Mixing
Are ultrafilters unique?
$begingroup$
I'm trying to get a feel for what an ultrafilter is by looking at some finite examples. Firstly, I took a look at this question: Example of a filter on a set, and it stoked the question: can sets such as the one in this question have several ultrafilters?
As an example of what I had been thinking: the answer to the aforementioned question states that the given ultrafilter is "centered at $3$", and so it follows that we should be able to center the filter at, for example, $4$, as well. In which case we would have that $f(S)=lbracelbrace4rbrace,lbrace3,4rbrace,lbrace4,5rbrace,lbrace3,4,5rbracerbrace$. Is this correct, or is there a flaw in my reasoning?
filters
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
|
show 4 more comments
$begingroup$
I'm trying to get a feel for what an ultrafilter is by looking at some finite examples. Firstly, I took a look at this question: Example of a filter on a set, and it stoked the question: can sets such as the one in this question have several ultrafilters?
As an example of what I had been thinking: the answer to the aforementioned question states that the given ultrafilter is "centered at $3$", and so it follows that we should be able to center the filter at, for example, $4$, as well. In which case we would have that $f(S)=lbracelbrace4rbrace,lbrace3,4rbrace,lbrace4,5rbrace,lbrace3,4,5rbracerbrace$. Is this correct, or is there a flaw in my reasoning?
filters
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
2
$begingroup$
No, your reasoning is fine. Ultrafilters are only unique on a singleton. But you can ask if they are equivalent, namely if $U$ and $V$ are two ultrafilters on a set $X$, is there a permutation of $X$ which maps $U$ to $V$ (by means of $Asubseteq X$ is mapped to ${f(a)mid ain A}$). And then the answer is a bit more complicated, but still is negative.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:14
$begingroup$
Meaning that, even if a set is infinite, its ultrafilter(s) may not be unique? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 12 at 8:15
$begingroup$
Where did you use the finiteness of the set? You just used the fact that $3$ and $4$ are two different things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:15
1
$begingroup$
When a set is finite all of its ultrafilters are principal. When a set is infinite, you have the principal ultrafilters, but you also have free filters, which can only be extended to free ultrafilters, but that requires the axiom of choice to get involved (and indeed it is consistent that all ultrafilters are principal). So to that end the word "constructing", if interpreted as providing an explicit definition, is generally accepted to not apply to free ultrafilters.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:20
1
$begingroup$
To some extent. But analysis itself also depends on the axiom of choice. The whole notion of measures, continuity, etc. can be formulated without choice, but end up nearly useless when doing so. There are a few papers you can find about some choice which is sufficient to develop NSA, those assumptions are not unreasonable in the general scheme of things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:40
|
show 4 more comments
$begingroup$
I'm trying to get a feel for what an ultrafilter is by looking at some finite examples. Firstly, I took a look at this question: Example of a filter on a set, and it stoked the question: can sets such as the one in this question have several ultrafilters?
As an example of what I had been thinking: the answer to the aforementioned question states that the given ultrafilter is "centered at $3$", and so it follows that we should be able to center the filter at, for example, $4$, as well. In which case we would have that $f(S)=lbracelbrace4rbrace,lbrace3,4rbrace,lbrace4,5rbrace,lbrace3,4,5rbracerbrace$. Is this correct, or is there a flaw in my reasoning?
filters
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
I'm trying to get a feel for what an ultrafilter is by looking at some finite examples. Firstly, I took a look at this question: Example of a filter on a set, and it stoked the question: can sets such as the one in this question have several ultrafilters?
As an example of what I had been thinking: the answer to the aforementioned question states that the given ultrafilter is "centered at $3$", and so it follows that we should be able to center the filter at, for example, $4$, as well. In which case we would have that $f(S)=lbracelbrace4rbrace,lbrace3,4rbrace,lbrace4,5rbrace,lbrace3,4,5rbracerbrace$. Is this correct, or is there a flaw in my reasoning?
filters
filters
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
edited Jan 12 at 8:16
Asaf Karagila♦
304k32430763
304k32430763
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 12 at 8:08
joshuaheckroodtjoshuaheckroodt
1,205622
1,205622
2
$begingroup$
No, your reasoning is fine. Ultrafilters are only unique on a singleton. But you can ask if they are equivalent, namely if $U$ and $V$ are two ultrafilters on a set $X$, is there a permutation of $X$ which maps $U$ to $V$ (by means of $Asubseteq X$ is mapped to ${f(a)mid ain A}$). And then the answer is a bit more complicated, but still is negative.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:14
$begingroup$
Meaning that, even if a set is infinite, its ultrafilter(s) may not be unique? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 12 at 8:15
$begingroup$
Where did you use the finiteness of the set? You just used the fact that $3$ and $4$ are two different things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:15
1
$begingroup$
When a set is finite all of its ultrafilters are principal. When a set is infinite, you have the principal ultrafilters, but you also have free filters, which can only be extended to free ultrafilters, but that requires the axiom of choice to get involved (and indeed it is consistent that all ultrafilters are principal). So to that end the word "constructing", if interpreted as providing an explicit definition, is generally accepted to not apply to free ultrafilters.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:20
1
$begingroup$
To some extent. But analysis itself also depends on the axiom of choice. The whole notion of measures, continuity, etc. can be formulated without choice, but end up nearly useless when doing so. There are a few papers you can find about some choice which is sufficient to develop NSA, those assumptions are not unreasonable in the general scheme of things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:40
|
show 4 more comments
2
$begingroup$
No, your reasoning is fine. Ultrafilters are only unique on a singleton. But you can ask if they are equivalent, namely if $U$ and $V$ are two ultrafilters on a set $X$, is there a permutation of $X$ which maps $U$ to $V$ (by means of $Asubseteq X$ is mapped to ${f(a)mid ain A}$). And then the answer is a bit more complicated, but still is negative.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:14
$begingroup$
Meaning that, even if a set is infinite, its ultrafilter(s) may not be unique? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 12 at 8:15
$begingroup$
Where did you use the finiteness of the set? You just used the fact that $3$ and $4$ are two different things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:15
1
$begingroup$
When a set is finite all of its ultrafilters are principal. When a set is infinite, you have the principal ultrafilters, but you also have free filters, which can only be extended to free ultrafilters, but that requires the axiom of choice to get involved (and indeed it is consistent that all ultrafilters are principal). So to that end the word "constructing", if interpreted as providing an explicit definition, is generally accepted to not apply to free ultrafilters.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:20
1
$begingroup$
To some extent. But analysis itself also depends on the axiom of choice. The whole notion of measures, continuity, etc. can be formulated without choice, but end up nearly useless when doing so. There are a few papers you can find about some choice which is sufficient to develop NSA, those assumptions are not unreasonable in the general scheme of things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:40
2
2
$begingroup$
No, your reasoning is fine. Ultrafilters are only unique on a singleton. But you can ask if they are equivalent, namely if $U$ and $V$ are two ultrafilters on a set $X$, is there a permutation of $X$ which maps $U$ to $V$ (by means of $Asubseteq X$ is mapped to ${f(a)mid ain A}$). And then the answer is a bit more complicated, but still is negative.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:14
$begingroup$
No, your reasoning is fine. Ultrafilters are only unique on a singleton. But you can ask if they are equivalent, namely if $U$ and $V$ are two ultrafilters on a set $X$, is there a permutation of $X$ which maps $U$ to $V$ (by means of $Asubseteq X$ is mapped to ${f(a)mid ain A}$). And then the answer is a bit more complicated, but still is negative.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:14
$begingroup$
Meaning that, even if a set is infinite, its ultrafilter(s) may not be unique? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 12 at 8:15
$begingroup$
Meaning that, even if a set is infinite, its ultrafilter(s) may not be unique? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 12 at 8:15
$begingroup$
Where did you use the finiteness of the set? You just used the fact that $3$ and $4$ are two different things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:15
$begingroup$
Where did you use the finiteness of the set? You just used the fact that $3$ and $4$ are two different things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:15
1
1
$begingroup$
When a set is finite all of its ultrafilters are principal. When a set is infinite, you have the principal ultrafilters, but you also have free filters, which can only be extended to free ultrafilters, but that requires the axiom of choice to get involved (and indeed it is consistent that all ultrafilters are principal). So to that end the word "constructing", if interpreted as providing an explicit definition, is generally accepted to not apply to free ultrafilters.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:20
$begingroup$
When a set is finite all of its ultrafilters are principal. When a set is infinite, you have the principal ultrafilters, but you also have free filters, which can only be extended to free ultrafilters, but that requires the axiom of choice to get involved (and indeed it is consistent that all ultrafilters are principal). So to that end the word "constructing", if interpreted as providing an explicit definition, is generally accepted to not apply to free ultrafilters.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:20
1
1
$begingroup$
To some extent. But analysis itself also depends on the axiom of choice. The whole notion of measures, continuity, etc. can be formulated without choice, but end up nearly useless when doing so. There are a few papers you can find about some choice which is sufficient to develop NSA, those assumptions are not unreasonable in the general scheme of things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:40
$begingroup$
To some extent. But analysis itself also depends on the axiom of choice. The whole notion of measures, continuity, etc. can be formulated without choice, but end up nearly useless when doing so. There are a few papers you can find about some choice which is sufficient to develop NSA, those assumptions are not unreasonable in the general scheme of things.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:40
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It's useful to have the following characterisations of ultrafilters on a set $X$:
For a non-trivial filter $mathcal{F}$ on $X$, the following statements are equivalent:
$mathcal{F}$ is an ultrafilter (i.e. maximal w.r.t. inclusion)- For all finitely many $A_1,ldots A_n subseteq X$ we have that $cup_{i=1}^n A_i in mathcal{F}$ iff there is some $1 le i le n$ such that $A_i in mathcal{F}$.
- For all subsets $A$ of $X$ either $A in mathcal{F}$ or $Xsetminus A in mathcal{F}$.
Equivalence 2. shows that if $mathcal{F}$ contains a finite subset $F={x_1,ldots,x_m}$ it contains one of the sets ${x_i}$, and then $mathcal{F} = {A: x_i in A}$ for that $x_i$, i.e. $mathcal{F}$ is a principal filter. IN particular on a finite set $X$ we only have the principal ultrafilters.
So a non-principal filter cannot contain any finite subset $F$, so by 3. $mathcal{F}$ contains all co-finite subsets, i.e. $mathcal{F}$ is an enlargement of the co-finite filter.
So ultrafilters are of two types: the principal ones (of which there are as many as points of $X$) and the free ones, that we can "get" by invoking choice and somehow extending the cofinite filter to an ultrafilter. There are $2^{2^{|X|}}$ of those for an infinite set $X$ (so many many more), assuming choice. But these are non-constructive and we can try to visualise them as some way of choosing for each infinite subset $A$ of $X$ whether to put $A$ in $mathcal{F}$ or its complement, but in a "filter-consistent" way (if we put in a set, we also have to put in all larger sets, and if we put in two, we also have to add the intersection) and no finite set is allowed in $mathcal{F}$ if we want a free ultrafilter, so their complements are surely in. It's not much but maybe it helps.
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
add a comment |
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$begingroup$
It's useful to have the following characterisations of ultrafilters on a set $X$:
For a non-trivial filter $mathcal{F}$ on $X$, the following statements are equivalent:
$mathcal{F}$ is an ultrafilter (i.e. maximal w.r.t. inclusion)- For all finitely many $A_1,ldots A_n subseteq X$ we have that $cup_{i=1}^n A_i in mathcal{F}$ iff there is some $1 le i le n$ such that $A_i in mathcal{F}$.
- For all subsets $A$ of $X$ either $A in mathcal{F}$ or $Xsetminus A in mathcal{F}$.
Equivalence 2. shows that if $mathcal{F}$ contains a finite subset $F={x_1,ldots,x_m}$ it contains one of the sets ${x_i}$, and then $mathcal{F} = {A: x_i in A}$ for that $x_i$, i.e. $mathcal{F}$ is a principal filter. IN particular on a finite set $X$ we only have the principal ultrafilters.
So a non-principal filter cannot contain any finite subset $F$, so by 3. $mathcal{F}$ contains all co-finite subsets, i.e. $mathcal{F}$ is an enlargement of the co-finite filter.
So ultrafilters are of two types: the principal ones (of which there are as many as points of $X$) and the free ones, that we can "get" by invoking choice and somehow extending the cofinite filter to an ultrafilter. There are $2^{2^{|X|}}$ of those for an infinite set $X$ (so many many more), assuming choice. But these are non-constructive and we can try to visualise them as some way of choosing for each infinite subset $A$ of $X$ whether to put $A$ in $mathcal{F}$ or its complement, but in a "filter-consistent" way (if we put in a set, we also have to put in all larger sets, and if we put in two, we also have to add the intersection) and no finite set is allowed in $mathcal{F}$ if we want a free ultrafilter, so their complements are surely in. It's not much but maybe it helps.
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
add a comment |
$begingroup$
It's useful to have the following characterisations of ultrafilters on a set $X$:
For a non-trivial filter $mathcal{F}$ on $X$, the following statements are equivalent:
$mathcal{F}$ is an ultrafilter (i.e. maximal w.r.t. inclusion)- For all finitely many $A_1,ldots A_n subseteq X$ we have that $cup_{i=1}^n A_i in mathcal{F}$ iff there is some $1 le i le n$ such that $A_i in mathcal{F}$.
- For all subsets $A$ of $X$ either $A in mathcal{F}$ or $Xsetminus A in mathcal{F}$.
Equivalence 2. shows that if $mathcal{F}$ contains a finite subset $F={x_1,ldots,x_m}$ it contains one of the sets ${x_i}$, and then $mathcal{F} = {A: x_i in A}$ for that $x_i$, i.e. $mathcal{F}$ is a principal filter. IN particular on a finite set $X$ we only have the principal ultrafilters.
So a non-principal filter cannot contain any finite subset $F$, so by 3. $mathcal{F}$ contains all co-finite subsets, i.e. $mathcal{F}$ is an enlargement of the co-finite filter.
So ultrafilters are of two types: the principal ones (of which there are as many as points of $X$) and the free ones, that we can "get" by invoking choice and somehow extending the cofinite filter to an ultrafilter. There are $2^{2^{|X|}}$ of those for an infinite set $X$ (so many many more), assuming choice. But these are non-constructive and we can try to visualise them as some way of choosing for each infinite subset $A$ of $X$ whether to put $A$ in $mathcal{F}$ or its complement, but in a "filter-consistent" way (if we put in a set, we also have to put in all larger sets, and if we put in two, we also have to add the intersection) and no finite set is allowed in $mathcal{F}$ if we want a free ultrafilter, so their complements are surely in. It's not much but maybe it helps.
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
add a comment |
$begingroup$
It's useful to have the following characterisations of ultrafilters on a set $X$:
For a non-trivial filter $mathcal{F}$ on $X$, the following statements are equivalent:
$mathcal{F}$ is an ultrafilter (i.e. maximal w.r.t. inclusion)- For all finitely many $A_1,ldots A_n subseteq X$ we have that $cup_{i=1}^n A_i in mathcal{F}$ iff there is some $1 le i le n$ such that $A_i in mathcal{F}$.
- For all subsets $A$ of $X$ either $A in mathcal{F}$ or $Xsetminus A in mathcal{F}$.
Equivalence 2. shows that if $mathcal{F}$ contains a finite subset $F={x_1,ldots,x_m}$ it contains one of the sets ${x_i}$, and then $mathcal{F} = {A: x_i in A}$ for that $x_i$, i.e. $mathcal{F}$ is a principal filter. IN particular on a finite set $X$ we only have the principal ultrafilters.
So a non-principal filter cannot contain any finite subset $F$, so by 3. $mathcal{F}$ contains all co-finite subsets, i.e. $mathcal{F}$ is an enlargement of the co-finite filter.
So ultrafilters are of two types: the principal ones (of which there are as many as points of $X$) and the free ones, that we can "get" by invoking choice and somehow extending the cofinite filter to an ultrafilter. There are $2^{2^{|X|}}$ of those for an infinite set $X$ (so many many more), assuming choice. But these are non-constructive and we can try to visualise them as some way of choosing for each infinite subset $A$ of $X$ whether to put $A$ in $mathcal{F}$ or its complement, but in a "filter-consistent" way (if we put in a set, we also have to put in all larger sets, and if we put in two, we also have to add the intersection) and no finite set is allowed in $mathcal{F}$ if we want a free ultrafilter, so their complements are surely in. It's not much but maybe it helps.
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
It's useful to have the following characterisations of ultrafilters on a set $X$:
For a non-trivial filter $mathcal{F}$ on $X$, the following statements are equivalent:
$mathcal{F}$ is an ultrafilter (i.e. maximal w.r.t. inclusion)- For all finitely many $A_1,ldots A_n subseteq X$ we have that $cup_{i=1}^n A_i in mathcal{F}$ iff there is some $1 le i le n$ such that $A_i in mathcal{F}$.
- For all subsets $A$ of $X$ either $A in mathcal{F}$ or $Xsetminus A in mathcal{F}$.
Equivalence 2. shows that if $mathcal{F}$ contains a finite subset $F={x_1,ldots,x_m}$ it contains one of the sets ${x_i}$, and then $mathcal{F} = {A: x_i in A}$ for that $x_i$, i.e. $mathcal{F}$ is a principal filter. IN particular on a finite set $X$ we only have the principal ultrafilters.
So a non-principal filter cannot contain any finite subset $F$, so by 3. $mathcal{F}$ contains all co-finite subsets, i.e. $mathcal{F}$ is an enlargement of the co-finite filter.
So ultrafilters are of two types: the principal ones (of which there are as many as points of $X$) and the free ones, that we can "get" by invoking choice and somehow extending the cofinite filter to an ultrafilter. There are $2^{2^{|X|}}$ of those for an infinite set $X$ (so many many more), assuming choice. But these are non-constructive and we can try to visualise them as some way of choosing for each infinite subset $A$ of $X$ whether to put $A$ in $mathcal{F}$ or its complement, but in a "filter-consistent" way (if we put in a set, we also have to put in all larger sets, and if we put in two, we also have to add the intersection) and no finite set is allowed in $mathcal{F}$ if we want a free ultrafilter, so their complements are surely in. It's not much but maybe it helps.
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
answered Jan 13 at 22:42
Henno BrandsmaHenno Brandsma
109k347114
109k347114
add a comment |
add a comment |
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$begingroup$
No, your reasoning is fine. Ultrafilters are only unique on a singleton. But you can ask if they are equivalent, namely if $U$ and $V$ are two ultrafilters on a set $X$, is there a permutation of $X$ which maps $U$ to $V$ (by means of $Asubseteq X$ is mapped to ${f(a)mid ain A}$). And then the answer is a bit more complicated, but still is negative.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:14
$begingroup$
Meaning that, even if a set is infinite, its ultrafilter(s) may not be unique? @AsafKaragila
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– joshuaheckroodt
Jan 12 at 8:15
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Where did you use the finiteness of the set? You just used the fact that $3$ and $4$ are two different things.
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– Asaf Karagila♦
Jan 12 at 8:15
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When a set is finite all of its ultrafilters are principal. When a set is infinite, you have the principal ultrafilters, but you also have free filters, which can only be extended to free ultrafilters, but that requires the axiom of choice to get involved (and indeed it is consistent that all ultrafilters are principal). So to that end the word "constructing", if interpreted as providing an explicit definition, is generally accepted to not apply to free ultrafilters.
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– Asaf Karagila♦
Jan 12 at 8:20
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To some extent. But analysis itself also depends on the axiom of choice. The whole notion of measures, continuity, etc. can be formulated without choice, but end up nearly useless when doing so. There are a few papers you can find about some choice which is sufficient to develop NSA, those assumptions are not unreasonable in the general scheme of things.
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– Asaf Karagila♦
Jan 12 at 8:40