find a free ultrafilter on $Bbb N$
Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=x_0$.
Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.
For any point $cin Bbb C$, can we construct a free ultrafilter $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to c$?
general-topology convergence filters
edited Dec 9 at 8:55
Gaby Alfonso
676315
asked Dec 9 at 8:05
mathrookie
804512
add a comment |
Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=x_0$.
Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.
For any point $cin Bbb C$, can we construct a free ultrafilter $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to c$?
general-topology convergence filters
edited Dec 9 at 8:55
Gaby Alfonso
676315
asked Dec 9 at 8:05
mathrookie
804512
I suppose that this is what you meant, but one should be a bit careful with formulation such as construct a free ultrafilter. Probably more appropriate formulation is prove that free ultrafilter with the required properties exists. (Since existence of a free ulftrafilter cannot be shown in ZF, you cannot expect an explicit description of such things. So any proof will contain some step which is non-contructive. For example, it might rely on Zorn's lemma.)
– Martin Sleziak
Dec 9 at 9:59
add a comment |
Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=x_0$.
Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.
For any point $cin Bbb C$, can we construct a free ultrafilter $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to c$?
general-topology convergence filters
edited Dec 9 at 8:55
Gaby Alfonso
676315
asked Dec 9 at 8:05
mathrookie
804512
Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=x_0$.
Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.
For any point $cin Bbb C$, can we construct a free ultrafilter $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to c$?
general-topology convergence filters
general-topology convergence filters
edited Dec 9 at 8:55
Gaby Alfonso
676315
asked Dec 9 at 8:05
mathrookie
804512
edited Dec 9 at 8:55
Gaby Alfonso
676315
asked Dec 9 at 8:05
mathrookie
804512
edited Dec 9 at 8:55
Gaby Alfonso
676315
edited Dec 9 at 8:55
Gaby Alfonso
676315
edited Dec 9 at 8:55
Gaby Alfonso
676315
676315
asked Dec 9 at 8:05
mathrookie
804512
asked Dec 9 at 8:05
mathrookie
804512
asked Dec 9 at 8:05
mathrookie
804512
804512
I suppose that this is what you meant, but one should be a bit careful with formulation such as construct a free ultrafilter. Probably more appropriate formulation is prove that free ultrafilter with the required properties exists. (Since existence of a free ulftrafilter cannot be shown in ZF, you cannot expect an explicit description of such things. So any proof will contain some step which is non-contructive. For example, it might rely on Zorn's lemma.)
– Martin Sleziak
Dec 9 at 9:59
add a comment |
I suppose that this is what you meant, but one should be a bit careful with formulation such as construct a free ultrafilter. Probably more appropriate formulation is prove that free ultrafilter with the required properties exists. (Since existence of a free ulftrafilter cannot be shown in ZF, you cannot expect an explicit description of such things. So any proof will contain some step which is non-contructive. For example, it might rely on Zorn's lemma.)
– Martin Sleziak
Dec 9 at 9:59
I suppose that this is what you meant, but one should be a bit careful with formulation such as construct a free ultrafilter. Probably more appropriate formulation is prove that free ultrafilter with the required properties exists. (Since existence of a free ulftrafilter cannot be shown in ZF, you cannot expect an explicit description of such things. So any proof will contain some step which is non-contructive. For example, it might rely on Zorn's lemma.)
– Martin Sleziak
Dec 9 at 9:59
I suppose that this is what you meant, but one should be a bit careful with formulation such as construct a free ultrafilter. Probably more appropriate formulation is prove that free ultrafilter with the required properties exists. (Since existence of a free ulftrafilter cannot be shown in ZF, you cannot expect an explicit description of such things. So any proof will contain some step which is non-contructive. For example, it might rely on Zorn's lemma.)
– Martin Sleziak
Dec 9 at 9:59
add a comment |
3 Answers
3
active
oldest
votes
No: if $(x_n)$ actually converges (as an ordinary sequence) to $x_0$, then it also converges to $x_0$ with respect to every free ultrafilter.
More generally, the set of limits of a sequence with respect to free ultrafilters is exactly the set of accumulation points of the sequence. You seem to already be aware of one direction of this implication; for the other direction, if $(x_n)$ converges to $c$ with respect to a free ultrafilter $omega$, then for every neighborhood $U$ of $c$ the set ${n:x_nin U}$ is in $omega$ and so in particular is infinite.
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
You mean the cofinte filter $mathcal{F_0}$ on $Bbb N$ is contained in any free ultrafilter $mathcal{F}$ on $Bbb N$,so if $x_nto x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have ${nin Bbb N:x_nin U_{x_0}}in mathcal{F_0}subsetmathcal{F}$.Is my thought correct?
– mathrookie
Dec 9 at 8:50
No. We know that ${ninmathbb{N}:x_nin U_{x_0}}inmathcal{F}$, not that it is in $mathcal{F}_0$. But every element of $mathcal{F}$ is infinite, so that's all we need.
– Eric Wofsey
Dec 9 at 9:02
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
– mathrookie
Dec 9 at 9:35
According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $mathcal{F}$ on $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
– mathrookie
Dec 9 at 9:44
add a comment |
Just to sketch a quick answer:
1) If your sequence happened to actually converge to $a$ (which is my notation for $x_0$, I dislike indices for objects which are not components of a certain family indexed by a certain set of indices), then the limit along any free ultrafilter will still be $a$.
2) You will only be able to find such ultrafilters for points $c notin overline{{x_n}_{n in mathbb{N}}}$. As pointed out in one of the answers above, the set of limits along ultrafilters coincides with the set of points adherent to the given sequence $x$.
answered Dec 9 at 8:33
ΑΘΩ
2363
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
– mathrookie
Dec 9 at 9:00
what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to ${x_{n}}_{n in mathbb{N}}$.
– ΑΘΩ
Dec 9 at 10:20
add a comment |
For 1, $x_0$ being an accumulation point of the sequence means that
all sets $hat{U}:={n in omega: x_n in U}$ are infinite, where $U$ ranges over the neighbourhoods of $x_0$, and obey the FIP. So they extend to some ultrafilter, and $(x_n)$ converges along this ultrafilter. So yes, if you really meant converge (as you should have).
An ultrafilter limit is unique. So if it converges to $c$ it won't converge to any other $c'$.
answered Dec 9 at 8:11
Henno Brandsma
105k346113
I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $mathcal{F^{prime}}$ such that $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
– mathrookie
Dec 9 at 23:39
@mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g.
– Henno Brandsma
Dec 10 at 4:38
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
– mathrookie
Dec 17 at 11:00
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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active
oldest
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active
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No: if $(x_n)$ actually converges (as an ordinary sequence) to $x_0$, then it also converges to $x_0$ with respect to every free ultrafilter.
More generally, the set of limits of a sequence with respect to free ultrafilters is exactly the set of accumulation points of the sequence. You seem to already be aware of one direction of this implication; for the other direction, if $(x_n)$ converges to $c$ with respect to a free ultrafilter $omega$, then for every neighborhood $U$ of $c$ the set ${n:x_nin U}$ is in $omega$ and so in particular is infinite.
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
You mean the cofinte filter $mathcal{F_0}$ on $Bbb N$ is contained in any free ultrafilter $mathcal{F}$ on $Bbb N$,so if $x_nto x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have ${nin Bbb N:x_nin U_{x_0}}in mathcal{F_0}subsetmathcal{F}$.Is my thought correct?
– mathrookie
Dec 9 at 8:50
No. We know that ${ninmathbb{N}:x_nin U_{x_0}}inmathcal{F}$, not that it is in $mathcal{F}_0$. But every element of $mathcal{F}$ is infinite, so that's all we need.
– Eric Wofsey
Dec 9 at 9:02
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
– mathrookie
Dec 9 at 9:35
According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $mathcal{F}$ on $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
– mathrookie
Dec 9 at 9:44
add a comment |
No: if $(x_n)$ actually converges (as an ordinary sequence) to $x_0$, then it also converges to $x_0$ with respect to every free ultrafilter.
More generally, the set of limits of a sequence with respect to free ultrafilters is exactly the set of accumulation points of the sequence. You seem to already be aware of one direction of this implication; for the other direction, if $(x_n)$ converges to $c$ with respect to a free ultrafilter $omega$, then for every neighborhood $U$ of $c$ the set ${n:x_nin U}$ is in $omega$ and so in particular is infinite.
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
You mean the cofinte filter $mathcal{F_0}$ on $Bbb N$ is contained in any free ultrafilter $mathcal{F}$ on $Bbb N$,so if $x_nto x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have ${nin Bbb N:x_nin U_{x_0}}in mathcal{F_0}subsetmathcal{F}$.Is my thought correct?
– mathrookie
Dec 9 at 8:50
No. We know that ${ninmathbb{N}:x_nin U_{x_0}}inmathcal{F}$, not that it is in $mathcal{F}_0$. But every element of $mathcal{F}$ is infinite, so that's all we need.
– Eric Wofsey
Dec 9 at 9:02
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
– mathrookie
Dec 9 at 9:35
According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $mathcal{F}$ on $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
– mathrookie
Dec 9 at 9:44
add a comment |
No: if $(x_n)$ actually converges (as an ordinary sequence) to $x_0$, then it also converges to $x_0$ with respect to every free ultrafilter.
More generally, the set of limits of a sequence with respect to free ultrafilters is exactly the set of accumulation points of the sequence. You seem to already be aware of one direction of this implication; for the other direction, if $(x_n)$ converges to $c$ with respect to a free ultrafilter $omega$, then for every neighborhood $U$ of $c$ the set ${n:x_nin U}$ is in $omega$ and so in particular is infinite.
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
No: if $(x_n)$ actually converges (as an ordinary sequence) to $x_0$, then it also converges to $x_0$ with respect to every free ultrafilter.
More generally, the set of limits of a sequence with respect to free ultrafilters is exactly the set of accumulation points of the sequence. You seem to already be aware of one direction of this implication; for the other direction, if $(x_n)$ converges to $c$ with respect to a free ultrafilter $omega$, then for every neighborhood $U$ of $c$ the set ${n:x_nin U}$ is in $omega$ and so in particular is infinite.
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
answered Dec 9 at 8:10
Eric Wofsey
179k12204331
179k12204331
You mean the cofinte filter $mathcal{F_0}$ on $Bbb N$ is contained in any free ultrafilter $mathcal{F}$ on $Bbb N$,so if $x_nto x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have ${nin Bbb N:x_nin U_{x_0}}in mathcal{F_0}subsetmathcal{F}$.Is my thought correct?
– mathrookie
Dec 9 at 8:50
No. We know that ${ninmathbb{N}:x_nin U_{x_0}}inmathcal{F}$, not that it is in $mathcal{F}_0$. But every element of $mathcal{F}$ is infinite, so that's all we need.
– Eric Wofsey
Dec 9 at 9:02
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
– mathrookie
Dec 9 at 9:35
According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $mathcal{F}$ on $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
– mathrookie
Dec 9 at 9:44
add a comment |
You mean the cofinte filter $mathcal{F_0}$ on $Bbb N$ is contained in any free ultrafilter $mathcal{F}$ on $Bbb N$,so if $x_nto x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have ${nin Bbb N:x_nin U_{x_0}}in mathcal{F_0}subsetmathcal{F}$.Is my thought correct?
– mathrookie
Dec 9 at 8:50
No. We know that ${ninmathbb{N}:x_nin U_{x_0}}inmathcal{F}$, not that it is in $mathcal{F}_0$. But every element of $mathcal{F}$ is infinite, so that's all we need.
– Eric Wofsey
Dec 9 at 9:02
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
– mathrookie
Dec 9 at 9:35
According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $mathcal{F}$ on $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
– mathrookie
Dec 9 at 9:44
You mean the cofinte filter $mathcal{F_0}$ on $Bbb N$ is contained in any free ultrafilter $mathcal{F}$ on $Bbb N$,so if $x_nto x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have ${nin Bbb N:x_nin U_{x_0}}in mathcal{F_0}subsetmathcal{F}$.Is my thought correct?
– mathrookie
Dec 9 at 8:50
You mean the cofinte filter $mathcal{F_0}$ on $Bbb N$ is contained in any free ultrafilter $mathcal{F}$ on $Bbb N$,so if $x_nto x_0$,then for any neighborhood $U_{x_0}$ of $x_0$,we have ${nin Bbb N:x_nin U_{x_0}}in mathcal{F_0}subsetmathcal{F}$.Is my thought correct?
– mathrookie
Dec 9 at 8:50
No. We know that ${ninmathbb{N}:x_nin U_{x_0}}inmathcal{F}$, not that it is in $mathcal{F}_0$. But every element of $mathcal{F}$ is infinite, so that's all we need.
– Eric Wofsey
Dec 9 at 9:02
No. We know that ${ninmathbb{N}:x_nin U_{x_0}}inmathcal{F}$, not that it is in $mathcal{F}_0$. But every element of $mathcal{F}$ is infinite, so that's all we need.
– Eric Wofsey
Dec 9 at 9:02
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
– mathrookie
Dec 9 at 9:35
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
– mathrookie
Dec 9 at 9:35
According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $mathcal{F}$ on $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
– mathrookie
Dec 9 at 9:44
According to the above answer,If point $c$ is not the acculation point of the sequence,can we deduce that for any free ultrafilter $mathcal{F}$ on $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
– mathrookie
Dec 9 at 9:44
add a comment |
Just to sketch a quick answer:
1) If your sequence happened to actually converge to $a$ (which is my notation for $x_0$, I dislike indices for objects which are not components of a certain family indexed by a certain set of indices), then the limit along any free ultrafilter will still be $a$.
2) You will only be able to find such ultrafilters for points $c notin overline{{x_n}_{n in mathbb{N}}}$. As pointed out in one of the answers above, the set of limits along ultrafilters coincides with the set of points adherent to the given sequence $x$.
answered Dec 9 at 8:33
ΑΘΩ
2363
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
– mathrookie
Dec 9 at 9:00
what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to ${x_{n}}_{n in mathbb{N}}$.
– ΑΘΩ
Dec 9 at 10:20
add a comment |
Just to sketch a quick answer:
1) If your sequence happened to actually converge to $a$ (which is my notation for $x_0$, I dislike indices for objects which are not components of a certain family indexed by a certain set of indices), then the limit along any free ultrafilter will still be $a$.
2) You will only be able to find such ultrafilters for points $c notin overline{{x_n}_{n in mathbb{N}}}$. As pointed out in one of the answers above, the set of limits along ultrafilters coincides with the set of points adherent to the given sequence $x$.
answered Dec 9 at 8:33
ΑΘΩ
2363
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
– mathrookie
Dec 9 at 9:00
what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to ${x_{n}}_{n in mathbb{N}}$.
– ΑΘΩ
Dec 9 at 10:20
add a comment |
Just to sketch a quick answer:
1) If your sequence happened to actually converge to $a$ (which is my notation for $x_0$, I dislike indices for objects which are not components of a certain family indexed by a certain set of indices), then the limit along any free ultrafilter will still be $a$.
2) You will only be able to find such ultrafilters for points $c notin overline{{x_n}_{n in mathbb{N}}}$. As pointed out in one of the answers above, the set of limits along ultrafilters coincides with the set of points adherent to the given sequence $x$.
answered Dec 9 at 8:33
ΑΘΩ
2363
Just to sketch a quick answer:
1) If your sequence happened to actually converge to $a$ (which is my notation for $x_0$, I dislike indices for objects which are not components of a certain family indexed by a certain set of indices), then the limit along any free ultrafilter will still be $a$.
2) You will only be able to find such ultrafilters for points $c notin overline{{x_n}_{n in mathbb{N}}}$. As pointed out in one of the answers above, the set of limits along ultrafilters coincides with the set of points adherent to the given sequence $x$.
answered Dec 9 at 8:33
ΑΘΩ
2363
edited Dec 9 at 8:39
answered Dec 9 at 8:33
ΑΘΩ
2363
answered Dec 9 at 8:33
ΑΘΩ
2363
answered Dec 9 at 8:33
ΑΘΩ
2363
2363
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
– mathrookie
Dec 9 at 9:00
what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to ${x_{n}}_{n in mathbb{N}}$.
– ΑΘΩ
Dec 9 at 10:20
add a comment |
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
– mathrookie
Dec 9 at 9:00
what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to ${x_{n}}_{n in mathbb{N}}$.
– ΑΘΩ
Dec 9 at 10:20
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
– mathrookie
Dec 9 at 9:00
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
– mathrookie
Dec 9 at 9:00
what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to ${x_{n}}_{n in mathbb{N}}$.
– ΑΘΩ
Dec 9 at 10:20
what kind of ultrafilter? do you mean such that convergence to $c$ along the ultrafilter will not occur? If so, matters are quite simple for the limit along any ultrafilter $mathscr{F}$ will necessarily exist (the sequence induces an ultrafilter in the compact space which is the closure of its bounded set of terms) and it necessarily belongs to the said closure! So it's not a matter of constructing a special ultrafilter to avoid convergence to $c$, that will automatically happen with any ultrafilter whatsoever, as long as $c$ is not adherent to ${x_{n}}_{n in mathbb{N}}$.
– ΑΘΩ
Dec 9 at 10:20
add a comment |
For 1, $x_0$ being an accumulation point of the sequence means that
all sets $hat{U}:={n in omega: x_n in U}$ are infinite, where $U$ ranges over the neighbourhoods of $x_0$, and obey the FIP. So they extend to some ultrafilter, and $(x_n)$ converges along this ultrafilter. So yes, if you really meant converge (as you should have).
An ultrafilter limit is unique. So if it converges to $c$ it won't converge to any other $c'$.
answered Dec 9 at 8:11
Henno Brandsma
105k346113
I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $mathcal{F^{prime}}$ such that $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
– mathrookie
Dec 9 at 23:39
@mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g.
– Henno Brandsma
Dec 10 at 4:38
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
– mathrookie
Dec 17 at 11:00
add a comment |
For 1, $x_0$ being an accumulation point of the sequence means that
all sets $hat{U}:={n in omega: x_n in U}$ are infinite, where $U$ ranges over the neighbourhoods of $x_0$, and obey the FIP. So they extend to some ultrafilter, and $(x_n)$ converges along this ultrafilter. So yes, if you really meant converge (as you should have).
An ultrafilter limit is unique. So if it converges to $c$ it won't converge to any other $c'$.
answered Dec 9 at 8:11
Henno Brandsma
105k346113
I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $mathcal{F^{prime}}$ such that $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
– mathrookie
Dec 9 at 23:39
@mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g.
– Henno Brandsma
Dec 10 at 4:38
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
– mathrookie
Dec 17 at 11:00
add a comment |
For 1, $x_0$ being an accumulation point of the sequence means that
all sets $hat{U}:={n in omega: x_n in U}$ are infinite, where $U$ ranges over the neighbourhoods of $x_0$, and obey the FIP. So they extend to some ultrafilter, and $(x_n)$ converges along this ultrafilter. So yes, if you really meant converge (as you should have).
An ultrafilter limit is unique. So if it converges to $c$ it won't converge to any other $c'$.
answered Dec 9 at 8:11
Henno Brandsma
105k346113
For 1, $x_0$ being an accumulation point of the sequence means that
all sets $hat{U}:={n in omega: x_n in U}$ are infinite, where $U$ ranges over the neighbourhoods of $x_0$, and obey the FIP. So they extend to some ultrafilter, and $(x_n)$ converges along this ultrafilter. So yes, if you really meant converge (as you should have).
An ultrafilter limit is unique. So if it converges to $c$ it won't converge to any other $c'$.
answered Dec 9 at 8:11
Henno Brandsma
105k346113
edited Dec 9 at 12:28
answered Dec 9 at 8:11
Henno Brandsma
105k346113
answered Dec 9 at 8:11
Henno Brandsma
105k346113
answered Dec 9 at 8:11
Henno Brandsma
105k346113
105k346113
I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $mathcal{F^{prime}}$ such that $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
– mathrookie
Dec 9 at 23:39
@mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g.
– Henno Brandsma
Dec 10 at 4:38
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
– mathrookie
Dec 17 at 11:00
add a comment |
I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $mathcal{F^{prime}}$ such that $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
– mathrookie
Dec 9 at 23:39
@mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g.
– Henno Brandsma
Dec 10 at 4:38
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
– mathrookie
Dec 17 at 11:00
I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $mathcal{F^{prime}}$ such that $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
– mathrookie
Dec 9 at 23:39
I know that an ultrafilter limit is unique.But if $(x_n)$ has a accumulation point $c$,there exists free ultrafilter $mathcal{F}$ on $Bbb N$ such that $lim_{mathcal{F}}x_n=c$.Is the free ultrafilter unique?Does there exists another free ultrfilter $mathcal{F^{prime}}$ such that $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
– mathrookie
Dec 9 at 23:39
@mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g.
– Henno Brandsma
Dec 10 at 4:38
@mathrookie there could be another one, it depends on the sequence sometimes. The sequence $0,1,0,1,0,1,ldots$ has many ultrafilters along which it converges to $0$, likewise for $1$, and none to other point, e.g.
– Henno Brandsma
Dec 10 at 4:38
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
– mathrookie
Dec 17 at 11:00
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
– mathrookie
Dec 17 at 11:00
add a comment |
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I suppose that this is what you meant, but one should be a bit careful with formulation such as construct a free ultrafilter. Probably more appropriate formulation is prove that free ultrafilter with the required properties exists. (Since existence of a free ulftrafilter cannot be shown in ZF, you cannot expect an explicit description of such things. So any proof will contain some step which is non-contructive. For example, it might rely on Zorn's lemma.)
– Martin Sleziak
Dec 9 at 9:59