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Does Every Set have 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 1 hour ago
Asaf Karagila♦
295k32410738
asked 1 hour ago
Beverlie
1,088318
add a comment |Â
up vote
1
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 1 hour ago
Asaf Karagila♦
295k32410738
asked 1 hour ago
Beverlie
1,088318
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
1 hour 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
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
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 1 hour ago
Asaf Karagila♦
295k32410738
asked 1 hour ago
Beverlie
1,088318
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 1 hour ago
Asaf Karagila♦
295k32410738
asked 1 hour ago
Beverlie
1,088318
edited 1 hour ago
Asaf Karagila♦
295k32410738
asked 1 hour ago
Beverlie
1,088318
edited 1 hour ago
Asaf Karagila♦
295k32410738
edited 1 hour ago
Asaf Karagila♦
295k32410738
edited 1 hour ago
Asaf Karagila♦
295k32410738
295k32410738
asked 1 hour ago
Beverlie
1,088318
asked 1 hour ago
Beverlie
1,088318
asked 1 hour ago
Beverlie
1,088318
1,088318
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
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 1 hour 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
1 hour ago
add a comment |Â
up vote
0
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 1 hour ago
ZempZemp
11
add a comment |Â
up vote
0
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 1 hour ago
Nuntractatuses Amável
2959
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
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 1 hour 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
1 hour ago
add a comment |Â
up vote
5
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 1 hour 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
1 hour ago
add a comment |Â
up vote
5
down vote
up vote
5
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 1 hour 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 1 hour ago
Mees de Vries
14.5k12348
answered 1 hour ago
Mees de Vries
14.5k12348
answered 1 hour ago
Mees de Vries
14.5k12348
answered 1 hour 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
1 hour 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
1 hour ago
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
1 hour ago
thanks for letting me the point that every set has a power set is an axiom.. never known that.
– Beverlie
1 hour ago
add a comment |Â
up vote
0
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 1 hour ago
ZempZemp
11
add a comment |Â
up vote
0
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 1 hour ago
ZempZemp
11
add a comment |Â
up vote
0
down vote
up vote
0
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 1 hour ago
ZempZemp
11
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 1 hour ago
ZempZemp
11
answered 1 hour ago
ZempZemp
11
answered 1 hour ago
ZempZemp
11
answered 1 hour ago
ZempZemp
11
11
add a comment |Â
add a comment |Â
up vote
0
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 1 hour ago
Nuntractatuses Amável
2959
add a comment |Â
up vote
0
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 1 hour ago
Nuntractatuses Amável
2959
add a comment |Â
up vote
0
down vote
up vote
0
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 1 hour ago
Nuntractatuses Amável
2959
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 1 hour ago
Nuntractatuses Amável
2959
answered 1 hour ago
Nuntractatuses Amável
2959
answered 1 hour ago
Nuntractatuses Amável
2959
answered 1 hour ago
Nuntractatuses Amável
2959
2959
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
1 hour 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
1 hour ago