Limit of sequence of sequences
$begingroup$
I got to thinking about sequences of Cauchy sequences. Here is a simple example. Let us define $b_n = (n,1,1,1,...)$ for $ninmathbb N$. So we have
begin{align}
b_1&=(1,1,1,1,...)\
b_2&=(2,1,1,1,...)\
b_3&=(3,1,1,1,...)\
dots
end{align}
Question: what is $lim_{ntoinfty} b_n$? It looks like that is not a valid sequence. Does this question even make sense? (I am not sure how such a limit would even be defined.)
functional-analysis cauchy-sequences
asked Jan 7 at 8:37
StefanieStefanie
597
$endgroup$
add a comment |
$begingroup$
I got to thinking about sequences of Cauchy sequences. Here is a simple example. Let us define $b_n = (n,1,1,1,...)$ for $ninmathbb N$. So we have
begin{align}
b_1&=(1,1,1,1,...)\
b_2&=(2,1,1,1,...)\
b_3&=(3,1,1,1,...)\
dots
end{align}
Question: what is $lim_{ntoinfty} b_n$? It looks like that is not a valid sequence. Does this question even make sense? (I am not sure how such a limit would even be defined.)
functional-analysis cauchy-sequences
asked Jan 7 at 8:37
StefanieStefanie
597
$endgroup$
add a comment |
$begingroup$
I got to thinking about sequences of Cauchy sequences. Here is a simple example. Let us define $b_n = (n,1,1,1,...)$ for $ninmathbb N$. So we have
begin{align}
b_1&=(1,1,1,1,...)\
b_2&=(2,1,1,1,...)\
b_3&=(3,1,1,1,...)\
dots
end{align}
Question: what is $lim_{ntoinfty} b_n$? It looks like that is not a valid sequence. Does this question even make sense? (I am not sure how such a limit would even be defined.)
functional-analysis cauchy-sequences
asked Jan 7 at 8:37
StefanieStefanie
597
$endgroup$
I got to thinking about sequences of Cauchy sequences. Here is a simple example. Let us define $b_n = (n,1,1,1,...)$ for $ninmathbb N$. So we have
begin{align}
b_1&=(1,1,1,1,...)\
b_2&=(2,1,1,1,...)\
b_3&=(3,1,1,1,...)\
dots
end{align}
Question: what is $lim_{ntoinfty} b_n$? It looks like that is not a valid sequence. Does this question even make sense? (I am not sure how such a limit would even be defined.)
functional-analysis cauchy-sequences
functional-analysis cauchy-sequences
asked Jan 7 at 8:37
StefanieStefanie
597
asked Jan 7 at 8:37
StefanieStefanie
597
asked Jan 7 at 8:37
StefanieStefanie
597
asked Jan 7 at 8:37
StefanieStefanie
597
asked Jan 7 at 8:37
StefanieStefanie
597
597
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A sequence in $X$ is actually a function $mathbb Nto X$ or equivalently an element of the set $X^mathbb N$.
If $X$ is a topological space then $X^{mathbb N}$ can be equipped with a topology that corresponds in some way with the original topology on $X$, which gives birth to the possibility of convergence of sequences.
There are several candidates for the topology on $X^{mathbb N}$.
In each of them a sequence of sequences might have a limit.
The question whether $lim_{ntoinfty}b_n$ exists (so that the expression makes sense) depends heavily on this context.
answered Jan 7 at 9:18
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you! So what about $X=mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$?
$endgroup$
– Stefanie
Jan 7 at 9:41
1
$begingroup$
$mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $rho((a_n)_n,(b_n)_n)=inf{overline d(a_n,b_n)mid ninmathbb N}$ where $overline d(a_n,b_n$ stands for $min{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge.
$endgroup$
– drhab
Jan 7 at 9:53
$begingroup$
In your comment to your A, I think you meant $sup$, not $inf,$ in the def'n of $rho$.
$endgroup$
– DanielWainfleet
Jan 7 at 19:30
$begingroup$
@DanielWainfleet Yes, you are right. Thank you for attending me.
$endgroup$
– drhab
Jan 7 at 20:32
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
A sequence in $X$ is actually a function $mathbb Nto X$ or equivalently an element of the set $X^mathbb N$.
If $X$ is a topological space then $X^{mathbb N}$ can be equipped with a topology that corresponds in some way with the original topology on $X$, which gives birth to the possibility of convergence of sequences.
There are several candidates for the topology on $X^{mathbb N}$.
In each of them a sequence of sequences might have a limit.
The question whether $lim_{ntoinfty}b_n$ exists (so that the expression makes sense) depends heavily on this context.
answered Jan 7 at 9:18
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you! So what about $X=mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$?
$endgroup$
– Stefanie
Jan 7 at 9:41
1
$begingroup$
$mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $rho((a_n)_n,(b_n)_n)=inf{overline d(a_n,b_n)mid ninmathbb N}$ where $overline d(a_n,b_n$ stands for $min{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge.
$endgroup$
– drhab
Jan 7 at 9:53
$begingroup$
In your comment to your A, I think you meant $sup$, not $inf,$ in the def'n of $rho$.
$endgroup$
– DanielWainfleet
Jan 7 at 19:30
$begingroup$
@DanielWainfleet Yes, you are right. Thank you for attending me.
$endgroup$
– drhab
Jan 7 at 20:32
add a comment |
$begingroup$
A sequence in $X$ is actually a function $mathbb Nto X$ or equivalently an element of the set $X^mathbb N$.
If $X$ is a topological space then $X^{mathbb N}$ can be equipped with a topology that corresponds in some way with the original topology on $X$, which gives birth to the possibility of convergence of sequences.
There are several candidates for the topology on $X^{mathbb N}$.
In each of them a sequence of sequences might have a limit.
The question whether $lim_{ntoinfty}b_n$ exists (so that the expression makes sense) depends heavily on this context.
answered Jan 7 at 9:18
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you! So what about $X=mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$?
$endgroup$
– Stefanie
Jan 7 at 9:41
1
$begingroup$
$mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $rho((a_n)_n,(b_n)_n)=inf{overline d(a_n,b_n)mid ninmathbb N}$ where $overline d(a_n,b_n$ stands for $min{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge.
$endgroup$
– drhab
Jan 7 at 9:53
$begingroup$
In your comment to your A, I think you meant $sup$, not $inf,$ in the def'n of $rho$.
$endgroup$
– DanielWainfleet
Jan 7 at 19:30
$begingroup$
@DanielWainfleet Yes, you are right. Thank you for attending me.
$endgroup$
– drhab
Jan 7 at 20:32
add a comment |
$begingroup$
A sequence in $X$ is actually a function $mathbb Nto X$ or equivalently an element of the set $X^mathbb N$.
If $X$ is a topological space then $X^{mathbb N}$ can be equipped with a topology that corresponds in some way with the original topology on $X$, which gives birth to the possibility of convergence of sequences.
There are several candidates for the topology on $X^{mathbb N}$.
In each of them a sequence of sequences might have a limit.
The question whether $lim_{ntoinfty}b_n$ exists (so that the expression makes sense) depends heavily on this context.
answered Jan 7 at 9:18
drhabdrhab
103k545136
$endgroup$
A sequence in $X$ is actually a function $mathbb Nto X$ or equivalently an element of the set $X^mathbb N$.
If $X$ is a topological space then $X^{mathbb N}$ can be equipped with a topology that corresponds in some way with the original topology on $X$, which gives birth to the possibility of convergence of sequences.
There are several candidates for the topology on $X^{mathbb N}$.
In each of them a sequence of sequences might have a limit.
The question whether $lim_{ntoinfty}b_n$ exists (so that the expression makes sense) depends heavily on this context.
answered Jan 7 at 9:18
drhabdrhab
103k545136
answered Jan 7 at 9:18
drhabdrhab
103k545136
answered Jan 7 at 9:18
drhabdrhab
103k545136
answered Jan 7 at 9:18
drhabdrhab
103k545136
103k545136
$begingroup$
Thank you! So what about $X=mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$?
$endgroup$
– Stefanie
Jan 7 at 9:41
1
$begingroup$
$mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $rho((a_n)_n,(b_n)_n)=inf{overline d(a_n,b_n)mid ninmathbb N}$ where $overline d(a_n,b_n$ stands for $min{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge.
$endgroup$
– drhab
Jan 7 at 9:53
$begingroup$
In your comment to your A, I think you meant $sup$, not $inf,$ in the def'n of $rho$.
$endgroup$
– DanielWainfleet
Jan 7 at 19:30
$begingroup$
@DanielWainfleet Yes, you are right. Thank you for attending me.
$endgroup$
– drhab
Jan 7 at 20:32
add a comment |
$begingroup$
Thank you! So what about $X=mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$?
$endgroup$
– Stefanie
Jan 7 at 9:41
1
$begingroup$
$mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $rho((a_n)_n,(b_n)_n)=inf{overline d(a_n,b_n)mid ninmathbb N}$ where $overline d(a_n,b_n$ stands for $min{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge.
$endgroup$
– drhab
Jan 7 at 9:53
$begingroup$
In your comment to your A, I think you meant $sup$, not $inf,$ in the def'n of $rho$.
$endgroup$
– DanielWainfleet
Jan 7 at 19:30
$begingroup$
@DanielWainfleet Yes, you are right. Thank you for attending me.
$endgroup$
– drhab
Jan 7 at 20:32
$begingroup$
Thank you! So what about $X=mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$?
$endgroup$
– Stefanie
Jan 7 at 9:41
$begingroup$
Thank you! So what about $X=mathbb Q$, which I suppose is the simplest case? (I am thinking about the context of a construction of the real numbers.) Is there a natural topology and if so what can we say about the limit of $b_n$?
$endgroup$
– Stefanie
Jan 7 at 9:41
1
1
$begingroup$
$mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $rho((a_n)_n,(b_n)_n)=inf{overline d(a_n,b_n)mid ninmathbb N}$ where $overline d(a_n,b_n$ stands for $min{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge.
$endgroup$
– drhab
Jan 7 at 9:53
$begingroup$
$mathbb Q$ with its usual topology is a metric space. Then a candidate is uniform topology based on metric $rho((a_n)_n,(b_n)_n)=inf{overline d(a_n,b_n)mid ninmathbb N}$ where $overline d(a_n,b_n$ stands for $min{|a_n-b_n|,1)$. Always there is the topology of pointwise convergence where sequence $(a_n)_n$ convergences iff $a_n$ converges for every $n$. And there are more (topology of compact convergence, compact-open topology, test-open topology, product topology). I cannot find one that makes your sequence $(b_n)_n$ converge.
$endgroup$
– drhab
Jan 7 at 9:53
$begingroup$
In your comment to your A, I think you meant $sup$, not $inf,$ in the def'n of $rho$.
$endgroup$
– DanielWainfleet
Jan 7 at 19:30
$begingroup$
In your comment to your A, I think you meant $sup$, not $inf,$ in the def'n of $rho$.
$endgroup$
– DanielWainfleet
Jan 7 at 19:30
$begingroup$
@DanielWainfleet Yes, you are right. Thank you for attending me.
$endgroup$
– drhab
Jan 7 at 20:32
$begingroup$
@DanielWainfleet Yes, you are right. Thank you for attending me.
$endgroup$
– drhab
Jan 7 at 20:32
add a comment |
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