What makes Fubini product (tensor product) of filters a natural operation?
Clash Royale CLAN TAG#URR8PPP
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In several texts you can encounter tensor product or Fubini product of two filters defined as follows: If $p$ is a filter on a set $X$, $q$ be a filter on a set $Y$, then
$$potimes q=Asubseteq Xtimes Y; x; y; (x,y)in Ain qin p$$
is a filter on $Xtimes Y$ which is called Fubini product or tensor product of $p$ and $q$.
Some authors define them only for ultrafilters. Some authors work dually with ideals. Quite often we work only with situation that $X=Y$. See, for example, [HS, Definition 11.1], [BS, page 125] or [Far, page 8].1
The definition can be roughly rephrased as: For almost every $x$ we have that almost all $y$'s belong to the $x$-section $A_x=yin Y; (x,y)in A$. Of course, we have to interpret almost every using the corresponding filters $p$ and $q$.
Question. What is the intuition behind this definition? Are there some viewpoints in which this operation arises naturally? Are there some important properties which makes this notion useful tool in some situations?
I was able to think at least of something:
- It seems that this operation is naturally related to iterated limit w.r.t. filter (ultrafilter). I have tried to expand on this in another post: Tensor product of ultrafilters corresponds to iterated limit.
- I have seen a video where Andreas Blass explains how this is related to Fubini's theorem and the corresponding measure if we view ultrafilters as finitely additive measures. It is a part of the video from Peter Krautzberger's talk Idempotent Ultrafilters: An Introduction approximately about 45:00.
- This might also be related to this question: Products of ultrafilters and products of measures. Although the connection as explained there seems a bit unclear to me. (I am not sure what authors of [BS] mean by "the usual definition of the product measure" - it is possible that this is the same explanation as in my previous bullet point, just put in another words.)
From the above three bullet points, I have feeling that I really understand only the first one. Still, I will be grateful if somebody expands on any of them. And certainly if somebody mentions another possibility how one might look at this type of product.
References
- [BS] J. L. Bell, A. B. Slomson: Models and Ultraproducts: An Introduction, North-Holland, 1974.
- [HS] N. Hindman, D. Strauss: Algebra in the Stone-Cech compactification, De Gruyter, 2012, Second edition.
- [Far] Farah I. Analytic Quotients. Theory of Liftings for Quotients over Analytic Ideals on the Integers, Memoirs of AMS, 2000.
1You may notice that the definition in [BS] reverses order compared to [HS]. As far as I can tell the definition which I have included is more frequent. Both [BS] and [HS] work only with ultrafilters. In [Far] this kind of product is defined dually for ideals. The definition given there is that Fubini product of ideals $mathcal I$ and $mathcal J$ is given by
$$Ainmathcal Itimes mathcal J qquadLeftrightarrowqquad i; A_inotinmathcal Iinmathcal J.$$
If we denote $p$ the dual filter to $mathcal I$ and $q$ the dual filter to $mathcal J$, and recall that dual filter contains exactly the complements of sets belonging to the ideal, then this basically says
beginalign*
Ain potimes q
&Leftrightarrow Xsetminus Ainmathcal Itimes mathcal J \
&Leftrightarrow i; (Xsetminus A)_inotinmathcal Iinmathcal J \
&Leftrightarrow i; (Xsetminus A)_iinmathcal Iin q \
&Leftrightarrow i; A_iinmathcal pin q \
endalign*
set-theory filters
add a comment |Â
up vote
13
down vote
favorite
In several texts you can encounter tensor product or Fubini product of two filters defined as follows: If $p$ is a filter on a set $X$, $q$ be a filter on a set $Y$, then
$$potimes q=Asubseteq Xtimes Y; x; y; (x,y)in Ain qin p$$
is a filter on $Xtimes Y$ which is called Fubini product or tensor product of $p$ and $q$.
Some authors define them only for ultrafilters. Some authors work dually with ideals. Quite often we work only with situation that $X=Y$. See, for example, [HS, Definition 11.1], [BS, page 125] or [Far, page 8].1
The definition can be roughly rephrased as: For almost every $x$ we have that almost all $y$'s belong to the $x$-section $A_x=yin Y; (x,y)in A$. Of course, we have to interpret almost every using the corresponding filters $p$ and $q$.
Question. What is the intuition behind this definition? Are there some viewpoints in which this operation arises naturally? Are there some important properties which makes this notion useful tool in some situations?
I was able to think at least of something:
- It seems that this operation is naturally related to iterated limit w.r.t. filter (ultrafilter). I have tried to expand on this in another post: Tensor product of ultrafilters corresponds to iterated limit.
- I have seen a video where Andreas Blass explains how this is related to Fubini's theorem and the corresponding measure if we view ultrafilters as finitely additive measures. It is a part of the video from Peter Krautzberger's talk Idempotent Ultrafilters: An Introduction approximately about 45:00.
- This might also be related to this question: Products of ultrafilters and products of measures. Although the connection as explained there seems a bit unclear to me. (I am not sure what authors of [BS] mean by "the usual definition of the product measure" - it is possible that this is the same explanation as in my previous bullet point, just put in another words.)
From the above three bullet points, I have feeling that I really understand only the first one. Still, I will be grateful if somebody expands on any of them. And certainly if somebody mentions another possibility how one might look at this type of product.
References
- [BS] J. L. Bell, A. B. Slomson: Models and Ultraproducts: An Introduction, North-Holland, 1974.
- [HS] N. Hindman, D. Strauss: Algebra in the Stone-Cech compactification, De Gruyter, 2012, Second edition.
- [Far] Farah I. Analytic Quotients. Theory of Liftings for Quotients over Analytic Ideals on the Integers, Memoirs of AMS, 2000.
1You may notice that the definition in [BS] reverses order compared to [HS]. As far as I can tell the definition which I have included is more frequent. Both [BS] and [HS] work only with ultrafilters. In [Far] this kind of product is defined dually for ideals. The definition given there is that Fubini product of ideals $mathcal I$ and $mathcal J$ is given by
$$Ainmathcal Itimes mathcal J qquadLeftrightarrowqquad i; A_inotinmathcal Iinmathcal J.$$
If we denote $p$ the dual filter to $mathcal I$ and $q$ the dual filter to $mathcal J$, and recall that dual filter contains exactly the complements of sets belonging to the ideal, then this basically says
beginalign*
Ain potimes q
&Leftrightarrow Xsetminus Ainmathcal Itimes mathcal J \
&Leftrightarrow i; (Xsetminus A)_inotinmathcal Iinmathcal J \
&Leftrightarrow i; (Xsetminus A)_iinmathcal Iin q \
&Leftrightarrow i; A_iinmathcal pin q \
endalign*
set-theory filters
add a comment |Â
up vote
13
down vote
favorite
up vote
13
down vote
favorite
In several texts you can encounter tensor product or Fubini product of two filters defined as follows: If $p$ is a filter on a set $X$, $q$ be a filter on a set $Y$, then
$$potimes q=Asubseteq Xtimes Y; x; y; (x,y)in Ain qin p$$
is a filter on $Xtimes Y$ which is called Fubini product or tensor product of $p$ and $q$.
Some authors define them only for ultrafilters. Some authors work dually with ideals. Quite often we work only with situation that $X=Y$. See, for example, [HS, Definition 11.1], [BS, page 125] or [Far, page 8].1
The definition can be roughly rephrased as: For almost every $x$ we have that almost all $y$'s belong to the $x$-section $A_x=yin Y; (x,y)in A$. Of course, we have to interpret almost every using the corresponding filters $p$ and $q$.
Question. What is the intuition behind this definition? Are there some viewpoints in which this operation arises naturally? Are there some important properties which makes this notion useful tool in some situations?
I was able to think at least of something:
- It seems that this operation is naturally related to iterated limit w.r.t. filter (ultrafilter). I have tried to expand on this in another post: Tensor product of ultrafilters corresponds to iterated limit.
- I have seen a video where Andreas Blass explains how this is related to Fubini's theorem and the corresponding measure if we view ultrafilters as finitely additive measures. It is a part of the video from Peter Krautzberger's talk Idempotent Ultrafilters: An Introduction approximately about 45:00.
- This might also be related to this question: Products of ultrafilters and products of measures. Although the connection as explained there seems a bit unclear to me. (I am not sure what authors of [BS] mean by "the usual definition of the product measure" - it is possible that this is the same explanation as in my previous bullet point, just put in another words.)
From the above three bullet points, I have feeling that I really understand only the first one. Still, I will be grateful if somebody expands on any of them. And certainly if somebody mentions another possibility how one might look at this type of product.
References
- [BS] J. L. Bell, A. B. Slomson: Models and Ultraproducts: An Introduction, North-Holland, 1974.
- [HS] N. Hindman, D. Strauss: Algebra in the Stone-Cech compactification, De Gruyter, 2012, Second edition.
- [Far] Farah I. Analytic Quotients. Theory of Liftings for Quotients over Analytic Ideals on the Integers, Memoirs of AMS, 2000.
1You may notice that the definition in [BS] reverses order compared to [HS]. As far as I can tell the definition which I have included is more frequent. Both [BS] and [HS] work only with ultrafilters. In [Far] this kind of product is defined dually for ideals. The definition given there is that Fubini product of ideals $mathcal I$ and $mathcal J$ is given by
$$Ainmathcal Itimes mathcal J qquadLeftrightarrowqquad i; A_inotinmathcal Iinmathcal J.$$
If we denote $p$ the dual filter to $mathcal I$ and $q$ the dual filter to $mathcal J$, and recall that dual filter contains exactly the complements of sets belonging to the ideal, then this basically says
beginalign*
Ain potimes q
&Leftrightarrow Xsetminus Ainmathcal Itimes mathcal J \
&Leftrightarrow i; (Xsetminus A)_inotinmathcal Iinmathcal J \
&Leftrightarrow i; (Xsetminus A)_iinmathcal Iin q \
&Leftrightarrow i; A_iinmathcal pin q \
endalign*
set-theory filters
In several texts you can encounter tensor product or Fubini product of two filters defined as follows: If $p$ is a filter on a set $X$, $q$ be a filter on a set $Y$, then
$$potimes q=Asubseteq Xtimes Y; x; y; (x,y)in Ain qin p$$
is a filter on $Xtimes Y$ which is called Fubini product or tensor product of $p$ and $q$.
Some authors define them only for ultrafilters. Some authors work dually with ideals. Quite often we work only with situation that $X=Y$. See, for example, [HS, Definition 11.1], [BS, page 125] or [Far, page 8].1
The definition can be roughly rephrased as: For almost every $x$ we have that almost all $y$'s belong to the $x$-section $A_x=yin Y; (x,y)in A$. Of course, we have to interpret almost every using the corresponding filters $p$ and $q$.
Question. What is the intuition behind this definition? Are there some viewpoints in which this operation arises naturally? Are there some important properties which makes this notion useful tool in some situations?
I was able to think at least of something:
- It seems that this operation is naturally related to iterated limit w.r.t. filter (ultrafilter). I have tried to expand on this in another post: Tensor product of ultrafilters corresponds to iterated limit.
- I have seen a video where Andreas Blass explains how this is related to Fubini's theorem and the corresponding measure if we view ultrafilters as finitely additive measures. It is a part of the video from Peter Krautzberger's talk Idempotent Ultrafilters: An Introduction approximately about 45:00.
- This might also be related to this question: Products of ultrafilters and products of measures. Although the connection as explained there seems a bit unclear to me. (I am not sure what authors of [BS] mean by "the usual definition of the product measure" - it is possible that this is the same explanation as in my previous bullet point, just put in another words.)
From the above three bullet points, I have feeling that I really understand only the first one. Still, I will be grateful if somebody expands on any of them. And certainly if somebody mentions another possibility how one might look at this type of product.
References
- [BS] J. L. Bell, A. B. Slomson: Models and Ultraproducts: An Introduction, North-Holland, 1974.
- [HS] N. Hindman, D. Strauss: Algebra in the Stone-Cech compactification, De Gruyter, 2012, Second edition.
- [Far] Farah I. Analytic Quotients. Theory of Liftings for Quotients over Analytic Ideals on the Integers, Memoirs of AMS, 2000.
1You may notice that the definition in [BS] reverses order compared to [HS]. As far as I can tell the definition which I have included is more frequent. Both [BS] and [HS] work only with ultrafilters. In [Far] this kind of product is defined dually for ideals. The definition given there is that Fubini product of ideals $mathcal I$ and $mathcal J$ is given by
$$Ainmathcal Itimes mathcal J qquadLeftrightarrowqquad i; A_inotinmathcal Iinmathcal J.$$
If we denote $p$ the dual filter to $mathcal I$ and $q$ the dual filter to $mathcal J$, and recall that dual filter contains exactly the complements of sets belonging to the ideal, then this basically says
beginalign*
Ain potimes q
&Leftrightarrow Xsetminus Ainmathcal Itimes mathcal J \
&Leftrightarrow i; (Xsetminus A)_inotinmathcal Iinmathcal J \
&Leftrightarrow i; (Xsetminus A)_iinmathcal Iin q \
&Leftrightarrow i; A_iinmathcal pin q \
endalign*
set-theory filters
edited May 29 at 8:38
asked May 28 at 18:10
Martin Sleziak
43.6k6113260
43.6k6113260
add a comment |Â
add a comment |Â
1 Answer
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In topology, filters help generalize concepts like limits. (See also: nets. For example, the Riemann integral is defined in terms of such a generalized limit.) Here's a topological characterization of the filter product:
Theorem: If $f:Xtimes Y rightarrow T$ is a function into a compact Hausdorff space $T$, and if $p$ and $q$ are ultrafilters on $X$ and $Y$ respectively, then there exists a unique product filter $potimes q$ such that:
$$lim_xrightarrow plim_yrightarrow q f(x,y) = lim_langle x,yranglerightarrow potimes q f(x,y)$$
in that both limits exist and are equal.
In measure theory, filters are (isomorphic with) a special kind of measure: if $F$ is a filter on $X$, then the function $p(A) equiv 1text if Ain F text else 0 $ is a 0-1 valued, finitely-additive measure on $X$. Conversely, if $p$ is any 0-1 valued, finitely-additive measure on $X$, then $A : p(A)=1$ is a filter.
Products of measures are defined in a straightforward way: if $mu$ and $nu$ are measures on spaces $X$ and $Y$, then $(mutimes nu)(A times B) equiv mu(A)cdot nu(B)$ is their product measure on the space $Xtimes Y$ generated by measureable rectangles. It looks like this isomorphism between 0-1 valued finitely-additive measures and filters preserves products: the filter corresponding to the product of two measures is the product filter for the corresponding two filters.
$(mutimes nu)(Atimes B) = 1 iff [mu(A) = 1 wedge mu(B) = 1] iff [A in F wedge B in G] iff [Atimes B] in (Fotimes G)$.
Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $Atimes B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.)
â Martin Sleziak
Sep 1 at 0:36
I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit.
â Martin Sleziak
Sep 1 at 0:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In topology, filters help generalize concepts like limits. (See also: nets. For example, the Riemann integral is defined in terms of such a generalized limit.) Here's a topological characterization of the filter product:
Theorem: If $f:Xtimes Y rightarrow T$ is a function into a compact Hausdorff space $T$, and if $p$ and $q$ are ultrafilters on $X$ and $Y$ respectively, then there exists a unique product filter $potimes q$ such that:
$$lim_xrightarrow plim_yrightarrow q f(x,y) = lim_langle x,yranglerightarrow potimes q f(x,y)$$
in that both limits exist and are equal.
In measure theory, filters are (isomorphic with) a special kind of measure: if $F$ is a filter on $X$, then the function $p(A) equiv 1text if Ain F text else 0 $ is a 0-1 valued, finitely-additive measure on $X$. Conversely, if $p$ is any 0-1 valued, finitely-additive measure on $X$, then $A : p(A)=1$ is a filter.
Products of measures are defined in a straightforward way: if $mu$ and $nu$ are measures on spaces $X$ and $Y$, then $(mutimes nu)(A times B) equiv mu(A)cdot nu(B)$ is their product measure on the space $Xtimes Y$ generated by measureable rectangles. It looks like this isomorphism between 0-1 valued finitely-additive measures and filters preserves products: the filter corresponding to the product of two measures is the product filter for the corresponding two filters.
$(mutimes nu)(Atimes B) = 1 iff [mu(A) = 1 wedge mu(B) = 1] iff [A in F wedge B in G] iff [Atimes B] in (Fotimes G)$.
Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $Atimes B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.)
â Martin Sleziak
Sep 1 at 0:36
I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit.
â Martin Sleziak
Sep 1 at 0:40
add a comment |Â
up vote
0
down vote
In topology, filters help generalize concepts like limits. (See also: nets. For example, the Riemann integral is defined in terms of such a generalized limit.) Here's a topological characterization of the filter product:
Theorem: If $f:Xtimes Y rightarrow T$ is a function into a compact Hausdorff space $T$, and if $p$ and $q$ are ultrafilters on $X$ and $Y$ respectively, then there exists a unique product filter $potimes q$ such that:
$$lim_xrightarrow plim_yrightarrow q f(x,y) = lim_langle x,yranglerightarrow potimes q f(x,y)$$
in that both limits exist and are equal.
In measure theory, filters are (isomorphic with) a special kind of measure: if $F$ is a filter on $X$, then the function $p(A) equiv 1text if Ain F text else 0 $ is a 0-1 valued, finitely-additive measure on $X$. Conversely, if $p$ is any 0-1 valued, finitely-additive measure on $X$, then $A : p(A)=1$ is a filter.
Products of measures are defined in a straightforward way: if $mu$ and $nu$ are measures on spaces $X$ and $Y$, then $(mutimes nu)(A times B) equiv mu(A)cdot nu(B)$ is their product measure on the space $Xtimes Y$ generated by measureable rectangles. It looks like this isomorphism between 0-1 valued finitely-additive measures and filters preserves products: the filter corresponding to the product of two measures is the product filter for the corresponding two filters.
$(mutimes nu)(Atimes B) = 1 iff [mu(A) = 1 wedge mu(B) = 1] iff [A in F wedge B in G] iff [Atimes B] in (Fotimes G)$.
Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $Atimes B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.)
â Martin Sleziak
Sep 1 at 0:36
I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit.
â Martin Sleziak
Sep 1 at 0:40
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In topology, filters help generalize concepts like limits. (See also: nets. For example, the Riemann integral is defined in terms of such a generalized limit.) Here's a topological characterization of the filter product:
Theorem: If $f:Xtimes Y rightarrow T$ is a function into a compact Hausdorff space $T$, and if $p$ and $q$ are ultrafilters on $X$ and $Y$ respectively, then there exists a unique product filter $potimes q$ such that:
$$lim_xrightarrow plim_yrightarrow q f(x,y) = lim_langle x,yranglerightarrow potimes q f(x,y)$$
in that both limits exist and are equal.
In measure theory, filters are (isomorphic with) a special kind of measure: if $F$ is a filter on $X$, then the function $p(A) equiv 1text if Ain F text else 0 $ is a 0-1 valued, finitely-additive measure on $X$. Conversely, if $p$ is any 0-1 valued, finitely-additive measure on $X$, then $A : p(A)=1$ is a filter.
Products of measures are defined in a straightforward way: if $mu$ and $nu$ are measures on spaces $X$ and $Y$, then $(mutimes nu)(A times B) equiv mu(A)cdot nu(B)$ is their product measure on the space $Xtimes Y$ generated by measureable rectangles. It looks like this isomorphism between 0-1 valued finitely-additive measures and filters preserves products: the filter corresponding to the product of two measures is the product filter for the corresponding two filters.
$(mutimes nu)(Atimes B) = 1 iff [mu(A) = 1 wedge mu(B) = 1] iff [A in F wedge B in G] iff [Atimes B] in (Fotimes G)$.
In topology, filters help generalize concepts like limits. (See also: nets. For example, the Riemann integral is defined in terms of such a generalized limit.) Here's a topological characterization of the filter product:
Theorem: If $f:Xtimes Y rightarrow T$ is a function into a compact Hausdorff space $T$, and if $p$ and $q$ are ultrafilters on $X$ and $Y$ respectively, then there exists a unique product filter $potimes q$ such that:
$$lim_xrightarrow plim_yrightarrow q f(x,y) = lim_langle x,yranglerightarrow potimes q f(x,y)$$
in that both limits exist and are equal.
In measure theory, filters are (isomorphic with) a special kind of measure: if $F$ is a filter on $X$, then the function $p(A) equiv 1text if Ain F text else 0 $ is a 0-1 valued, finitely-additive measure on $X$. Conversely, if $p$ is any 0-1 valued, finitely-additive measure on $X$, then $A : p(A)=1$ is a filter.
Products of measures are defined in a straightforward way: if $mu$ and $nu$ are measures on spaces $X$ and $Y$, then $(mutimes nu)(A times B) equiv mu(A)cdot nu(B)$ is their product measure on the space $Xtimes Y$ generated by measureable rectangles. It looks like this isomorphism between 0-1 valued finitely-additive measures and filters preserves products: the filter corresponding to the product of two measures is the product filter for the corresponding two filters.
$(mutimes nu)(Atimes B) = 1 iff [mu(A) = 1 wedge mu(B) = 1] iff [A in F wedge B in G] iff [Atimes B] in (Fotimes G)$.
answered Sep 1 at 0:27
user326210
8,707625
8,707625
Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $Atimes B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.)
â Martin Sleziak
Sep 1 at 0:36
I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit.
â Martin Sleziak
Sep 1 at 0:40
add a comment |Â
Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $Atimes B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.)
â Martin Sleziak
Sep 1 at 0:36
I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit.
â Martin Sleziak
Sep 1 at 0:40
Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $Atimes B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.)
â Martin Sleziak
Sep 1 at 0:36
Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $Atimes B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.)
â Martin Sleziak
Sep 1 at 0:36
I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit.
â Martin Sleziak
Sep 1 at 0:40
I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit.
â Martin Sleziak
Sep 1 at 0:40
add a comment |Â
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