difference between convergence along free ultrafilter and common convergence
$begingroup$
Suppose $mathcal{F}$ is any free ultrafilter on $betamathbb{N}setminusmathbb{N}$,$(x_n)$ is a sequence of complex numbers.
My question is:What is the deference between the limit along $mathcal{F}$ ,$lim_{mathcal{F}}x_n$ and $lim_{nto infty}x_n$?If $lim_{ntoinfty}x_nnot to 0$,can we conclude that $lim_{mathcal{F}}x_nnotto 0$?
general-topology convergence filters
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
$endgroup$
add a comment |
$begingroup$
Suppose $mathcal{F}$ is any free ultrafilter on $betamathbb{N}setminusmathbb{N}$,$(x_n)$ is a sequence of complex numbers.
My question is:What is the deference between the limit along $mathcal{F}$ ,$lim_{mathcal{F}}x_n$ and $lim_{nto infty}x_n$?If $lim_{ntoinfty}x_nnot to 0$,can we conclude that $lim_{mathcal{F}}x_nnotto 0$?
general-topology convergence filters
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
$endgroup$
2
$begingroup$
The sequence $(0,1,0,1,0,1,0,1,0,1,dots)$ doesn't converge. It converges with respect to every ultrafilter $mathcal F$ on $mathbb N$. You should figure out what it converges to, in terms of $mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $mathbb N$.
$endgroup$
– Andreas Blass
Dec 7 '18 at 2:23
add a comment |
$begingroup$
Suppose $mathcal{F}$ is any free ultrafilter on $betamathbb{N}setminusmathbb{N}$,$(x_n)$ is a sequence of complex numbers.
My question is:What is the deference between the limit along $mathcal{F}$ ,$lim_{mathcal{F}}x_n$ and $lim_{nto infty}x_n$?If $lim_{ntoinfty}x_nnot to 0$,can we conclude that $lim_{mathcal{F}}x_nnotto 0$?
general-topology convergence filters
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
$endgroup$
Suppose $mathcal{F}$ is any free ultrafilter on $betamathbb{N}setminusmathbb{N}$,$(x_n)$ is a sequence of complex numbers.
My question is:What is the deference between the limit along $mathcal{F}$ ,$lim_{mathcal{F}}x_n$ and $lim_{nto infty}x_n$?If $lim_{ntoinfty}x_nnot to 0$,can we conclude that $lim_{mathcal{F}}x_nnotto 0$?
general-topology convergence filters
general-topology convergence filters
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
edited Dec 6 '18 at 22:09
Eric Wofsey
181k12208336
181k12208336
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
asked Dec 6 '18 at 21:59
mathrookiemathrookie
832512
832512
2
$begingroup$
The sequence $(0,1,0,1,0,1,0,1,0,1,dots)$ doesn't converge. It converges with respect to every ultrafilter $mathcal F$ on $mathbb N$. You should figure out what it converges to, in terms of $mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $mathbb N$.
$endgroup$
– Andreas Blass
Dec 7 '18 at 2:23
add a comment |
2
$begingroup$
The sequence $(0,1,0,1,0,1,0,1,0,1,dots)$ doesn't converge. It converges with respect to every ultrafilter $mathcal F$ on $mathbb N$. You should figure out what it converges to, in terms of $mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $mathbb N$.
$endgroup$
– Andreas Blass
Dec 7 '18 at 2:23
2
2
$begingroup$
The sequence $(0,1,0,1,0,1,0,1,0,1,dots)$ doesn't converge. It converges with respect to every ultrafilter $mathcal F$ on $mathbb N$. You should figure out what it converges to, in terms of $mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $mathbb N$.
$endgroup$
– Andreas Blass
Dec 7 '18 at 2:23
$begingroup$
The sequence $(0,1,0,1,0,1,0,1,0,1,dots)$ doesn't converge. It converges with respect to every ultrafilter $mathcal F$ on $mathbb N$. You should figure out what it converges to, in terms of $mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $mathbb N$.
$endgroup$
– Andreas Blass
Dec 7 '18 at 2:23
add a comment |
1 Answer
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$begingroup$
No, I'll use Andreas' elementary example from the comments: the sequence $x_n = 0$ for $n$ even, $x_n= 1$ for $n$ odd, does not converge in $mathbb{R}$ in the usual topology.
However, if $mathcal{F}$ is a free ultrafilter on $mathbb{N}$, then either $E= {2n: n in mathbb{N}}$ or $O = {2n+1: n in mathbb{N}}$ is in $mathcal{F}$ (as $O cup E = mathbb{N}$ and we have an ultrafilter). If the former, then for any neighbourhood $U$ of $0$ we have that ${n : x_n in U} supseteq E$ so ${n : x_n in U} in mathcal{F}$, and so $lim_mathcal{F} x_n = 0$ and if the latter we similarly see that $lim_mathcal{F} x_n = 1$. In any case, the non-convergent sequence $(x_n)_n$ does converge along any free ultrafilter. So the proposed implication does not hold.
In fact one can show that any bounded sequence in the reals or complex numbers has a limit along any ultrafilter whether it converges (usually) or not.
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
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$begingroup$
No, I'll use Andreas' elementary example from the comments: the sequence $x_n = 0$ for $n$ even, $x_n= 1$ for $n$ odd, does not converge in $mathbb{R}$ in the usual topology.
However, if $mathcal{F}$ is a free ultrafilter on $mathbb{N}$, then either $E= {2n: n in mathbb{N}}$ or $O = {2n+1: n in mathbb{N}}$ is in $mathcal{F}$ (as $O cup E = mathbb{N}$ and we have an ultrafilter). If the former, then for any neighbourhood $U$ of $0$ we have that ${n : x_n in U} supseteq E$ so ${n : x_n in U} in mathcal{F}$, and so $lim_mathcal{F} x_n = 0$ and if the latter we similarly see that $lim_mathcal{F} x_n = 1$. In any case, the non-convergent sequence $(x_n)_n$ does converge along any free ultrafilter. So the proposed implication does not hold.
In fact one can show that any bounded sequence in the reals or complex numbers has a limit along any ultrafilter whether it converges (usually) or not.
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
No, I'll use Andreas' elementary example from the comments: the sequence $x_n = 0$ for $n$ even, $x_n= 1$ for $n$ odd, does not converge in $mathbb{R}$ in the usual topology.
However, if $mathcal{F}$ is a free ultrafilter on $mathbb{N}$, then either $E= {2n: n in mathbb{N}}$ or $O = {2n+1: n in mathbb{N}}$ is in $mathcal{F}$ (as $O cup E = mathbb{N}$ and we have an ultrafilter). If the former, then for any neighbourhood $U$ of $0$ we have that ${n : x_n in U} supseteq E$ so ${n : x_n in U} in mathcal{F}$, and so $lim_mathcal{F} x_n = 0$ and if the latter we similarly see that $lim_mathcal{F} x_n = 1$. In any case, the non-convergent sequence $(x_n)_n$ does converge along any free ultrafilter. So the proposed implication does not hold.
In fact one can show that any bounded sequence in the reals or complex numbers has a limit along any ultrafilter whether it converges (usually) or not.
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
No, I'll use Andreas' elementary example from the comments: the sequence $x_n = 0$ for $n$ even, $x_n= 1$ for $n$ odd, does not converge in $mathbb{R}$ in the usual topology.
However, if $mathcal{F}$ is a free ultrafilter on $mathbb{N}$, then either $E= {2n: n in mathbb{N}}$ or $O = {2n+1: n in mathbb{N}}$ is in $mathcal{F}$ (as $O cup E = mathbb{N}$ and we have an ultrafilter). If the former, then for any neighbourhood $U$ of $0$ we have that ${n : x_n in U} supseteq E$ so ${n : x_n in U} in mathcal{F}$, and so $lim_mathcal{F} x_n = 0$ and if the latter we similarly see that $lim_mathcal{F} x_n = 1$. In any case, the non-convergent sequence $(x_n)_n$ does converge along any free ultrafilter. So the proposed implication does not hold.
In fact one can show that any bounded sequence in the reals or complex numbers has a limit along any ultrafilter whether it converges (usually) or not.
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
No, I'll use Andreas' elementary example from the comments: the sequence $x_n = 0$ for $n$ even, $x_n= 1$ for $n$ odd, does not converge in $mathbb{R}$ in the usual topology.
However, if $mathcal{F}$ is a free ultrafilter on $mathbb{N}$, then either $E= {2n: n in mathbb{N}}$ or $O = {2n+1: n in mathbb{N}}$ is in $mathcal{F}$ (as $O cup E = mathbb{N}$ and we have an ultrafilter). If the former, then for any neighbourhood $U$ of $0$ we have that ${n : x_n in U} supseteq E$ so ${n : x_n in U} in mathcal{F}$, and so $lim_mathcal{F} x_n = 0$ and if the latter we similarly see that $lim_mathcal{F} x_n = 1$. In any case, the non-convergent sequence $(x_n)_n$ does converge along any free ultrafilter. So the proposed implication does not hold.
In fact one can show that any bounded sequence in the reals or complex numbers has a limit along any ultrafilter whether it converges (usually) or not.
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
edited Dec 9 '18 at 6:45
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 8 '18 at 22:32
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
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$begingroup$
The sequence $(0,1,0,1,0,1,0,1,0,1,dots)$ doesn't converge. It converges with respect to every ultrafilter $mathcal F$ on $mathbb N$. You should figure out what it converges to, in terms of $mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $mathbb N$.
$endgroup$
– Andreas Blass
Dec 7 '18 at 2:23