Defining powerset of $Omega = [0, 1]$ in measure theory context
$begingroup$
I am confused about the concept of powerset in the following context.
I am reading the Bandit Algorithm book (http://banditalgs.com/) by Csaba Szepesvari. There it gives the following theorem:
Theorem 2.1: Let $Omega = [0,1]$ and $mathcal{F}$ is the powerset of $Omega$. Then there does not exist a measure $mathbb{P}$ on $(Omega, mathcal{F})$ such that $mathbb{P}([a,b]) = b-a$ for all $0leq a leq b leq 1$.
What does the powerset of $Omega$ look like since there are infinitely many real numbers in it?
probability-theory measure-theory terminology
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
asked Dec 12 '18 at 15:55
hi15hi15
1586
$endgroup$
add a comment |
$begingroup$
I am confused about the concept of powerset in the following context.
I am reading the Bandit Algorithm book (http://banditalgs.com/) by Csaba Szepesvari. There it gives the following theorem:
Theorem 2.1: Let $Omega = [0,1]$ and $mathcal{F}$ is the powerset of $Omega$. Then there does not exist a measure $mathbb{P}$ on $(Omega, mathcal{F})$ such that $mathbb{P}([a,b]) = b-a$ for all $0leq a leq b leq 1$.
What does the powerset of $Omega$ look like since there are infinitely many real numbers in it?
probability-theory measure-theory terminology
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
asked Dec 12 '18 at 15:55
hi15hi15
1586
$endgroup$
4
$begingroup$
That's an odd question. The power set is the collection of all subsets.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 16:18
$begingroup$
It seems you're just asking about the powerset of the unit interval. I understand your concern since each element of the set $S$ is drawn from a continuum, so there are uncountably many elements in it, it's doubly difficult to imagine every possible subset of $S$. But this powerset contains every single isolated point, every segment, every union of those, and every possible combination of finitely many, countably many, or uncountably many of those.
$endgroup$
– user334732
Dec 12 '18 at 18:28
$begingroup$
It is hard to "visualize". A version of Cantor's diagonal argument shows that there are an infinity of subsets of $[0,1]$, with this infinity being much larger than the infinite number of real numbers in the set $[0,1]$.
$endgroup$
– hardmath
Dec 12 '18 at 20:04
add a comment |
$begingroup$
I am confused about the concept of powerset in the following context.
I am reading the Bandit Algorithm book (http://banditalgs.com/) by Csaba Szepesvari. There it gives the following theorem:
Theorem 2.1: Let $Omega = [0,1]$ and $mathcal{F}$ is the powerset of $Omega$. Then there does not exist a measure $mathbb{P}$ on $(Omega, mathcal{F})$ such that $mathbb{P}([a,b]) = b-a$ for all $0leq a leq b leq 1$.
What does the powerset of $Omega$ look like since there are infinitely many real numbers in it?
probability-theory measure-theory terminology
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
asked Dec 12 '18 at 15:55
hi15hi15
1586
$endgroup$
I am confused about the concept of powerset in the following context.
I am reading the Bandit Algorithm book (http://banditalgs.com/) by Csaba Szepesvari. There it gives the following theorem:
Theorem 2.1: Let $Omega = [0,1]$ and $mathcal{F}$ is the powerset of $Omega$. Then there does not exist a measure $mathbb{P}$ on $(Omega, mathcal{F})$ such that $mathbb{P}([a,b]) = b-a$ for all $0leq a leq b leq 1$.
What does the powerset of $Omega$ look like since there are infinitely many real numbers in it?
probability-theory measure-theory terminology
probability-theory measure-theory terminology
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
asked Dec 12 '18 at 15:55
hi15hi15
1586
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
asked Dec 12 '18 at 15:55
hi15hi15
1586
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
edited Dec 12 '18 at 16:28
Andrés E. Caicedo
65.8k8160252
65.8k8160252
asked Dec 12 '18 at 15:55
hi15hi15
1586
asked Dec 12 '18 at 15:55
hi15hi15
1586
asked Dec 12 '18 at 15:55
hi15hi15
1586
1586
4
$begingroup$
That's an odd question. The power set is the collection of all subsets.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 16:18
$begingroup$
It seems you're just asking about the powerset of the unit interval. I understand your concern since each element of the set $S$ is drawn from a continuum, so there are uncountably many elements in it, it's doubly difficult to imagine every possible subset of $S$. But this powerset contains every single isolated point, every segment, every union of those, and every possible combination of finitely many, countably many, or uncountably many of those.
$endgroup$
– user334732
Dec 12 '18 at 18:28
$begingroup$
It is hard to "visualize". A version of Cantor's diagonal argument shows that there are an infinity of subsets of $[0,1]$, with this infinity being much larger than the infinite number of real numbers in the set $[0,1]$.
$endgroup$
– hardmath
Dec 12 '18 at 20:04
add a comment |
4
$begingroup$
That's an odd question. The power set is the collection of all subsets.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 16:18
$begingroup$
It seems you're just asking about the powerset of the unit interval. I understand your concern since each element of the set $S$ is drawn from a continuum, so there are uncountably many elements in it, it's doubly difficult to imagine every possible subset of $S$. But this powerset contains every single isolated point, every segment, every union of those, and every possible combination of finitely many, countably many, or uncountably many of those.
$endgroup$
– user334732
Dec 12 '18 at 18:28
$begingroup$
It is hard to "visualize". A version of Cantor's diagonal argument shows that there are an infinity of subsets of $[0,1]$, with this infinity being much larger than the infinite number of real numbers in the set $[0,1]$.
$endgroup$
– hardmath
Dec 12 '18 at 20:04
4
4
$begingroup$
That's an odd question. The power set is the collection of all subsets.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 16:18
$begingroup$
That's an odd question. The power set is the collection of all subsets.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 16:18
$begingroup$
It seems you're just asking about the powerset of the unit interval. I understand your concern since each element of the set $S$ is drawn from a continuum, so there are uncountably many elements in it, it's doubly difficult to imagine every possible subset of $S$. But this powerset contains every single isolated point, every segment, every union of those, and every possible combination of finitely many, countably many, or uncountably many of those.
$endgroup$
– user334732
Dec 12 '18 at 18:28
$begingroup$
It seems you're just asking about the powerset of the unit interval. I understand your concern since each element of the set $S$ is drawn from a continuum, so there are uncountably many elements in it, it's doubly difficult to imagine every possible subset of $S$. But this powerset contains every single isolated point, every segment, every union of those, and every possible combination of finitely many, countably many, or uncountably many of those.
$endgroup$
– user334732
Dec 12 '18 at 18:28
$begingroup$
It is hard to "visualize". A version of Cantor's diagonal argument shows that there are an infinity of subsets of $[0,1]$, with this infinity being much larger than the infinite number of real numbers in the set $[0,1]$.
$endgroup$
– hardmath
Dec 12 '18 at 20:04
$begingroup$
It is hard to "visualize". A version of Cantor's diagonal argument shows that there are an infinity of subsets of $[0,1]$, with this infinity being much larger than the infinite number of real numbers in the set $[0,1]$.
$endgroup$
– hardmath
Dec 12 '18 at 20:04
add a comment |
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4
$begingroup$
That's an odd question. The power set is the collection of all subsets.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 16:18
$begingroup$
It seems you're just asking about the powerset of the unit interval. I understand your concern since each element of the set $S$ is drawn from a continuum, so there are uncountably many elements in it, it's doubly difficult to imagine every possible subset of $S$. But this powerset contains every single isolated point, every segment, every union of those, and every possible combination of finitely many, countably many, or uncountably many of those.
$endgroup$
– user334732
Dec 12 '18 at 18:28
$begingroup$
It is hard to "visualize". A version of Cantor's diagonal argument shows that there are an infinity of subsets of $[0,1]$, with this infinity being much larger than the infinite number of real numbers in the set $[0,1]$.
$endgroup$
– hardmath
Dec 12 '18 at 20:04