Product $sigma$-algebra of power sets
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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
asked Dec 7 '18 at 0:55
AlephNullAlephNull
2799
$endgroup$
add a comment |
$begingroup$
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
asked Dec 7 '18 at 0:55
AlephNullAlephNull
2799
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 7 '18 at 0:55
AlephNullAlephNull
2799
$endgroup$
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
asked Dec 7 '18 at 0:55
AlephNullAlephNull
2799
asked Dec 7 '18 at 0:55
AlephNullAlephNull
2799
asked Dec 7 '18 at 0:55
AlephNullAlephNull
2799
asked Dec 7 '18 at 0:55
AlephNullAlephNull
2799
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 7 '18 at 7:23
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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)$.
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
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add a comment |
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2 Answers
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2 Answers
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active
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votes
$begingroup$
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.
answered Dec 7 '18 at 7:23
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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.
answered Dec 7 '18 at 7:23
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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.
answered Dec 7 '18 at 7:23
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
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.
answered Dec 7 '18 at 7:23
Henno BrandsmaHenno Brandsma
108k347114
edited Dec 7 '18 at 22:25
answered Dec 7 '18 at 7:23
Henno BrandsmaHenno Brandsma
108k347114
answered Dec 7 '18 at 7:23
Henno BrandsmaHenno Brandsma
108k347114
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
|
show 3 more comments
$begingroup$
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.
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– Daniel Schepler
Dec 7 '18 at 16:59
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@DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
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– Henno Brandsma
Dec 7 '18 at 17:11
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@DanielSchepler I know it for $aleph_1$, what is the reference for $mathbb{R}$? without CH of course....
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– Henno Brandsma
Dec 7 '18 at 17:11
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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 $mathfrak{c}^{aleph_0} = mathfrak{c}$.
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– Daniel Schepler
Dec 7 '18 at 18:09
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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 $mathfrak{c}^{aleph_0} = mathfrak{c}$.
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– Daniel Schepler
Dec 7 '18 at 18:09
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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 $mathbb{N}^{mathbb{N}}$.)
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– Daniel Schepler
Dec 7 '18 at 18:19
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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 $mathbb{N}^{mathbb{N}}$.)
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– Daniel Schepler
Dec 7 '18 at 18:19
|
show 3 more comments
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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)$.
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
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add a comment |
$begingroup$
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)$.
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
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)$.
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
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)$.
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 1:13
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
add a comment |
add a comment |
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In the special case mentioned, every subset of $mathbb{N} times mathbb{N}$ must be a countable union of singleton sets ${ (m_0, n_0) } = { m_0 } times { n_0 }$.
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– Daniel Schepler
Dec 7 '18 at 1:02
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Of course, a silly oversight...
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– AlephNull
Dec 7 '18 at 11:00