find a free ultrafilter on $Bbb N$












0












$begingroup$


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 $lim_{mathcal{F}}x_n=x_0$.




  1. Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.


  2. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.)
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 9:59
















0












$begingroup$


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 $lim_{mathcal{F}}x_n=x_0$.




  1. Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.


  2. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.)
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 9:59














0












0








0


1



$begingroup$


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 $lim_{mathcal{F}}x_n=x_0$.




  1. Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.


  2. 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







share|cite|improve this question











$endgroup$




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 $lim_{mathcal{F}}x_n=x_0$.




  1. Can we find a free ultrafiler $omega$ on $Bbb N$ such that $lim_{omega}x_nnot to x_0$.


  2. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 8:55









Gaby Alfonso

911317




911317










asked Dec 9 '18 at 8:05









mathrookiemathrookie

911512




911512












  • $begingroup$
    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.)
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 9:59


















  • $begingroup$
    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.)
    $endgroup$
    – Martin Sleziak
    Dec 9 '18 at 9:59
















$begingroup$
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.)
$endgroup$
– Martin Sleziak
Dec 9 '18 at 9:59




$begingroup$
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.)
$endgroup$
– Martin Sleziak
Dec 9 '18 at 9:59










3 Answers
3






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 8:50












  • $begingroup$
    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.
    $endgroup$
    – Eric Wofsey
    Dec 9 '18 at 9:02










  • $begingroup$
    What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:35










  • $begingroup$
    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 $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:44



















2












$begingroup$

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 $x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:00










  • $begingroup$
    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 ${x_{n}}_{n in mathbb{N}}$.
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:20



















1












$begingroup$

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 $c'$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 23:39










  • $begingroup$
    @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.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 4:38










  • $begingroup$
    .I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
    $endgroup$
    – mathrookie
    Dec 17 '18 at 11:00











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 8:50












  • $begingroup$
    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.
    $endgroup$
    – Eric Wofsey
    Dec 9 '18 at 9:02










  • $begingroup$
    What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:35










  • $begingroup$
    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 $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:44
















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 8:50












  • $begingroup$
    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.
    $endgroup$
    – Eric Wofsey
    Dec 9 '18 at 9:02










  • $begingroup$
    What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:35










  • $begingroup$
    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 $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:44














4












4








4





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 8:10









Eric WofseyEric Wofsey

186k14215342




186k14215342












  • $begingroup$
    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?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 8:50












  • $begingroup$
    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.
    $endgroup$
    – Eric Wofsey
    Dec 9 '18 at 9:02










  • $begingroup$
    What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:35










  • $begingroup$
    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 $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:44


















  • $begingroup$
    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?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 8:50












  • $begingroup$
    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.
    $endgroup$
    – Eric Wofsey
    Dec 9 '18 at 9:02










  • $begingroup$
    What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:35










  • $begingroup$
    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 $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:44
















$begingroup$
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?
$endgroup$
– mathrookie
Dec 9 '18 at 8:50






$begingroup$
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?
$endgroup$
– mathrookie
Dec 9 '18 at 8:50














$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 9 '18 at 9:02




$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 9 '18 at 9:02












$begingroup$
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
$endgroup$
– mathrookie
Dec 9 '18 at 9:35




$begingroup$
What I siad is about the case when $(x_n)$ converges to $x_0$ as a ordinary sequence,is my proof correct?
$endgroup$
– mathrookie
Dec 9 '18 at 9:35












$begingroup$
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 $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
$endgroup$
– mathrookie
Dec 9 '18 at 9:44




$begingroup$
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 $Bbb N$,$lim_{mathcal{F}}x_n not to c$?
$endgroup$
– mathrookie
Dec 9 '18 at 9:44











2












$begingroup$

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 $x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:00










  • $begingroup$
    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 ${x_{n}}_{n in mathbb{N}}$.
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:20
















2












$begingroup$

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 $x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:00










  • $begingroup$
    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 ${x_{n}}_{n in mathbb{N}}$.
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:20














2












2








2





$begingroup$

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 $x$.






share|cite|improve this answer











$endgroup$



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 $x$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 8:39

























answered Dec 9 '18 at 8:33









ΑΘΩΑΘΩ

2563




2563












  • $begingroup$
    If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:00










  • $begingroup$
    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 ${x_{n}}_{n in mathbb{N}}$.
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:20


















  • $begingroup$
    If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 9:00










  • $begingroup$
    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 ${x_{n}}_{n in mathbb{N}}$.
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:20
















$begingroup$
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
$endgroup$
– mathrookie
Dec 9 '18 at 9:00




$begingroup$
If $c$ is not a accumulation point ,can we find the free ultrafilter $mathcal{F}$ on $Bbb N$ ?how to construct it?
$endgroup$
– mathrookie
Dec 9 '18 at 9:00












$begingroup$
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 ${x_{n}}_{n in mathbb{N}}$.
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:20




$begingroup$
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 ${x_{n}}_{n in mathbb{N}}$.
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:20











1












$begingroup$

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 $c'$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 23:39










  • $begingroup$
    @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.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 4:38










  • $begingroup$
    .I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
    $endgroup$
    – mathrookie
    Dec 17 '18 at 11:00
















1












$begingroup$

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 $c'$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 23:39










  • $begingroup$
    @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.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 4:38










  • $begingroup$
    .I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
    $endgroup$
    – mathrookie
    Dec 17 '18 at 11:00














1












1








1





$begingroup$

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 $c'$.






share|cite|improve this answer











$endgroup$



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 $c'$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 12:28

























answered Dec 9 '18 at 8:11









Henno BrandsmaHenno Brandsma

110k347116




110k347116












  • $begingroup$
    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 $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 23:39










  • $begingroup$
    @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.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 4:38










  • $begingroup$
    .I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
    $endgroup$
    – mathrookie
    Dec 17 '18 at 11:00


















  • $begingroup$
    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 $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
    $endgroup$
    – mathrookie
    Dec 9 '18 at 23:39










  • $begingroup$
    @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.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 4:38










  • $begingroup$
    .I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
    $endgroup$
    – mathrookie
    Dec 17 '18 at 11:00
















$begingroup$
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 $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
$endgroup$
– mathrookie
Dec 9 '18 at 23:39




$begingroup$
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 $lim_{mathcal{F^{prime}}}x_n=lim_{mathcal{F}}x_n=c$?
$endgroup$
– mathrookie
Dec 9 '18 at 23:39












$begingroup$
@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.
$endgroup$
– Henno Brandsma
Dec 10 '18 at 4:38




$begingroup$
@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.
$endgroup$
– Henno Brandsma
Dec 10 '18 at 4:38












$begingroup$
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
$endgroup$
– mathrookie
Dec 17 '18 at 11:00




$begingroup$
.I have another question:If $x_nnot to c$,does there exist free untafiler $omega$ on $Bbb N$ such that $lim_{nto omega}x_nnot to c$?
$endgroup$
– mathrookie
Dec 17 '18 at 11:00


















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