How to show that the image of a complete metric space under an isometry is closed?

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Let $f,,:,, (M,d) rightarrow (N,sigma)$ be an isometry, that is
$$
sigma(f(x),f(y))=d(x,y)
$$

for all $x,y in M$.



If $(M,d)$ is complete, show that $f(M)$ is closed in $(N,sigma)$.










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    Let $f,,:,, (M,d) rightarrow (N,sigma)$ be an isometry, that is
    $$
    sigma(f(x),f(y))=d(x,y)
    $$

    for all $x,y in M$.



    If $(M,d)$ is complete, show that $f(M)$ is closed in $(N,sigma)$.










    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $f,,:,, (M,d) rightarrow (N,sigma)$ be an isometry, that is
      $$
      sigma(f(x),f(y))=d(x,y)
      $$

      for all $x,y in M$.



      If $(M,d)$ is complete, show that $f(M)$ is closed in $(N,sigma)$.










      share|cite|improve this question













      Let $f,,:,, (M,d) rightarrow (N,sigma)$ be an isometry, that is
      $$
      sigma(f(x),f(y))=d(x,y)
      $$

      for all $x,y in M$.



      If $(M,d)$ is complete, show that $f(M)$ is closed in $(N,sigma)$.







      real-analysis






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      asked Dec 9 at 6:46









      Sepide

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          Let $yin N$ be a limit point of $f(M)$. Then there's a sequence $(f(x_n))in f(M)$ converging to $y$. Then $(f(x_n))$ is Cauchy in $(N,sigma)$. Since $f$ is an isometry, $(x_n)$ is Cauchy in $(M,d)$. Since $(M,d)$ is complete, $(x_n)$ converges to $xin M$. Then $f(x)=lim_ntoinftyf(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.






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            Let $yin N$. If $x_nin M$ such that $f(x_n)to y$, then $f(x_n)_n$ is convergent, hence is Cauchy, hence $x_n_n$ is Cauchy, hence is convergent to some $x$. Then $f(x_n)to f(x)$ and so $y=f(x)$.






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              2 Answers
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              active

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              up vote
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              accepted










              Let $yin N$ be a limit point of $f(M)$. Then there's a sequence $(f(x_n))in f(M)$ converging to $y$. Then $(f(x_n))$ is Cauchy in $(N,sigma)$. Since $f$ is an isometry, $(x_n)$ is Cauchy in $(M,d)$. Since $(M,d)$ is complete, $(x_n)$ converges to $xin M$. Then $f(x)=lim_ntoinftyf(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.






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                up vote
                4
                down vote



                accepted










                Let $yin N$ be a limit point of $f(M)$. Then there's a sequence $(f(x_n))in f(M)$ converging to $y$. Then $(f(x_n))$ is Cauchy in $(N,sigma)$. Since $f$ is an isometry, $(x_n)$ is Cauchy in $(M,d)$. Since $(M,d)$ is complete, $(x_n)$ converges to $xin M$. Then $f(x)=lim_ntoinftyf(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.






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                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  Let $yin N$ be a limit point of $f(M)$. Then there's a sequence $(f(x_n))in f(M)$ converging to $y$. Then $(f(x_n))$ is Cauchy in $(N,sigma)$. Since $f$ is an isometry, $(x_n)$ is Cauchy in $(M,d)$. Since $(M,d)$ is complete, $(x_n)$ converges to $xin M$. Then $f(x)=lim_ntoinftyf(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.






                  share|cite|improve this answer












                  Let $yin N$ be a limit point of $f(M)$. Then there's a sequence $(f(x_n))in f(M)$ converging to $y$. Then $(f(x_n))$ is Cauchy in $(N,sigma)$. Since $f$ is an isometry, $(x_n)$ is Cauchy in $(M,d)$. Since $(M,d)$ is complete, $(x_n)$ converges to $xin M$. Then $f(x)=lim_ntoinftyf(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.







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                  answered Dec 9 at 7:16









                  Chris Custer

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                      Let $yin N$. If $x_nin M$ such that $f(x_n)to y$, then $f(x_n)_n$ is convergent, hence is Cauchy, hence $x_n_n$ is Cauchy, hence is convergent to some $x$. Then $f(x_n)to f(x)$ and so $y=f(x)$.






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                        Let $yin N$. If $x_nin M$ such that $f(x_n)to y$, then $f(x_n)_n$ is convergent, hence is Cauchy, hence $x_n_n$ is Cauchy, hence is convergent to some $x$. Then $f(x_n)to f(x)$ and so $y=f(x)$.






                        share|cite|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          Let $yin N$. If $x_nin M$ such that $f(x_n)to y$, then $f(x_n)_n$ is convergent, hence is Cauchy, hence $x_n_n$ is Cauchy, hence is convergent to some $x$. Then $f(x_n)to f(x)$ and so $y=f(x)$.






                          share|cite|improve this answer












                          Let $yin N$. If $x_nin M$ such that $f(x_n)to y$, then $f(x_n)_n$ is convergent, hence is Cauchy, hence $x_n_n$ is Cauchy, hence is convergent to some $x$. Then $f(x_n)to f(x)$ and so $y=f(x)$.







                          share|cite|improve this answer












                          share|cite|improve this answer



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                          answered Dec 9 at 6:56









                          Hagen von Eitzen

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