Mixing
Is $hat{G}$ is complete with respect to the induced topology of $G$?
For a topological group $G$ and a given fundamental system of neighbourhoods of $G$ we can define the completion of G and we call it $hat{G}$. The induced fundamental system of neighbourhoods of $hat{G}$ is given by this Topology induced by the completion of a topological group.
Then Can we say that $hat{G}$ is complete ? i.e., every Cauchy sequence in $hat{G}$ is convergent ?
For this we assume that ${z_n}$ be any Cauchy sequence in $hat{G},$ then given any open neighbourhood $tilde{N}$ of $hat{G}$ there exists an integer $k$ such that whenever $m,n geq k,$ $z_m-z_n in tilde{N}.$ Then how can I show that ${z_n}$ is convergent ? That is we are looking for an element $s in hat{G}$ such that for any neighbourhood $hat{P}$ of $hat{G},$ $z_n in s+hat{P}$ for large $n$. In general will it hold ? I need some help. Thanks.
abstract-algebra general-topology commutative-algebra topological-groups formal-completions
asked Nov 20 '18 at 17:03
user371231
665511
add a comment |
For a topological group $G$ and a given fundamental system of neighbourhoods of $G$ we can define the completion of G and we call it $hat{G}$. The induced fundamental system of neighbourhoods of $hat{G}$ is given by this Topology induced by the completion of a topological group.
Then Can we say that $hat{G}$ is complete ? i.e., every Cauchy sequence in $hat{G}$ is convergent ?
For this we assume that ${z_n}$ be any Cauchy sequence in $hat{G},$ then given any open neighbourhood $tilde{N}$ of $hat{G}$ there exists an integer $k$ such that whenever $m,n geq k,$ $z_m-z_n in tilde{N}.$ Then how can I show that ${z_n}$ is convergent ? That is we are looking for an element $s in hat{G}$ such that for any neighbourhood $hat{P}$ of $hat{G},$ $z_n in s+hat{P}$ for large $n$. In general will it hold ? I need some help. Thanks.
abstract-algebra general-topology commutative-algebra topological-groups formal-completions
asked Nov 20 '18 at 17:03
user371231
665511
Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness?
– Henno Brandsma
Nov 20 '18 at 17:34
Is the completion not the completion of the uniform structure on $G$?
– Robert Thingum
Nov 20 '18 at 19:39
Yes, but if $G$ is not first-countable then the completion of $G$, as a uniform space, in general is not determined by Cauchy sequences alone. Sequential completeness only suffices if $G$ is first-countable.
– Monstrous Moonshiner
Nov 20 '18 at 20:52
Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was.
– Robert Thingum
Nov 20 '18 at 20:57
add a comment |
For a topological group $G$ and a given fundamental system of neighbourhoods of $G$ we can define the completion of G and we call it $hat{G}$. The induced fundamental system of neighbourhoods of $hat{G}$ is given by this Topology induced by the completion of a topological group.
Then Can we say that $hat{G}$ is complete ? i.e., every Cauchy sequence in $hat{G}$ is convergent ?
For this we assume that ${z_n}$ be any Cauchy sequence in $hat{G},$ then given any open neighbourhood $tilde{N}$ of $hat{G}$ there exists an integer $k$ such that whenever $m,n geq k,$ $z_m-z_n in tilde{N}.$ Then how can I show that ${z_n}$ is convergent ? That is we are looking for an element $s in hat{G}$ such that for any neighbourhood $hat{P}$ of $hat{G},$ $z_n in s+hat{P}$ for large $n$. In general will it hold ? I need some help. Thanks.
abstract-algebra general-topology commutative-algebra topological-groups formal-completions
asked Nov 20 '18 at 17:03
user371231
665511
For a topological group $G$ and a given fundamental system of neighbourhoods of $G$ we can define the completion of G and we call it $hat{G}$. The induced fundamental system of neighbourhoods of $hat{G}$ is given by this Topology induced by the completion of a topological group.
Then Can we say that $hat{G}$ is complete ? i.e., every Cauchy sequence in $hat{G}$ is convergent ?
For this we assume that ${z_n}$ be any Cauchy sequence in $hat{G},$ then given any open neighbourhood $tilde{N}$ of $hat{G}$ there exists an integer $k$ such that whenever $m,n geq k,$ $z_m-z_n in tilde{N}.$ Then how can I show that ${z_n}$ is convergent ? That is we are looking for an element $s in hat{G}$ such that for any neighbourhood $hat{P}$ of $hat{G},$ $z_n in s+hat{P}$ for large $n$. In general will it hold ? I need some help. Thanks.
abstract-algebra general-topology commutative-algebra topological-groups formal-completions
abstract-algebra general-topology commutative-algebra topological-groups formal-completions
asked Nov 20 '18 at 17:03
user371231
665511
asked Nov 20 '18 at 17:03
user371231
665511
edited Nov 20 '18 at 19:53
asked Nov 20 '18 at 17:03
user371231
665511
asked Nov 20 '18 at 17:03
user371231
665511
asked Nov 20 '18 at 17:03
user371231
665511
665511
Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness?
– Henno Brandsma
Nov 20 '18 at 17:34
Is the completion not the completion of the uniform structure on $G$?
– Robert Thingum
Nov 20 '18 at 19:39
Yes, but if $G$ is not first-countable then the completion of $G$, as a uniform space, in general is not determined by Cauchy sequences alone. Sequential completeness only suffices if $G$ is first-countable.
– Monstrous Moonshiner
Nov 20 '18 at 20:52
Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was.
– Robert Thingum
Nov 20 '18 at 20:57
add a comment |
Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness?
– Henno Brandsma
Nov 20 '18 at 17:34
Is the completion not the completion of the uniform structure on $G$?
– Robert Thingum
Nov 20 '18 at 19:39
Yes, but if $G$ is not first-countable then the completion of $G$, as a uniform space, in general is not determined by Cauchy sequences alone. Sequential completeness only suffices if $G$ is first-countable.
– Monstrous Moonshiner
Nov 20 '18 at 20:52
Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was.
– Robert Thingum
Nov 20 '18 at 20:57
Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness?
– Henno Brandsma
Nov 20 '18 at 17:34
Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness?
– Henno Brandsma
Nov 20 '18 at 17:34
Is the completion not the completion of the uniform structure on $G$?
– Robert Thingum
Nov 20 '18 at 19:39
Is the completion not the completion of the uniform structure on $G$?
– Robert Thingum
Nov 20 '18 at 19:39
Yes, but if $G$ is not first-countable then the completion of $G$, as a uniform space, in general is not determined by Cauchy sequences alone. Sequential completeness only suffices if $G$ is first-countable.
– Monstrous Moonshiner
Nov 20 '18 at 20:52
Yes, but if $G$ is not first-countable then the completion of $G$, as a uniform space, in general is not determined by Cauchy sequences alone. Sequential completeness only suffices if $G$ is first-countable.
– Monstrous Moonshiner
Nov 20 '18 at 20:52
Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was.
– Robert Thingum
Nov 20 '18 at 20:57
Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was.
– Robert Thingum
Nov 20 '18 at 20:57
add a comment |
1 Answer
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For simplicity, I'll ignore the very accurate point made in the comments that defining completeness in terms of sequences isn't sufficient in general, and assume that the topology on $G$ is first-countable, and therefore $hat G$ is first-countable as well.
First, we need to correct your definition of Cauchy sequences and of convergence. For example, in $Bbb R, left{frac 1nright}$ is a Cauchy sequence (as is any convergent sequence), and $(1,2)$ is a neighborhood in $Bbb R$, but there is not a $k in Bbb N$ such that for $n, m > k, left(frac 1m - frac 1nright) in (1,2)$.
$tilde N$ needs to be a neighborhood of $0$, not just any neighborhood of $hat G$. Similarly, $hat P$ is also a neighborhood of $0$ (and so $s + hat P$ is a neighborhood of $s$).
Since ${z_n}_{ninBbb N} subset hat G$, for each $n, z_n$ is some Cauchy sequence ${z_{nm}}_{minBbb N}$ in $G$. You can use the fact that ${z_n}_{ninBbb N}$ is Cauchy in $hat G$ to show that the diagonal sequence ${z_{nn}}_{nin Bbb N}$ is Cauchy in $G$.
Then $s = {z_{nn}}_{nin Bbb N}$.
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
I need to check if it works, though looks like it will work.
– user371231
Nov 21 '18 at 7:02
add a comment |
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1 Answer
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For simplicity, I'll ignore the very accurate point made in the comments that defining completeness in terms of sequences isn't sufficient in general, and assume that the topology on $G$ is first-countable, and therefore $hat G$ is first-countable as well.
First, we need to correct your definition of Cauchy sequences and of convergence. For example, in $Bbb R, left{frac 1nright}$ is a Cauchy sequence (as is any convergent sequence), and $(1,2)$ is a neighborhood in $Bbb R$, but there is not a $k in Bbb N$ such that for $n, m > k, left(frac 1m - frac 1nright) in (1,2)$.
$tilde N$ needs to be a neighborhood of $0$, not just any neighborhood of $hat G$. Similarly, $hat P$ is also a neighborhood of $0$ (and so $s + hat P$ is a neighborhood of $s$).
Since ${z_n}_{ninBbb N} subset hat G$, for each $n, z_n$ is some Cauchy sequence ${z_{nm}}_{minBbb N}$ in $G$. You can use the fact that ${z_n}_{ninBbb N}$ is Cauchy in $hat G$ to show that the diagonal sequence ${z_{nn}}_{nin Bbb N}$ is Cauchy in $G$.
Then $s = {z_{nn}}_{nin Bbb N}$.
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
I need to check if it works, though looks like it will work.
– user371231
Nov 21 '18 at 7:02
add a comment |
For simplicity, I'll ignore the very accurate point made in the comments that defining completeness in terms of sequences isn't sufficient in general, and assume that the topology on $G$ is first-countable, and therefore $hat G$ is first-countable as well.
First, we need to correct your definition of Cauchy sequences and of convergence. For example, in $Bbb R, left{frac 1nright}$ is a Cauchy sequence (as is any convergent sequence), and $(1,2)$ is a neighborhood in $Bbb R$, but there is not a $k in Bbb N$ such that for $n, m > k, left(frac 1m - frac 1nright) in (1,2)$.
$tilde N$ needs to be a neighborhood of $0$, not just any neighborhood of $hat G$. Similarly, $hat P$ is also a neighborhood of $0$ (and so $s + hat P$ is a neighborhood of $s$).
Since ${z_n}_{ninBbb N} subset hat G$, for each $n, z_n$ is some Cauchy sequence ${z_{nm}}_{minBbb N}$ in $G$. You can use the fact that ${z_n}_{ninBbb N}$ is Cauchy in $hat G$ to show that the diagonal sequence ${z_{nn}}_{nin Bbb N}$ is Cauchy in $G$.
Then $s = {z_{nn}}_{nin Bbb N}$.
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
I need to check if it works, though looks like it will work.
– user371231
Nov 21 '18 at 7:02
add a comment |
For simplicity, I'll ignore the very accurate point made in the comments that defining completeness in terms of sequences isn't sufficient in general, and assume that the topology on $G$ is first-countable, and therefore $hat G$ is first-countable as well.
First, we need to correct your definition of Cauchy sequences and of convergence. For example, in $Bbb R, left{frac 1nright}$ is a Cauchy sequence (as is any convergent sequence), and $(1,2)$ is a neighborhood in $Bbb R$, but there is not a $k in Bbb N$ such that for $n, m > k, left(frac 1m - frac 1nright) in (1,2)$.
$tilde N$ needs to be a neighborhood of $0$, not just any neighborhood of $hat G$. Similarly, $hat P$ is also a neighborhood of $0$ (and so $s + hat P$ is a neighborhood of $s$).
Since ${z_n}_{ninBbb N} subset hat G$, for each $n, z_n$ is some Cauchy sequence ${z_{nm}}_{minBbb N}$ in $G$. You can use the fact that ${z_n}_{ninBbb N}$ is Cauchy in $hat G$ to show that the diagonal sequence ${z_{nn}}_{nin Bbb N}$ is Cauchy in $G$.
Then $s = {z_{nn}}_{nin Bbb N}$.
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
For simplicity, I'll ignore the very accurate point made in the comments that defining completeness in terms of sequences isn't sufficient in general, and assume that the topology on $G$ is first-countable, and therefore $hat G$ is first-countable as well.
First, we need to correct your definition of Cauchy sequences and of convergence. For example, in $Bbb R, left{frac 1nright}$ is a Cauchy sequence (as is any convergent sequence), and $(1,2)$ is a neighborhood in $Bbb R$, but there is not a $k in Bbb N$ such that for $n, m > k, left(frac 1m - frac 1nright) in (1,2)$.
$tilde N$ needs to be a neighborhood of $0$, not just any neighborhood of $hat G$. Similarly, $hat P$ is also a neighborhood of $0$ (and so $s + hat P$ is a neighborhood of $s$).
Since ${z_n}_{ninBbb N} subset hat G$, for each $n, z_n$ is some Cauchy sequence ${z_{nm}}_{minBbb N}$ in $G$. You can use the fact that ${z_n}_{ninBbb N}$ is Cauchy in $hat G$ to show that the diagonal sequence ${z_{nn}}_{nin Bbb N}$ is Cauchy in $G$.
Then $s = {z_{nn}}_{nin Bbb N}$.
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
answered Nov 21 '18 at 1:17
Paul Sinclair
19.3k21441
19.3k21441
I need to check if it works, though looks like it will work.
– user371231
Nov 21 '18 at 7:02
add a comment |
I need to check if it works, though looks like it will work.
– user371231
Nov 21 '18 at 7:02
I need to check if it works, though looks like it will work.
– user371231
Nov 21 '18 at 7:02
I need to check if it works, though looks like it will work.
– user371231
Nov 21 '18 at 7:02
add a comment |
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Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness?
– Henno Brandsma
Nov 20 '18 at 17:34
Is the completion not the completion of the uniform structure on $G$?
– Robert Thingum
Nov 20 '18 at 19:39
Yes, but if $G$ is not first-countable then the completion of $G$, as a uniform space, in general is not determined by Cauchy sequences alone. Sequential completeness only suffices if $G$ is first-countable.
– Monstrous Moonshiner
Nov 20 '18 at 20:52
Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was.
– Robert Thingum
Nov 20 '18 at 20:57