Mixing
A Prime $mathcal P$-filter is contained in a unique $mathcal P$-ultrafilter?
Some backround:
Let $mathcal P$ be a class of subsets of a topological space such that if $P_1$ and $P_2$ are sets from $mathcal P$ then $P_1cap P_2$ and $P_1cup P_2$ belong to $mathcal P$. A $mathcal P$-filter $mathcal F$ is a collection of nonempty elements of $mathcal P$ closed for finite intersections and such that for any $P_1in mathcal F$ and $P_1subseteq P_2in mathcal P$ we have $P_2in mathcal F$.
A $mathcal P$-filter $mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $mathcal P$ and $P_1cup P_2in mathcal F$, then $P_1in mathcal F$ or $P_2in mathcal F$. A $mathcal P$-ultrafilter is just a maximal $mathcal P$-filter.
My question is, is every prime $mathcal P$-filter contained in a unique $mathcal P$-ultrafilter?; this is exercise 12E.6 of Willard's General Topology.
I have proved that if $mathcal F$ is a $mathcal P$-ultrafilter and $Pin mathcal P$ is such that $Pcap Fneq emptyset$ for all $Fin mathcal F$, then $Pin mathcal F$. I think this must be used in the proof but I don't know how.
All hints are appreciated.
general-topology filters
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
add a comment |
Some backround:
Let $mathcal P$ be a class of subsets of a topological space such that if $P_1$ and $P_2$ are sets from $mathcal P$ then $P_1cap P_2$ and $P_1cup P_2$ belong to $mathcal P$. A $mathcal P$-filter $mathcal F$ is a collection of nonempty elements of $mathcal P$ closed for finite intersections and such that for any $P_1in mathcal F$ and $P_1subseteq P_2in mathcal P$ we have $P_2in mathcal F$.
A $mathcal P$-filter $mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $mathcal P$ and $P_1cup P_2in mathcal F$, then $P_1in mathcal F$ or $P_2in mathcal F$. A $mathcal P$-ultrafilter is just a maximal $mathcal P$-filter.
My question is, is every prime $mathcal P$-filter contained in a unique $mathcal P$-ultrafilter?; this is exercise 12E.6 of Willard's General Topology.
I have proved that if $mathcal F$ is a $mathcal P$-ultrafilter and $Pin mathcal P$ is such that $Pcap Fneq emptyset$ for all $Fin mathcal F$, then $Pin mathcal F$. I think this must be used in the proof but I don't know how.
All hints are appreciated.
general-topology filters
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
Have you proved that every ultrafilter is prime?
– Asaf Karagila♦
May 21 '13 at 2:53
yes, this follows easily from the condition at the end of my question.
– Camilo Arosemena-Serrato
May 21 '13 at 2:55
I know it does, I asked if you proved that. Did you also prove the slightly stronger proposition: If $cal F$ is a filter, and $P$ is such that for all $Fincal F$, $Pcap Fneqvarnothing$, then there exists $cal F'$ extending $cal F$ such that $Pincal F'$? (The statement about ultrafilters follows trivially form this.)
– Asaf Karagila♦
May 21 '13 at 4:09
yes, I did. I've also proved the question for when $mathcal P$ is the set of all zero-sets of the space. In there you have to use the following property: Given any two disjoint zero-sets $A,B$, there are zero-sets $C,D$ such that $Asubseteq C^c$, $Bsubseteq D^c$ and $C^ccap D^c=emptyset$. Perhaps the exercise is false, and we can come up with a counterexample with this.
– Camilo Arosemena-Serrato
May 21 '13 at 4:42
@CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC.
– Henno Brandsma
May 21 '13 at 6:51
add a comment |
Some backround:
Let $mathcal P$ be a class of subsets of a topological space such that if $P_1$ and $P_2$ are sets from $mathcal P$ then $P_1cap P_2$ and $P_1cup P_2$ belong to $mathcal P$. A $mathcal P$-filter $mathcal F$ is a collection of nonempty elements of $mathcal P$ closed for finite intersections and such that for any $P_1in mathcal F$ and $P_1subseteq P_2in mathcal P$ we have $P_2in mathcal F$.
A $mathcal P$-filter $mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $mathcal P$ and $P_1cup P_2in mathcal F$, then $P_1in mathcal F$ or $P_2in mathcal F$. A $mathcal P$-ultrafilter is just a maximal $mathcal P$-filter.
My question is, is every prime $mathcal P$-filter contained in a unique $mathcal P$-ultrafilter?; this is exercise 12E.6 of Willard's General Topology.
I have proved that if $mathcal F$ is a $mathcal P$-ultrafilter and $Pin mathcal P$ is such that $Pcap Fneq emptyset$ for all $Fin mathcal F$, then $Pin mathcal F$. I think this must be used in the proof but I don't know how.
All hints are appreciated.
general-topology filters
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
Some backround:
Let $mathcal P$ be a class of subsets of a topological space such that if $P_1$ and $P_2$ are sets from $mathcal P$ then $P_1cap P_2$ and $P_1cup P_2$ belong to $mathcal P$. A $mathcal P$-filter $mathcal F$ is a collection of nonempty elements of $mathcal P$ closed for finite intersections and such that for any $P_1in mathcal F$ and $P_1subseteq P_2in mathcal P$ we have $P_2in mathcal F$.
A $mathcal P$-filter $mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $mathcal P$ and $P_1cup P_2in mathcal F$, then $P_1in mathcal F$ or $P_2in mathcal F$. A $mathcal P$-ultrafilter is just a maximal $mathcal P$-filter.
My question is, is every prime $mathcal P$-filter contained in a unique $mathcal P$-ultrafilter?; this is exercise 12E.6 of Willard's General Topology.
I have proved that if $mathcal F$ is a $mathcal P$-ultrafilter and $Pin mathcal P$ is such that $Pcap Fneq emptyset$ for all $Fin mathcal F$, then $Pin mathcal F$. I think this must be used in the proof but I don't know how.
All hints are appreciated.
general-topology filters
general-topology filters
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
edited May 21 '13 at 2:51
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
asked May 20 '13 at 21:53
Camilo Arosemena-Serrato
5,63611848
5,63611848
Have you proved that every ultrafilter is prime?
– Asaf Karagila♦
May 21 '13 at 2:53
yes, this follows easily from the condition at the end of my question.
– Camilo Arosemena-Serrato
May 21 '13 at 2:55
I know it does, I asked if you proved that. Did you also prove the slightly stronger proposition: If $cal F$ is a filter, and $P$ is such that for all $Fincal F$, $Pcap Fneqvarnothing$, then there exists $cal F'$ extending $cal F$ such that $Pincal F'$? (The statement about ultrafilters follows trivially form this.)
– Asaf Karagila♦
May 21 '13 at 4:09
yes, I did. I've also proved the question for when $mathcal P$ is the set of all zero-sets of the space. In there you have to use the following property: Given any two disjoint zero-sets $A,B$, there are zero-sets $C,D$ such that $Asubseteq C^c$, $Bsubseteq D^c$ and $C^ccap D^c=emptyset$. Perhaps the exercise is false, and we can come up with a counterexample with this.
– Camilo Arosemena-Serrato
May 21 '13 at 4:42
@CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC.
– Henno Brandsma
May 21 '13 at 6:51
add a comment |
Have you proved that every ultrafilter is prime?
– Asaf Karagila♦
May 21 '13 at 2:53
yes, this follows easily from the condition at the end of my question.
– Camilo Arosemena-Serrato
May 21 '13 at 2:55
I know it does, I asked if you proved that. Did you also prove the slightly stronger proposition: If $cal F$ is a filter, and $P$ is such that for all $Fincal F$, $Pcap Fneqvarnothing$, then there exists $cal F'$ extending $cal F$ such that $Pincal F'$? (The statement about ultrafilters follows trivially form this.)
– Asaf Karagila♦
May 21 '13 at 4:09
yes, I did. I've also proved the question for when $mathcal P$ is the set of all zero-sets of the space. In there you have to use the following property: Given any two disjoint zero-sets $A,B$, there are zero-sets $C,D$ such that $Asubseteq C^c$, $Bsubseteq D^c$ and $C^ccap D^c=emptyset$. Perhaps the exercise is false, and we can come up with a counterexample with this.
– Camilo Arosemena-Serrato
May 21 '13 at 4:42
@CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC.
– Henno Brandsma
May 21 '13 at 6:51
Have you proved that every ultrafilter is prime?
– Asaf Karagila♦
May 21 '13 at 2:53
Have you proved that every ultrafilter is prime?
– Asaf Karagila♦
May 21 '13 at 2:53
yes, this follows easily from the condition at the end of my question.
– Camilo Arosemena-Serrato
May 21 '13 at 2:55
yes, this follows easily from the condition at the end of my question.
– Camilo Arosemena-Serrato
May 21 '13 at 2:55
I know it does, I asked if you proved that. Did you also prove the slightly stronger proposition: If $cal F$ is a filter, and $P$ is such that for all $Fincal F$, $Pcap Fneqvarnothing$, then there exists $cal F'$ extending $cal F$ such that $Pincal F'$? (The statement about ultrafilters follows trivially form this.)
– Asaf Karagila♦
May 21 '13 at 4:09
I know it does, I asked if you proved that. Did you also prove the slightly stronger proposition: If $cal F$ is a filter, and $P$ is such that for all $Fincal F$, $Pcap Fneqvarnothing$, then there exists $cal F'$ extending $cal F$ such that $Pincal F'$? (The statement about ultrafilters follows trivially form this.)
– Asaf Karagila♦
May 21 '13 at 4:09
yes, I did. I've also proved the question for when $mathcal P$ is the set of all zero-sets of the space. In there you have to use the following property: Given any two disjoint zero-sets $A,B$, there are zero-sets $C,D$ such that $Asubseteq C^c$, $Bsubseteq D^c$ and $C^ccap D^c=emptyset$. Perhaps the exercise is false, and we can come up with a counterexample with this.
– Camilo Arosemena-Serrato
May 21 '13 at 4:42
yes, I did. I've also proved the question for when $mathcal P$ is the set of all zero-sets of the space. In there you have to use the following property: Given any two disjoint zero-sets $A,B$, there are zero-sets $C,D$ such that $Asubseteq C^c$, $Bsubseteq D^c$ and $C^ccap D^c=emptyset$. Perhaps the exercise is false, and we can come up with a counterexample with this.
– Camilo Arosemena-Serrato
May 21 '13 at 4:42
@CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC.
– Henno Brandsma
May 21 '13 at 6:51
@CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC.
– Henno Brandsma
May 21 '13 at 6:51
add a comment |
2 Answers
2
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In general it's false, I think, even for finite collections.
The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate
such an example as a collection of subsets of a topological space..:
Let $X = mathbb{R}$, say, and $mathcal{P} = {[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]}$ (the lattice diagram is a "double diamond").
Then $mathcal{F} = {[0,9], [3,9]}$ is a prime filter, but both $mathcal{U} = mathcal{P}setminus {[3,5],[4,5]}$ and $mathcal{U}' = mathcal{P} setminus {[4,6],[4,5]}$ are ultrafilters extending $mathcal{F}$.
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
add a comment |
It's false in general, and even for cases of interest to Williard.
Let ${cal P}$ be the collection of open subsets of the real line. Let ${cal F}$ be the filter of open sets containing 0. This is clearly a prime filter.
Now, every set of the form $(0,frac{1}{n})$ intersects every element of ${cal F}$, so there is an open ultrafilter ${cal F}_1$ containing ${cal F}$ and also every set of this form.
But, in the same way, every set of the form $(-frac{1}{n},0)$ intersects every element of ${cal F}$, so there is an ultrafilter ${cal F}_2$ containing ${cal F}$ and all of these.
Clearly, ${cal F}_1$ and ${cal F}_2$ are different and ${cal F}$ is contained in both.
answered Nov 20 '18 at 19:05
polymath257
141
add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
oldest
votes
In general it's false, I think, even for finite collections.
The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate
such an example as a collection of subsets of a topological space..:
Let $X = mathbb{R}$, say, and $mathcal{P} = {[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]}$ (the lattice diagram is a "double diamond").
Then $mathcal{F} = {[0,9], [3,9]}$ is a prime filter, but both $mathcal{U} = mathcal{P}setminus {[3,5],[4,5]}$ and $mathcal{U}' = mathcal{P} setminus {[4,6],[4,5]}$ are ultrafilters extending $mathcal{F}$.
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
add a comment |
In general it's false, I think, even for finite collections.
The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate
such an example as a collection of subsets of a topological space..:
Let $X = mathbb{R}$, say, and $mathcal{P} = {[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]}$ (the lattice diagram is a "double diamond").
Then $mathcal{F} = {[0,9], [3,9]}$ is a prime filter, but both $mathcal{U} = mathcal{P}setminus {[3,5],[4,5]}$ and $mathcal{U}' = mathcal{P} setminus {[4,6],[4,5]}$ are ultrafilters extending $mathcal{F}$.
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
add a comment |
In general it's false, I think, even for finite collections.
The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate
such an example as a collection of subsets of a topological space..:
Let $X = mathbb{R}$, say, and $mathcal{P} = {[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]}$ (the lattice diagram is a "double diamond").
Then $mathcal{F} = {[0,9], [3,9]}$ is a prime filter, but both $mathcal{U} = mathcal{P}setminus {[3,5],[4,5]}$ and $mathcal{U}' = mathcal{P} setminus {[4,6],[4,5]}$ are ultrafilters extending $mathcal{F}$.
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
In general it's false, I think, even for finite collections.
The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate
such an example as a collection of subsets of a topological space..:
Let $X = mathbb{R}$, say, and $mathcal{P} = {[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]}$ (the lattice diagram is a "double diamond").
Then $mathcal{F} = {[0,9], [3,9]}$ is a prime filter, but both $mathcal{U} = mathcal{P}setminus {[3,5],[4,5]}$ and $mathcal{U}' = mathcal{P} setminus {[4,6],[4,5]}$ are ultrafilters extending $mathcal{F}$.
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
answered May 21 '13 at 6:59
Henno Brandsma
105k347114
105k347114
add a comment |
add a comment |
It's false in general, and even for cases of interest to Williard.
Let ${cal P}$ be the collection of open subsets of the real line. Let ${cal F}$ be the filter of open sets containing 0. This is clearly a prime filter.
Now, every set of the form $(0,frac{1}{n})$ intersects every element of ${cal F}$, so there is an open ultrafilter ${cal F}_1$ containing ${cal F}$ and also every set of this form.
But, in the same way, every set of the form $(-frac{1}{n},0)$ intersects every element of ${cal F}$, so there is an ultrafilter ${cal F}_2$ containing ${cal F}$ and all of these.
Clearly, ${cal F}_1$ and ${cal F}_2$ are different and ${cal F}$ is contained in both.
answered Nov 20 '18 at 19:05
polymath257
141
add a comment |
It's false in general, and even for cases of interest to Williard.
Let ${cal P}$ be the collection of open subsets of the real line. Let ${cal F}$ be the filter of open sets containing 0. This is clearly a prime filter.
Now, every set of the form $(0,frac{1}{n})$ intersects every element of ${cal F}$, so there is an open ultrafilter ${cal F}_1$ containing ${cal F}$ and also every set of this form.
But, in the same way, every set of the form $(-frac{1}{n},0)$ intersects every element of ${cal F}$, so there is an ultrafilter ${cal F}_2$ containing ${cal F}$ and all of these.
Clearly, ${cal F}_1$ and ${cal F}_2$ are different and ${cal F}$ is contained in both.
answered Nov 20 '18 at 19:05
polymath257
141
add a comment |
It's false in general, and even for cases of interest to Williard.
Let ${cal P}$ be the collection of open subsets of the real line. Let ${cal F}$ be the filter of open sets containing 0. This is clearly a prime filter.
Now, every set of the form $(0,frac{1}{n})$ intersects every element of ${cal F}$, so there is an open ultrafilter ${cal F}_1$ containing ${cal F}$ and also every set of this form.
But, in the same way, every set of the form $(-frac{1}{n},0)$ intersects every element of ${cal F}$, so there is an ultrafilter ${cal F}_2$ containing ${cal F}$ and all of these.
Clearly, ${cal F}_1$ and ${cal F}_2$ are different and ${cal F}$ is contained in both.
answered Nov 20 '18 at 19:05
polymath257
141
It's false in general, and even for cases of interest to Williard.
Let ${cal P}$ be the collection of open subsets of the real line. Let ${cal F}$ be the filter of open sets containing 0. This is clearly a prime filter.
Now, every set of the form $(0,frac{1}{n})$ intersects every element of ${cal F}$, so there is an open ultrafilter ${cal F}_1$ containing ${cal F}$ and also every set of this form.
But, in the same way, every set of the form $(-frac{1}{n},0)$ intersects every element of ${cal F}$, so there is an ultrafilter ${cal F}_2$ containing ${cal F}$ and all of these.
Clearly, ${cal F}_1$ and ${cal F}_2$ are different and ${cal F}$ is contained in both.
answered Nov 20 '18 at 19:05
polymath257
141
answered Nov 20 '18 at 19:05
polymath257
141
answered Nov 20 '18 at 19:05
polymath257
141
answered Nov 20 '18 at 19:05
polymath257
141
141
add a comment |
add a comment |
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Have you proved that every ultrafilter is prime?
– Asaf Karagila♦
May 21 '13 at 2:53
yes, this follows easily from the condition at the end of my question.
– Camilo Arosemena-Serrato
May 21 '13 at 2:55
I know it does, I asked if you proved that. Did you also prove the slightly stronger proposition: If $cal F$ is a filter, and $P$ is such that for all $Fincal F$, $Pcap Fneqvarnothing$, then there exists $cal F'$ extending $cal F$ such that $Pincal F'$? (The statement about ultrafilters follows trivially form this.)
– Asaf Karagila♦
May 21 '13 at 4:09
yes, I did. I've also proved the question for when $mathcal P$ is the set of all zero-sets of the space. In there you have to use the following property: Given any two disjoint zero-sets $A,B$, there are zero-sets $C,D$ such that $Asubseteq C^c$, $Bsubseteq D^c$ and $C^ccap D^c=emptyset$. Perhaps the exercise is false, and we can come up with a counterexample with this.
– Camilo Arosemena-Serrato
May 21 '13 at 4:42
@CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC.
– Henno Brandsma
May 21 '13 at 6:51