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 $(N,sigma)$.
real-analysis
asked Dec 9 at 6:46
Sepide
2687
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up vote
2
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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 $(N,sigma)$.
real-analysis
asked Dec 9 at 6:46
Sepide
2687
add a comment |
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 $(N,sigma)$.
real-analysis
asked Dec 9 at 6:46
Sepide
2687
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 at 6:46
Sepide
2687
asked Dec 9 at 6:46
Sepide
2687
asked Dec 9 at 6:46
Sepide
2687
asked Dec 9 at 6:46
Sepide
2687
asked Dec 9 at 6:46
Sepide
2687
2687
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Dec 9 at 7:16
Chris Custer
10.3k3724
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up vote
2
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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 $y=f(x)$.
answered Dec 9 at 6:56
Hagen von Eitzen
275k21268495
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Dec 9 at 7:16
Chris Custer
10.3k3724
add a comment |
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.
answered Dec 9 at 7:16
Chris Custer
10.3k3724
add a comment |
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.
answered Dec 9 at 7:16
Chris Custer
10.3k3724
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.
answered Dec 9 at 7:16
Chris Custer
10.3k3724
answered Dec 9 at 7:16
Chris Custer
10.3k3724
answered Dec 9 at 7:16
Chris Custer
10.3k3724
answered Dec 9 at 7:16
Chris Custer
10.3k3724
10.3k3724
add a comment |
add a comment |
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 $y=f(x)$.
answered Dec 9 at 6:56
Hagen von Eitzen
275k21268495
add a comment |
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 $y=f(x)$.
answered Dec 9 at 6:56
Hagen von Eitzen
275k21268495
add a comment |
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 $y=f(x)$.
answered Dec 9 at 6:56
Hagen von Eitzen
275k21268495
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 at 6:56
Hagen von Eitzen
275k21268495
answered Dec 9 at 6:56
Hagen von Eitzen
275k21268495
answered Dec 9 at 6:56
Hagen von Eitzen
275k21268495
answered Dec 9 at 6:56
Hagen von Eitzen
275k21268495
275k21268495
add a comment |
add a comment |
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