How to show that the image of a complete metric space under an isometry is closed?
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
asked Dec 9 '18 at 6:46
Sepide
2918
add a comment |
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
asked Dec 9 '18 at 6:46
Sepide
2918
add a comment |
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
asked Dec 9 '18 at 6:46
Sepide
2918
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
real-analysis
asked Dec 9 '18 at 6:46
Sepide
2918
asked Dec 9 '18 at 6:46
Sepide
2918
asked Dec 9 '18 at 6:46
Sepide
2918
asked Dec 9 '18 at 6:46
Sepide
2918
asked Dec 9 '18 at 6:46
Sepide
2918
2918
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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 $f(x)=lim_{ntoinfty}f(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
add a comment |
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 $y=f(x)$.
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
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 $f(x)=lim_{ntoinfty}f(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
add a comment |
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 $f(x)=lim_{ntoinfty}f(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
add a comment |
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 $f(x)=lim_{ntoinfty}f(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
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 $f(x)=lim_{ntoinfty}f(x_n)=y$. Thus $yin f(M)$. So $f(M)$ is closed.
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
answered Dec 9 '18 at 7:16
Chris Custer
10.8k3724
10.8k3724
add a comment |
add a comment |
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 $y=f(x)$.
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
add a comment |
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 $y=f(x)$.
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
add a comment |
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 $y=f(x)$.
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
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 $y=f(x)$.
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
answered Dec 9 '18 at 6:56
Hagen von Eitzen
276k21269496
276k21269496
add a comment |
add a comment |
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