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?










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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?










    share|cite|improve this question
























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      up vote
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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?










      share|cite|improve this question













      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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      asked Nov 14 at 18:49









      user573597

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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 $phi(x_r) = lambda_n$






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

              oldest

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              up vote
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              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.






              share|cite|improve this answer

























                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.






                share|cite|improve this answer























                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  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 $phi(x_r) = lambda_n$






                      share|cite|improve this answer

























                        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 $phi(x_r) = lambda_n$






                        share|cite|improve this answer























                          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 $phi(x_r) = lambda_n$






                          share|cite|improve this answer












                          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 $phi(x_r) = lambda_n$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 15 at 14:06









                          user573597

                          103




                          103






























                               

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