? This is not true in general, and I can't find where the Wikipedia article on probability spaces says that.
– joriki
2 hours ago
@joriki your link is what I exactly intended to refer in OP. as you mentioned, I thought myself intuitively reasonable to regard it as a power set.. but it's not. Thanks for the correction point!
– Beverlie
2 hours ago
add a comment |Â
Could you say more specifically where you read that $mathcal F$ is the power set of $Omega$? This is not true in general, and I can't find where the Wikipedia article on probability spaces says that.
– joriki
2 hours ago
@joriki your link is what I exactly intended to refer in OP. as you mentioned, I thought myself intuitively reasonable to regard it as a power set.. but it's not. Thanks for the correction point!
– Beverlie
2 hours ago
Could you say more specifically where you read that $mathcal F$ is the power set of $Omega$? This is not true in general, and I can't find where the Wikipedia article on probability spaces says that.
– joriki
2 hours ago
Could you say more specifically where you read that $mathcal F$ is the power set of $Omega$? This is not true in general, and I can't find where the Wikipedia article on probability spaces says that.
– joriki
2 hours ago
@joriki your link is what I exactly intended to refer in OP. as you mentioned, I thought myself intuitively reasonable to regard it as a power set.. but it's not. Thanks for the correction point!
– Beverlie
2 hours ago
@joriki your link is what I exactly intended to refer in OP. as you mentioned, I thought myself intuitively reasonable to regard it as a power set.. but it's not. Thanks for the correction point!
– Beverlie
2 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
8
down vote
In standard mathematics, every set has a power set. This is encoded in the Axiom of Power Set. However, your confusion lies with the definition of a probability space, not with set theory.
The set $mathcal F$ in a probability space $(Omega, mathcal F, P)$ is not necessarily the power set of $Omega$. The set $mathcal F$ is a subset of the power set $mathcal P(Omega)$. This $mathcal F$ is required to be a so-called sigma algebra, which tells you that it shares some properties in common with the full power set, but it need not be the full power set at all.
In particular, for any $Omega$, you can take $mathcal F = emptyset, Omega$, and this will be a sigma algebra on $Omega$. Unless $|Omega| leq 1$, it will not be the power set.
answered 2 hours ago
Mees de Vries
14.5k12348
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
2 hours ago
1
I think it is worth pointing out that in many cases we do not have a good way of using the power set of $Omega$ for the set $mathcal F$, so the practice of allowing the use of a subset of the power set is not just something we arbitrarily choose to do, it's practically a necessity.
– David K
38 mins ago
add a comment |Â
up vote
2
down vote
This $mathcalF$ is actually not the power set, but a sigma algebra. The power set is a sigma algebra and is often used, but sometimes probability theory requires smaller subsets of the power set in order to properly define the problem.
answered 2 hours ago
ZempZemp
212
add a comment |Â
up vote
1
down vote
I think the notation here is due to the fact that a probability space is, is particular, a finite measure space. So, in the more general sense, the set $mathcalF$ does not need to be the power set, it is sufficient that the set $mathcalF$ satisfy the axioms of a $sigma$-algebra of subsets of $Omega$. And yes, every set has a power set, it is guaranteed by the power set axiom in set theory. If you're interested, study some measure theory, which provides the mathematical tools for developing rigorous probability.
answered 2 hours ago
Nuntractatuses Amável
3059
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
In standard mathematics, every set has a power set. This is encoded in the Axiom of Power Set. However, your confusion lies with the definition of a probability space, not with set theory.
The set $mathcal F$ in a probability space $(Omega, mathcal F, P)$ is not necessarily the power set of $Omega$. The set $mathcal F$ is a subset of the power set $mathcal P(Omega)$. This $mathcal F$ is required to be a so-called sigma algebra, which tells you that it shares some properties in common with the full power set, but it need not be the full power set at all.
In particular, for any $Omega$, you can take $mathcal F = emptyset, Omega$, and this will be a sigma algebra on $Omega$. Unless $|Omega| leq 1$, it will not be the power set.
answered 2 hours ago
Mees de Vries
14.5k12348
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
2 hours ago
1
I think it is worth pointing out that in many cases we do not have a good way of using the power set of $Omega$ for the set $mathcal F$, so the practice of allowing the use of a subset of the power set is not just something we arbitrarily choose to do, it's practically a necessity.
– David K
38 mins ago
add a comment |Â
up vote
8
down vote
In standard mathematics, every set has a power set. This is encoded in the Axiom of Power Set. However, your confusion lies with the definition of a probability space, not with set theory.
The set $mathcal F$ in a probability space $(Omega, mathcal F, P)$ is not necessarily the power set of $Omega$. The set $mathcal F$ is a subset of the power set $mathcal P(Omega)$. This $mathcal F$ is required to be a so-called sigma algebra, which tells you that it shares some properties in common with the full power set, but it need not be the full power set at all.
In particular, for any $Omega$, you can take $mathcal F = emptyset, Omega$, and this will be a sigma algebra on $Omega$. Unless $|Omega| leq 1$, it will not be the power set.
answered 2 hours ago
Mees de Vries
14.5k12348
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
2 hours ago
1
I think it is worth pointing out that in many cases we do not have a good way of using the power set of $Omega$ for the set $mathcal F$, so the practice of allowing the use of a subset of the power set is not just something we arbitrarily choose to do, it's practically a necessity.
– David K
38 mins ago
add a comment |Â
up vote
8
down vote
up vote
8
down vote
In standard mathematics, every set has a power set. This is encoded in the Axiom of Power Set. However, your confusion lies with the definition of a probability space, not with set theory.
The set $mathcal F$ in a probability space $(Omega, mathcal F, P)$ is not necessarily the power set of $Omega$. The set $mathcal F$ is a subset of the power set $mathcal P(Omega)$. This $mathcal F$ is required to be a so-called sigma algebra, which tells you that it shares some properties in common with the full power set, but it need not be the full power set at all.
In particular, for any $Omega$, you can take $mathcal F = emptyset, Omega$, and this will be a sigma algebra on $Omega$. Unless $|Omega| leq 1$, it will not be the power set.
answered 2 hours ago
Mees de Vries
14.5k12348
In standard mathematics, every set has a power set. This is encoded in the Axiom of Power Set. However, your confusion lies with the definition of a probability space, not with set theory.
The set $mathcal F$ in a probability space $(Omega, mathcal F, P)$ is not necessarily the power set of $Omega$. The set $mathcal F$ is a subset of the power set $mathcal P(Omega)$. This $mathcal F$ is required to be a so-called sigma algebra, which tells you that it shares some properties in common with the full power set, but it need not be the full power set at all.
In particular, for any $Omega$, you can take $mathcal F = emptyset, Omega$, and this will be a sigma algebra on $Omega$. Unless $|Omega| leq 1$, it will not be the power set.
answered 2 hours ago
Mees de Vries
14.5k12348
answered 2 hours ago
Mees de Vries
14.5k12348
answered 2 hours ago
Mees de Vries
14.5k12348
answered 2 hours ago
Mees de Vries
14.5k12348
14.5k12348
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
2 hours ago
1
I think it is worth pointing out that in many cases we do not have a good way of using the power set of $Omega$ for the set $mathcal F$, so the practice of allowing the use of a subset of the power set is not just something we arbitrarily choose to do, it's practically a necessity.
– David K
38 mins ago
add a comment |Â
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
2 hours ago
1
I think it is worth pointing out that in many cases we do not have a good way of using the power set of $Omega$ for the set $mathcal F$, so the practice of allowing the use of a subset of the power set is not just something we arbitrarily choose to do, it's practically a necessity.
– David K
38 mins ago
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
2 hours ago
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
2 hours ago
1
1
I think it is worth pointing out that in many cases we do not have a good way of using the power set of $Omega$ for the set $mathcal F$, so the practice of allowing the use of a subset of the power set is not just something we arbitrarily choose to do, it's practically a necessity.
– David K
38 mins ago
I think it is worth pointing out that in many cases we do not have a good way of using the power set of $Omega$ for the set $mathcal F$, so the practice of allowing the use of a subset of the power set is not just something we arbitrarily choose to do, it's practically a necessity.
– David K
38 mins ago
add a comment |Â
up vote
2
down vote
This $mathcalF$ is actually not the power set, but a sigma algebra. The power set is a sigma algebra and is often used, but sometimes probability theory requires smaller subsets of the power set in order to properly define the problem.
answered 2 hours ago
ZempZemp
212
add a comment |Â
up vote
2
down vote
This $mathcalF$ is actually not the power set, but a sigma algebra. The power set is a sigma algebra and is often used, but sometimes probability theory requires smaller subsets of the power set in order to properly define the problem.
answered 2 hours ago
ZempZemp
212
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This $mathcalF$ is actually not the power set, but a sigma algebra. The power set is a sigma algebra and is often used, but sometimes probability theory requires smaller subsets of the power set in order to properly define the problem.
answered 2 hours ago
ZempZemp
212
This $mathcalF$ is actually not the power set, but a sigma algebra. The power set is a sigma algebra and is often used, but sometimes probability theory requires smaller subsets of the power set in order to properly define the problem.
answered 2 hours ago
ZempZemp
212
answered 2 hours ago
ZempZemp
212
answered 2 hours ago
ZempZemp
212
answered 2 hours ago
ZempZemp
212
212
add a comment |Â
add a comment |Â
up vote
1
down vote
I think the notation here is due to the fact that a probability space is, is particular, a finite measure space. So, in the more general sense, the set $mathcalF$ does not need to be the power set, it is sufficient that the set $mathcalF$ satisfy the axioms of a $sigma$-algebra of subsets of $Omega$. And yes, every set has a power set, it is guaranteed by the power set axiom in set theory. If you're interested, study some measure theory, which provides the mathematical tools for developing rigorous probability.
answered 2 hours ago
Nuntractatuses Amável
3059
add a comment |Â
up vote
1
down vote
I think the notation here is due to the fact that a probability space is, is particular, a finite measure space. So, in the more general sense, the set $mathcalF$ does not need to be the power set, it is sufficient that the set $mathcalF$ satisfy the axioms of a $sigma$-algebra of subsets of $Omega$. And yes, every set has a power set, it is guaranteed by the power set axiom in set theory. If you're interested, study some measure theory, which provides the mathematical tools for developing rigorous probability.
answered 2 hours ago
Nuntractatuses Amável
3059
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think the notation here is due to the fact that a probability space is, is particular, a finite measure space. So, in the more general sense, the set $mathcalF$ does not need to be the power set, it is sufficient that the set $mathcalF$ satisfy the axioms of a $sigma$-algebra of subsets of $Omega$. And yes, every set has a power set, it is guaranteed by the power set axiom in set theory. If you're interested, study some measure theory, which provides the mathematical tools for developing rigorous probability.
answered 2 hours ago
Nuntractatuses Amável
3059
I think the notation here is due to the fact that a probability space is, is particular, a finite measure space. So, in the more general sense, the set $mathcalF$ does not need to be the power set, it is sufficient that the set $mathcalF$ satisfy the axioms of a $sigma$-algebra of subsets of $Omega$. And yes, every set has a power set, it is guaranteed by the power set axiom in set theory. If you're interested, study some measure theory, which provides the mathematical tools for developing rigorous probability.
answered 2 hours ago
Nuntractatuses Amável
3059
answered 2 hours ago
Nuntractatuses Amável
3059
answered 2 hours ago
Nuntractatuses Amável
3059
answered 2 hours ago
Nuntractatuses Amável
3059
3059
add a comment |Â
add a comment |Â
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Could you say more specifically where you read that $mathcal F$ is the power set of $Omega$? This is not true in general, and I can't find where the Wikipedia article on probability spaces says that.
– joriki
2 hours ago
@joriki your link is what I exactly intended to refer in OP. as you mentioned, I thought myself intuitively reasonable to regard it as a power set.. but it's not. Thanks for the correction point!
– Beverlie
2 hours ago