Product $sigma$-algebra of power sets












1












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Let $X,Y$ be any two sets. In general, what is the product $sigma$-algebra $mathcal{P}(X) times mathcal{P}(Y)$? In the context (using Fubini's Theorem to prove that one can reverse the order of summation for absolutely convergent double series) the author makes an implicit assumption that, at least in the case $X=Y=mathbb{N}$, we have $mathcal{P}(X) times mathcal{P}(Y)=mathcal{P}(X times Y)$. But I do not see why is this true or in what cases it holds. Here it seems that every subset of $mathbb{N}^2$ must somehow be a union (possibly with complements) of Cartesian products of subsets of $mathbb{N}$.






measure-theory set-theory natural-numbers







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$endgroup$












  • $begingroup$
    In the special case mentioned, every subset of $mathbb{N} times mathbb{N}$ must be a countable union of singleton sets ${ (m_0, n_0) } = { m_0 } times { n_0 }$.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 1:02










  • $begingroup$
    Of course, a silly oversight...
    $endgroup$
    – AlephNull
    Dec 7 '18 at 11:00
















1












$begingroup$


Let $X,Y$ be any two sets. In general, what is the product $sigma$-algebra $mathcal{P}(X) times mathcal{P}(Y)$? In the context (using Fubini's Theorem to prove that one can reverse the order of summation for absolutely convergent double series) the author makes an implicit assumption that, at least in the case $X=Y=mathbb{N}$, we have $mathcal{P}(X) times mathcal{P}(Y)=mathcal{P}(X times Y)$. But I do not see why is this true or in what cases it holds. Here it seems that every subset of $mathbb{N}^2$ must somehow be a union (possibly with complements) of Cartesian products of subsets of $mathbb{N}$.






measure-theory set-theory natural-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    In the special case mentioned, every subset of $mathbb{N} times mathbb{N}$ must be a countable union of singleton sets ${ (m_0, n_0) } = { m_0 } times { n_0 }$.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 1:02










  • $begingroup$
    Of course, a silly oversight...
    $endgroup$
    – AlephNull
    Dec 7 '18 at 11:00














1












1








1





$begingroup$


Let $X,Y$ be any two sets. In general, what is the product $sigma$-algebra $mathcal{P}(X) times mathcal{P}(Y)$? In the context (using Fubini's Theorem to prove that one can reverse the order of summation for absolutely convergent double series) the author makes an implicit assumption that, at least in the case $X=Y=mathbb{N}$, we have $mathcal{P}(X) times mathcal{P}(Y)=mathcal{P}(X times Y)$. But I do not see why is this true or in what cases it holds. Here it seems that every subset of $mathbb{N}^2$ must somehow be a union (possibly with complements) of Cartesian products of subsets of $mathbb{N}$.






measure-theory set-theory natural-numbers







share|cite|improve this question









$endgroup$




Let $X,Y$ be any two sets. In general, what is the product $sigma$-algebra $mathcal{P}(X) times mathcal{P}(Y)$? In the context (using Fubini's Theorem to prove that one can reverse the order of summation for absolutely convergent double series) the author makes an implicit assumption that, at least in the case $X=Y=mathbb{N}$, we have $mathcal{P}(X) times mathcal{P}(Y)=mathcal{P}(X times Y)$. But I do not see why is this true or in what cases it holds. Here it seems that every subset of $mathbb{N}^2$ must somehow be a union (possibly with complements) of Cartesian products of subsets of $mathbb{N}$.






measure-theory set-theory natural-numbers




measure-theory set-theory natural-numbers






share|cite|improve this question













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asked Dec 7 '18 at 0:55









AlephNullAlephNull

2799




2799












  • $begingroup$
    In the special case mentioned, every subset of $mathbb{N} times mathbb{N}$ must be a countable union of singleton sets ${ (m_0, n_0) } = { m_0 } times { n_0 }$.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 1:02










  • $begingroup$
    Of course, a silly oversight...
    $endgroup$
    – AlephNull
    Dec 7 '18 at 11:00


















  • $begingroup$
    In the special case mentioned, every subset of $mathbb{N} times mathbb{N}$ must be a countable union of singleton sets ${ (m_0, n_0) } = { m_0 } times { n_0 }$.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 1:02










  • $begingroup$
    Of course, a silly oversight...
    $endgroup$
    – AlephNull
    Dec 7 '18 at 11:00
















$begingroup$
In the special case mentioned, every subset of $mathbb{N} times mathbb{N}$ must be a countable union of singleton sets ${ (m_0, n_0) } = { m_0 } times { n_0 }$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 1:02




$begingroup$
In the special case mentioned, every subset of $mathbb{N} times mathbb{N}$ must be a countable union of singleton sets ${ (m_0, n_0) } = { m_0 } times { n_0 }$.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 1:02












$begingroup$
Of course, a silly oversight...
$endgroup$
– AlephNull
Dec 7 '18 at 11:00




$begingroup$
Of course, a silly oversight...
$endgroup$
– AlephNull
Dec 7 '18 at 11:00










2 Answers
2






active

oldest

votes


















1












$begingroup$

You cannot show that $mathcal{P}(X times Y) = mathcal{P}(X) times mathcal{P}(Y)$, and this is not what your author claims.



For countable $X$ and $Y$, the sigma-algebra generated by all sets $A times B$, where $A in mathcal{P}(X)$ and $B in mathcal{P}(Y)$ contains all singletons ${(x,y)} = {x} times {y}$ and thus all subsets of the countable set $X times Y$, so the generated sigma-algebra for the product is indeed $mathcal{P}(X times Y)$.



This no longer holds for uncountable sets though. In that case it's a classic fact that the diagonal $D= {(x,x): x in X}$ is not in the generated sigma-algebra by $mathcal{P}(X) times mathcal{P}(X)$ when $|X| > mathfrak{c}$ (see Nedoma's pathology), so we generate a proper sub-sigma-algebra of $mathcal{P}(X times Y)$. For $|X| =aleph_1$ we do get the desired result. (due to V. Rao.) Daniel Schepler (in the comments below) shows that this also extends to sets of size $mathfrak{c}$, so we have a division of uncountable sizes here: small uncountable (where we do get $mathbb{P}(X times Y) = mathcal{P}(X) otimes mathcal{P}(Y)$) and large uncountable where this fails to be the case. The countable case is straightforward, as we saw.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not sure if it was clear enough (although I did say this in the first sentence and imply it in the last), but when I refer to $mathcal{P}(X) times mathcal{P}(Y)$, I mean the product $sigma$-algebra generated by those $sigma$-algebras, not just the usual Cartesian product. The (trivial) countable case is clear now, but I don't quite understand the argument for the uncountable case. If it is complicated could you provide a link to a proof?
    $endgroup$
    – AlephNull
    Dec 7 '18 at 11:08












  • $begingroup$
    The diagonal is actually in the product $sigma$-algebra also for $mathbb{R}$ - but it's not in the product $sigma$-algebra if $|X| > mathfrak{c}$. e.g. if $X$ is the set of sequences $mathbb{N} to mathbb{N}$ then for each $n$, ${ x mid (x_1)_n = (x_2)_n }$ is a countable union of products, and then the countable intersection of these sets is the diagonal.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 16:59










  • $begingroup$
    @DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
    $endgroup$
    – Henno Brandsma
    Dec 7 '18 at 17:11










  • $begingroup$
    I already showed the diagonal is in the product $sigma$-algebra in the case $X = mathbb{N}^{mathbb{N}}$ which is bijective with $mathbb{R}$. A possible proof that the diagonal isn't in the product $sigma$-algebra if $|X| > mathfrak{c}$ is that the subsets of $X times X$ with at most $mathfrak{c}$ distinct "vertical fibers" form a $sigma$-algebra containing product sets - using the fact that $mathfrak{c}^{aleph_0} = mathfrak{c}$.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 18:09










  • $begingroup$
    Also, the Borel $sigma$-algebra on $mathbb{R} times mathbb{R}$ already contains the diagonal and it's certainly contained in the $sigma$-algebra generated by $mathcal{P}(mathbb{R}) times mathcal{P}(mathbb{R})$. (Unfolding this proof a bit gets to essentially the proof I presented for $mathbb{N}^{mathbb{N}}$.)
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 18:19



















0












$begingroup$

It's about $sigma$-algebras.



Take any $Rsubset mathcal P(mathbb Ntimesmathbb N)$. Define
$$
R^x={yinmathbb N: (x,y)in R}.
$$
Then
$$
R=bigcup_{xinmathbb N} {x}times R^x.
$$
As each of the sets ${x}times R^xinmathcal P(mathbb N)times mathcal P(mathbb N)$, their (countable) union is in the $sigma$-algebra they generate. Thus the $sigma$-algebra generated by $ mathcal P(mathbb N)times mathcal P(mathbb N)$ is $ mathcal P(mathbb Ntimesmathbb N)$.






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    2 Answers
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    2 Answers
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    1












    $begingroup$

    You cannot show that $mathcal{P}(X times Y) = mathcal{P}(X) times mathcal{P}(Y)$, and this is not what your author claims.



    For countable $X$ and $Y$, the sigma-algebra generated by all sets $A times B$, where $A in mathcal{P}(X)$ and $B in mathcal{P}(Y)$ contains all singletons ${(x,y)} = {x} times {y}$ and thus all subsets of the countable set $X times Y$, so the generated sigma-algebra for the product is indeed $mathcal{P}(X times Y)$.



    This no longer holds for uncountable sets though. In that case it's a classic fact that the diagonal $D= {(x,x): x in X}$ is not in the generated sigma-algebra by $mathcal{P}(X) times mathcal{P}(X)$ when $|X| > mathfrak{c}$ (see Nedoma's pathology), so we generate a proper sub-sigma-algebra of $mathcal{P}(X times Y)$. For $|X| =aleph_1$ we do get the desired result. (due to V. Rao.) Daniel Schepler (in the comments below) shows that this also extends to sets of size $mathfrak{c}$, so we have a division of uncountable sizes here: small uncountable (where we do get $mathbb{P}(X times Y) = mathcal{P}(X) otimes mathcal{P}(Y)$) and large uncountable where this fails to be the case. The countable case is straightforward, as we saw.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Not sure if it was clear enough (although I did say this in the first sentence and imply it in the last), but when I refer to $mathcal{P}(X) times mathcal{P}(Y)$, I mean the product $sigma$-algebra generated by those $sigma$-algebras, not just the usual Cartesian product. The (trivial) countable case is clear now, but I don't quite understand the argument for the uncountable case. If it is complicated could you provide a link to a proof?
      $endgroup$
      – AlephNull
      Dec 7 '18 at 11:08












    • $begingroup$
      The diagonal is actually in the product $sigma$-algebra also for $mathbb{R}$ - but it's not in the product $sigma$-algebra if $|X| > mathfrak{c}$. e.g. if $X$ is the set of sequences $mathbb{N} to mathbb{N}$ then for each $n$, ${ x mid (x_1)_n = (x_2)_n }$ is a countable union of products, and then the countable intersection of these sets is the diagonal.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 16:59










    • $begingroup$
      @DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
      $endgroup$
      – Henno Brandsma
      Dec 7 '18 at 17:11










    • $begingroup$
      I already showed the diagonal is in the product $sigma$-algebra in the case $X = mathbb{N}^{mathbb{N}}$ which is bijective with $mathbb{R}$. A possible proof that the diagonal isn't in the product $sigma$-algebra if $|X| > mathfrak{c}$ is that the subsets of $X times X$ with at most $mathfrak{c}$ distinct "vertical fibers" form a $sigma$-algebra containing product sets - using the fact that $mathfrak{c}^{aleph_0} = mathfrak{c}$.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:09










    • $begingroup$
      Also, the Borel $sigma$-algebra on $mathbb{R} times mathbb{R}$ already contains the diagonal and it's certainly contained in the $sigma$-algebra generated by $mathcal{P}(mathbb{R}) times mathcal{P}(mathbb{R})$. (Unfolding this proof a bit gets to essentially the proof I presented for $mathbb{N}^{mathbb{N}}$.)
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:19
















    1












    $begingroup$

    You cannot show that $mathcal{P}(X times Y) = mathcal{P}(X) times mathcal{P}(Y)$, and this is not what your author claims.



    For countable $X$ and $Y$, the sigma-algebra generated by all sets $A times B$, where $A in mathcal{P}(X)$ and $B in mathcal{P}(Y)$ contains all singletons ${(x,y)} = {x} times {y}$ and thus all subsets of the countable set $X times Y$, so the generated sigma-algebra for the product is indeed $mathcal{P}(X times Y)$.



    This no longer holds for uncountable sets though. In that case it's a classic fact that the diagonal $D= {(x,x): x in X}$ is not in the generated sigma-algebra by $mathcal{P}(X) times mathcal{P}(X)$ when $|X| > mathfrak{c}$ (see Nedoma's pathology), so we generate a proper sub-sigma-algebra of $mathcal{P}(X times Y)$. For $|X| =aleph_1$ we do get the desired result. (due to V. Rao.) Daniel Schepler (in the comments below) shows that this also extends to sets of size $mathfrak{c}$, so we have a division of uncountable sizes here: small uncountable (where we do get $mathbb{P}(X times Y) = mathcal{P}(X) otimes mathcal{P}(Y)$) and large uncountable where this fails to be the case. The countable case is straightforward, as we saw.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Not sure if it was clear enough (although I did say this in the first sentence and imply it in the last), but when I refer to $mathcal{P}(X) times mathcal{P}(Y)$, I mean the product $sigma$-algebra generated by those $sigma$-algebras, not just the usual Cartesian product. The (trivial) countable case is clear now, but I don't quite understand the argument for the uncountable case. If it is complicated could you provide a link to a proof?
      $endgroup$
      – AlephNull
      Dec 7 '18 at 11:08












    • $begingroup$
      The diagonal is actually in the product $sigma$-algebra also for $mathbb{R}$ - but it's not in the product $sigma$-algebra if $|X| > mathfrak{c}$. e.g. if $X$ is the set of sequences $mathbb{N} to mathbb{N}$ then for each $n$, ${ x mid (x_1)_n = (x_2)_n }$ is a countable union of products, and then the countable intersection of these sets is the diagonal.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 16:59










    • $begingroup$
      @DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
      $endgroup$
      – Henno Brandsma
      Dec 7 '18 at 17:11










    • $begingroup$
      I already showed the diagonal is in the product $sigma$-algebra in the case $X = mathbb{N}^{mathbb{N}}$ which is bijective with $mathbb{R}$. A possible proof that the diagonal isn't in the product $sigma$-algebra if $|X| > mathfrak{c}$ is that the subsets of $X times X$ with at most $mathfrak{c}$ distinct "vertical fibers" form a $sigma$-algebra containing product sets - using the fact that $mathfrak{c}^{aleph_0} = mathfrak{c}$.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:09










    • $begingroup$
      Also, the Borel $sigma$-algebra on $mathbb{R} times mathbb{R}$ already contains the diagonal and it's certainly contained in the $sigma$-algebra generated by $mathcal{P}(mathbb{R}) times mathcal{P}(mathbb{R})$. (Unfolding this proof a bit gets to essentially the proof I presented for $mathbb{N}^{mathbb{N}}$.)
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:19














    1












    1








    1





    $begingroup$

    You cannot show that $mathcal{P}(X times Y) = mathcal{P}(X) times mathcal{P}(Y)$, and this is not what your author claims.



    For countable $X$ and $Y$, the sigma-algebra generated by all sets $A times B$, where $A in mathcal{P}(X)$ and $B in mathcal{P}(Y)$ contains all singletons ${(x,y)} = {x} times {y}$ and thus all subsets of the countable set $X times Y$, so the generated sigma-algebra for the product is indeed $mathcal{P}(X times Y)$.



    This no longer holds for uncountable sets though. In that case it's a classic fact that the diagonal $D= {(x,x): x in X}$ is not in the generated sigma-algebra by $mathcal{P}(X) times mathcal{P}(X)$ when $|X| > mathfrak{c}$ (see Nedoma's pathology), so we generate a proper sub-sigma-algebra of $mathcal{P}(X times Y)$. For $|X| =aleph_1$ we do get the desired result. (due to V. Rao.) Daniel Schepler (in the comments below) shows that this also extends to sets of size $mathfrak{c}$, so we have a division of uncountable sizes here: small uncountable (where we do get $mathbb{P}(X times Y) = mathcal{P}(X) otimes mathcal{P}(Y)$) and large uncountable where this fails to be the case. The countable case is straightforward, as we saw.






    share|cite|improve this answer











    $endgroup$



    You cannot show that $mathcal{P}(X times Y) = mathcal{P}(X) times mathcal{P}(Y)$, and this is not what your author claims.



    For countable $X$ and $Y$, the sigma-algebra generated by all sets $A times B$, where $A in mathcal{P}(X)$ and $B in mathcal{P}(Y)$ contains all singletons ${(x,y)} = {x} times {y}$ and thus all subsets of the countable set $X times Y$, so the generated sigma-algebra for the product is indeed $mathcal{P}(X times Y)$.



    This no longer holds for uncountable sets though. In that case it's a classic fact that the diagonal $D= {(x,x): x in X}$ is not in the generated sigma-algebra by $mathcal{P}(X) times mathcal{P}(X)$ when $|X| > mathfrak{c}$ (see Nedoma's pathology), so we generate a proper sub-sigma-algebra of $mathcal{P}(X times Y)$. For $|X| =aleph_1$ we do get the desired result. (due to V. Rao.) Daniel Schepler (in the comments below) shows that this also extends to sets of size $mathfrak{c}$, so we have a division of uncountable sizes here: small uncountable (where we do get $mathbb{P}(X times Y) = mathcal{P}(X) otimes mathcal{P}(Y)$) and large uncountable where this fails to be the case. The countable case is straightforward, as we saw.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 7 '18 at 22:25

























    answered Dec 7 '18 at 7:23









    Henno BrandsmaHenno Brandsma

    108k347114




    108k347114












    • $begingroup$
      Not sure if it was clear enough (although I did say this in the first sentence and imply it in the last), but when I refer to $mathcal{P}(X) times mathcal{P}(Y)$, I mean the product $sigma$-algebra generated by those $sigma$-algebras, not just the usual Cartesian product. The (trivial) countable case is clear now, but I don't quite understand the argument for the uncountable case. If it is complicated could you provide a link to a proof?
      $endgroup$
      – AlephNull
      Dec 7 '18 at 11:08












    • $begingroup$
      The diagonal is actually in the product $sigma$-algebra also for $mathbb{R}$ - but it's not in the product $sigma$-algebra if $|X| > mathfrak{c}$. e.g. if $X$ is the set of sequences $mathbb{N} to mathbb{N}$ then for each $n$, ${ x mid (x_1)_n = (x_2)_n }$ is a countable union of products, and then the countable intersection of these sets is the diagonal.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 16:59










    • $begingroup$
      @DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
      $endgroup$
      – Henno Brandsma
      Dec 7 '18 at 17:11










    • $begingroup$
      I already showed the diagonal is in the product $sigma$-algebra in the case $X = mathbb{N}^{mathbb{N}}$ which is bijective with $mathbb{R}$. A possible proof that the diagonal isn't in the product $sigma$-algebra if $|X| > mathfrak{c}$ is that the subsets of $X times X$ with at most $mathfrak{c}$ distinct "vertical fibers" form a $sigma$-algebra containing product sets - using the fact that $mathfrak{c}^{aleph_0} = mathfrak{c}$.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:09










    • $begingroup$
      Also, the Borel $sigma$-algebra on $mathbb{R} times mathbb{R}$ already contains the diagonal and it's certainly contained in the $sigma$-algebra generated by $mathcal{P}(mathbb{R}) times mathcal{P}(mathbb{R})$. (Unfolding this proof a bit gets to essentially the proof I presented for $mathbb{N}^{mathbb{N}}$.)
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:19


















    • $begingroup$
      Not sure if it was clear enough (although I did say this in the first sentence and imply it in the last), but when I refer to $mathcal{P}(X) times mathcal{P}(Y)$, I mean the product $sigma$-algebra generated by those $sigma$-algebras, not just the usual Cartesian product. The (trivial) countable case is clear now, but I don't quite understand the argument for the uncountable case. If it is complicated could you provide a link to a proof?
      $endgroup$
      – AlephNull
      Dec 7 '18 at 11:08












    • $begingroup$
      The diagonal is actually in the product $sigma$-algebra also for $mathbb{R}$ - but it's not in the product $sigma$-algebra if $|X| > mathfrak{c}$. e.g. if $X$ is the set of sequences $mathbb{N} to mathbb{N}$ then for each $n$, ${ x mid (x_1)_n = (x_2)_n }$ is a countable union of products, and then the countable intersection of these sets is the diagonal.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 16:59










    • $begingroup$
      @DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
      $endgroup$
      – Henno Brandsma
      Dec 7 '18 at 17:11










    • $begingroup$
      I already showed the diagonal is in the product $sigma$-algebra in the case $X = mathbb{N}^{mathbb{N}}$ which is bijective with $mathbb{R}$. A possible proof that the diagonal isn't in the product $sigma$-algebra if $|X| > mathfrak{c}$ is that the subsets of $X times X$ with at most $mathfrak{c}$ distinct "vertical fibers" form a $sigma$-algebra containing product sets - using the fact that $mathfrak{c}^{aleph_0} = mathfrak{c}$.
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:09










    • $begingroup$
      Also, the Borel $sigma$-algebra on $mathbb{R} times mathbb{R}$ already contains the diagonal and it's certainly contained in the $sigma$-algebra generated by $mathcal{P}(mathbb{R}) times mathcal{P}(mathbb{R})$. (Unfolding this proof a bit gets to essentially the proof I presented for $mathbb{N}^{mathbb{N}}$.)
      $endgroup$
      – Daniel Schepler
      Dec 7 '18 at 18:19
















    $begingroup$
    Not sure if it was clear enough (although I did say this in the first sentence and imply it in the last), but when I refer to $mathcal{P}(X) times mathcal{P}(Y)$, I mean the product $sigma$-algebra generated by those $sigma$-algebras, not just the usual Cartesian product. The (trivial) countable case is clear now, but I don't quite understand the argument for the uncountable case. If it is complicated could you provide a link to a proof?
    $endgroup$
    – AlephNull
    Dec 7 '18 at 11:08






    $begingroup$
    Not sure if it was clear enough (although I did say this in the first sentence and imply it in the last), but when I refer to $mathcal{P}(X) times mathcal{P}(Y)$, I mean the product $sigma$-algebra generated by those $sigma$-algebras, not just the usual Cartesian product. The (trivial) countable case is clear now, but I don't quite understand the argument for the uncountable case. If it is complicated could you provide a link to a proof?
    $endgroup$
    – AlephNull
    Dec 7 '18 at 11:08














    $begingroup$
    The diagonal is actually in the product $sigma$-algebra also for $mathbb{R}$ - but it's not in the product $sigma$-algebra if $|X| > mathfrak{c}$. e.g. if $X$ is the set of sequences $mathbb{N} to mathbb{N}$ then for each $n$, ${ x mid (x_1)_n = (x_2)_n }$ is a countable union of products, and then the countable intersection of these sets is the diagonal.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 16:59




    $begingroup$
    The diagonal is actually in the product $sigma$-algebra also for $mathbb{R}$ - but it's not in the product $sigma$-algebra if $|X| > mathfrak{c}$. e.g. if $X$ is the set of sequences $mathbb{N} to mathbb{N}$ then for each $n$, ${ x mid (x_1)_n = (x_2)_n }$ is a countable union of products, and then the countable intersection of these sets is the diagonal.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 16:59












    $begingroup$
    @DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
    $endgroup$
    – Henno Brandsma
    Dec 7 '18 at 17:11




    $begingroup$
    @DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
    $endgroup$
    – Henno Brandsma
    Dec 7 '18 at 17:11












    $begingroup$
    I already showed the diagonal is in the product $sigma$-algebra in the case $X = mathbb{N}^{mathbb{N}}$ which is bijective with $mathbb{R}$. A possible proof that the diagonal isn't in the product $sigma$-algebra if $|X| > mathfrak{c}$ is that the subsets of $X times X$ with at most $mathfrak{c}$ distinct "vertical fibers" form a $sigma$-algebra containing product sets - using the fact that $mathfrak{c}^{aleph_0} = mathfrak{c}$.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 18:09




    $begingroup$
    I already showed the diagonal is in the product $sigma$-algebra in the case $X = mathbb{N}^{mathbb{N}}$ which is bijective with $mathbb{R}$. A possible proof that the diagonal isn't in the product $sigma$-algebra if $|X| > mathfrak{c}$ is that the subsets of $X times X$ with at most $mathfrak{c}$ distinct "vertical fibers" form a $sigma$-algebra containing product sets - using the fact that $mathfrak{c}^{aleph_0} = mathfrak{c}$.
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 18:09












    $begingroup$
    Also, the Borel $sigma$-algebra on $mathbb{R} times mathbb{R}$ already contains the diagonal and it's certainly contained in the $sigma$-algebra generated by $mathcal{P}(mathbb{R}) times mathcal{P}(mathbb{R})$. (Unfolding this proof a bit gets to essentially the proof I presented for $mathbb{N}^{mathbb{N}}$.)
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 18:19




    $begingroup$
    Also, the Borel $sigma$-algebra on $mathbb{R} times mathbb{R}$ already contains the diagonal and it's certainly contained in the $sigma$-algebra generated by $mathcal{P}(mathbb{R}) times mathcal{P}(mathbb{R})$. (Unfolding this proof a bit gets to essentially the proof I presented for $mathbb{N}^{mathbb{N}}$.)
    $endgroup$
    – Daniel Schepler
    Dec 7 '18 at 18:19











    0












    $begingroup$

    It's about $sigma$-algebras.



    Take any $Rsubset mathcal P(mathbb Ntimesmathbb N)$. Define
    $$
    R^x={yinmathbb N: (x,y)in R}.
    $$
    Then
    $$
    R=bigcup_{xinmathbb N} {x}times R^x.
    $$
    As each of the sets ${x}times R^xinmathcal P(mathbb N)times mathcal P(mathbb N)$, their (countable) union is in the $sigma$-algebra they generate. Thus the $sigma$-algebra generated by $ mathcal P(mathbb N)times mathcal P(mathbb N)$ is $ mathcal P(mathbb Ntimesmathbb N)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It's about $sigma$-algebras.



      Take any $Rsubset mathcal P(mathbb Ntimesmathbb N)$. Define
      $$
      R^x={yinmathbb N: (x,y)in R}.
      $$
      Then
      $$
      R=bigcup_{xinmathbb N} {x}times R^x.
      $$
      As each of the sets ${x}times R^xinmathcal P(mathbb N)times mathcal P(mathbb N)$, their (countable) union is in the $sigma$-algebra they generate. Thus the $sigma$-algebra generated by $ mathcal P(mathbb N)times mathcal P(mathbb N)$ is $ mathcal P(mathbb Ntimesmathbb N)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It's about $sigma$-algebras.



        Take any $Rsubset mathcal P(mathbb Ntimesmathbb N)$. Define
        $$
        R^x={yinmathbb N: (x,y)in R}.
        $$
        Then
        $$
        R=bigcup_{xinmathbb N} {x}times R^x.
        $$
        As each of the sets ${x}times R^xinmathcal P(mathbb N)times mathcal P(mathbb N)$, their (countable) union is in the $sigma$-algebra they generate. Thus the $sigma$-algebra generated by $ mathcal P(mathbb N)times mathcal P(mathbb N)$ is $ mathcal P(mathbb Ntimesmathbb N)$.






        share|cite|improve this answer









        $endgroup$



        It's about $sigma$-algebras.



        Take any $Rsubset mathcal P(mathbb Ntimesmathbb N)$. Define
        $$
        R^x={yinmathbb N: (x,y)in R}.
        $$
        Then
        $$
        R=bigcup_{xinmathbb N} {x}times R^x.
        $$
        As each of the sets ${x}times R^xinmathcal P(mathbb N)times mathcal P(mathbb N)$, their (countable) union is in the $sigma$-algebra they generate. Thus the $sigma$-algebra generated by $ mathcal P(mathbb N)times mathcal P(mathbb N)$ is $ mathcal P(mathbb Ntimesmathbb N)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 1:13









        Martin ArgeramiMartin Argerami

        126k1182181




        126k1182181






























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