Completion (construction Atiyah MacDonald chapter 10)
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Following Atiyah MacDonald (Chapter 10: Completions), let $G$ be a topological abelian group. We assume that $0 in G$ has a fundamental system of neighborhoods consisting of subgroups $G = G_0 supseteq G_1 supseteq … supseteq G_n supseteq … $
Suppose $(x_k)$ is a Cauchy sequence in $G$.
Then it is claimed, that the image of $x_k$ in $G/G_n$ is ultimately constant. Why is this the case?
commutative-algebra
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Following Atiyah MacDonald (Chapter 10: Completions), let $G$ be a topological abelian group. We assume that $0 in G$ has a fundamental system of neighborhoods consisting of subgroups $G = G_0 supseteq G_1 supseteq … supseteq G_n supseteq … $
Suppose $(x_k)$ is a Cauchy sequence in $G$.
Then it is claimed, that the image of $x_k$ in $G/G_n$ is ultimately constant. Why is this the case?
commutative-algebra
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Following Atiyah MacDonald (Chapter 10: Completions), let $G$ be a topological abelian group. We assume that $0 in G$ has a fundamental system of neighborhoods consisting of subgroups $G = G_0 supseteq G_1 supseteq … supseteq G_n supseteq … $
Suppose $(x_k)$ is a Cauchy sequence in $G$.
Then it is claimed, that the image of $x_k$ in $G/G_n$ is ultimately constant. Why is this the case?
commutative-algebra
Following Atiyah MacDonald (Chapter 10: Completions), let $G$ be a topological abelian group. We assume that $0 in G$ has a fundamental system of neighborhoods consisting of subgroups $G = G_0 supseteq G_1 supseteq … supseteq G_n supseteq … $
Suppose $(x_k)$ is a Cauchy sequence in $G$.
Then it is claimed, that the image of $x_k$ in $G/G_n$ is ultimately constant. Why is this the case?
commutative-algebra
commutative-algebra
asked Nov 14 at 18:49
user573597
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2 Answers
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Saying that $(x_n)$ is Cauchy means that, for every $k$, there exists $N$ such that, for $m,n>N$, $x_m-x_nin G_k$.
Thus, for $m,n>N$, $x_m+G_k=x_n+G_k$. This is the same as saying that the image of $(x_n)$ in $G/G_k$ is eventually constant.
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Definition: A Cauchy sequence in $G$ is a sequence $(x_k)$ of elements of $G$ such that, for any neighborhood $U$ of $0$, there exists an integer $s(U)$ with the property that $x_m-x_l in U$ for all $m,l geq s(U)$.
My idea:
Let $phi: G rightarrow G/G_n$ denote the projection sending $x_k$ to $x_k +G_n$.
Choose $U = G_n$ $Rightarrow x_m-x_l in G_n$ for all $m,l geq s(G_n)$
$Rightarrow x_m+G_n = x_l +G_n$ for all $m,l geq s(G_n)$
$Rightarrow phi(x_m) = phi(x_l)$ for all $m,l geq s(G_n)$
$Rightarrow$ For all $r geq s(G_n)$ we have $phi(x_r) = lambda_n$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Saying that $(x_n)$ is Cauchy means that, for every $k$, there exists $N$ such that, for $m,n>N$, $x_m-x_nin G_k$.
Thus, for $m,n>N$, $x_m+G_k=x_n+G_k$. This is the same as saying that the image of $(x_n)$ in $G/G_k$ is eventually constant.
add a comment |
up vote
0
down vote
accepted
Saying that $(x_n)$ is Cauchy means that, for every $k$, there exists $N$ such that, for $m,n>N$, $x_m-x_nin G_k$.
Thus, for $m,n>N$, $x_m+G_k=x_n+G_k$. This is the same as saying that the image of $(x_n)$ in $G/G_k$ is eventually constant.
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Saying that $(x_n)$ is Cauchy means that, for every $k$, there exists $N$ such that, for $m,n>N$, $x_m-x_nin G_k$.
Thus, for $m,n>N$, $x_m+G_k=x_n+G_k$. This is the same as saying that the image of $(x_n)$ in $G/G_k$ is eventually constant.
Saying that $(x_n)$ is Cauchy means that, for every $k$, there exists $N$ such that, for $m,n>N$, $x_m-x_nin G_k$.
Thus, for $m,n>N$, $x_m+G_k=x_n+G_k$. This is the same as saying that the image of $(x_n)$ in $G/G_k$ is eventually constant.
answered Nov 15 at 14:26
egreg
174k1383198
174k1383198
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Definition: A Cauchy sequence in $G$ is a sequence $(x_k)$ of elements of $G$ such that, for any neighborhood $U$ of $0$, there exists an integer $s(U)$ with the property that $x_m-x_l in U$ for all $m,l geq s(U)$.
My idea:
Let $phi: G rightarrow G/G_n$ denote the projection sending $x_k$ to $x_k +G_n$.
Choose $U = G_n$ $Rightarrow x_m-x_l in G_n$ for all $m,l geq s(G_n)$
$Rightarrow x_m+G_n = x_l +G_n$ for all $m,l geq s(G_n)$
$Rightarrow phi(x_m) = phi(x_l)$ for all $m,l geq s(G_n)$
$Rightarrow$ For all $r geq s(G_n)$ we have $phi(x_r) = lambda_n$
add a comment |
up vote
0
down vote
Definition: A Cauchy sequence in $G$ is a sequence $(x_k)$ of elements of $G$ such that, for any neighborhood $U$ of $0$, there exists an integer $s(U)$ with the property that $x_m-x_l in U$ for all $m,l geq s(U)$.
My idea:
Let $phi: G rightarrow G/G_n$ denote the projection sending $x_k$ to $x_k +G_n$.
Choose $U = G_n$ $Rightarrow x_m-x_l in G_n$ for all $m,l geq s(G_n)$
$Rightarrow x_m+G_n = x_l +G_n$ for all $m,l geq s(G_n)$
$Rightarrow phi(x_m) = phi(x_l)$ for all $m,l geq s(G_n)$
$Rightarrow$ For all $r geq s(G_n)$ we have $phi(x_r) = lambda_n$
add a comment |
up vote
0
down vote
up vote
0
down vote
Definition: A Cauchy sequence in $G$ is a sequence $(x_k)$ of elements of $G$ such that, for any neighborhood $U$ of $0$, there exists an integer $s(U)$ with the property that $x_m-x_l in U$ for all $m,l geq s(U)$.
My idea:
Let $phi: G rightarrow G/G_n$ denote the projection sending $x_k$ to $x_k +G_n$.
Choose $U = G_n$ $Rightarrow x_m-x_l in G_n$ for all $m,l geq s(G_n)$
$Rightarrow x_m+G_n = x_l +G_n$ for all $m,l geq s(G_n)$
$Rightarrow phi(x_m) = phi(x_l)$ for all $m,l geq s(G_n)$
$Rightarrow$ For all $r geq s(G_n)$ we have $phi(x_r) = lambda_n$
Definition: A Cauchy sequence in $G$ is a sequence $(x_k)$ of elements of $G$ such that, for any neighborhood $U$ of $0$, there exists an integer $s(U)$ with the property that $x_m-x_l in U$ for all $m,l geq s(U)$.
My idea:
Let $phi: G rightarrow G/G_n$ denote the projection sending $x_k$ to $x_k +G_n$.
Choose $U = G_n$ $Rightarrow x_m-x_l in G_n$ for all $m,l geq s(G_n)$
$Rightarrow x_m+G_n = x_l +G_n$ for all $m,l geq s(G_n)$
$Rightarrow phi(x_m) = phi(x_l)$ for all $m,l geq s(G_n)$
$Rightarrow$ For all $r geq s(G_n)$ we have $phi(x_r) = lambda_n$
answered Nov 15 at 14:06
user573597
103
103
add a comment |
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