Defining powerset of $Omega = [0, 1]$ in measure theory context












0












$begingroup$


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?










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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
















0












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 16:28









Andrés E. Caicedo

65.8k8160252




65.8k8160252










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














  • 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










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