Mixing
Isometry $f:Mmapsto M$ is surjective if $M$ is compact: proof by composing $f$ with itself
$begingroup$
Let $(M,d)$ be a compact space and $f:Mmapsto M$ an isometry. To prove that $f$ is surjective we take an arbitrary $ain M$ and consider $f^n:Mmapsto M$ the function obtained by composing $f$ $n$ times with itself.
Since $M$ is compact $(f^n(a))_{ninBbb N}$ has a subsequence $(f^{n_k}(a))_{kinBbb N}$ that converges in $(M,d)$. Let's call its limit $a^*in M$
We get $d(a,f^{n_{k+1}-n_k}(a))=d(f^{n_k}(a),f^{n_{k+1}}(a))xrightarrow[krightarrowinfty]{}d(a^*,a^*)=0$
Why does that prove that $ain f(M)$? (and since $a$ is arbitrary $Msubset f(M)implies M=f(M))$
metric-spaces compactness
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
Let $(M,d)$ be a compact space and $f:Mmapsto M$ an isometry. To prove that $f$ is surjective we take an arbitrary $ain M$ and consider $f^n:Mmapsto M$ the function obtained by composing $f$ $n$ times with itself.
Since $M$ is compact $(f^n(a))_{ninBbb N}$ has a subsequence $(f^{n_k}(a))_{kinBbb N}$ that converges in $(M,d)$. Let's call its limit $a^*in M$
We get $d(a,f^{n_{k+1}-n_k}(a))=d(f^{n_k}(a),f^{n_{k+1}}(a))xrightarrow[krightarrowinfty]{}d(a^*,a^*)=0$
Why does that prove that $ain f(M)$? (and since $a$ is arbitrary $Msubset f(M)implies M=f(M))$
metric-spaces compactness
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
$endgroup$
1
$begingroup$
It shows $a=limlimits_{ktoinfty}f^{n_{k+1}-n_k}(a).$ Note that $f^{n_{k+1}-n_k}(a)in f(M)$ and $f(M)$ is closed.
$endgroup$
– Song
Jan 21 at 15:52
$begingroup$
@Song Why not an official answer?
$endgroup$
– Paul Frost
Jan 21 at 17:37
$begingroup$
@PaulFrost I saw someone had already answered it, but he/she deleted his/her answer now. Thank you for suggestion :)
$endgroup$
– Song
Jan 21 at 17:53
add a comment |
$begingroup$
Let $(M,d)$ be a compact space and $f:Mmapsto M$ an isometry. To prove that $f$ is surjective we take an arbitrary $ain M$ and consider $f^n:Mmapsto M$ the function obtained by composing $f$ $n$ times with itself.
Since $M$ is compact $(f^n(a))_{ninBbb N}$ has a subsequence $(f^{n_k}(a))_{kinBbb N}$ that converges in $(M,d)$. Let's call its limit $a^*in M$
We get $d(a,f^{n_{k+1}-n_k}(a))=d(f^{n_k}(a),f^{n_{k+1}}(a))xrightarrow[krightarrowinfty]{}d(a^*,a^*)=0$
Why does that prove that $ain f(M)$? (and since $a$ is arbitrary $Msubset f(M)implies M=f(M))$
metric-spaces compactness
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
$endgroup$
Let $(M,d)$ be a compact space and $f:Mmapsto M$ an isometry. To prove that $f$ is surjective we take an arbitrary $ain M$ and consider $f^n:Mmapsto M$ the function obtained by composing $f$ $n$ times with itself.
Since $M$ is compact $(f^n(a))_{ninBbb N}$ has a subsequence $(f^{n_k}(a))_{kinBbb N}$ that converges in $(M,d)$. Let's call its limit $a^*in M$
We get $d(a,f^{n_{k+1}-n_k}(a))=d(f^{n_k}(a),f^{n_{k+1}}(a))xrightarrow[krightarrowinfty]{}d(a^*,a^*)=0$
Why does that prove that $ain f(M)$? (and since $a$ is arbitrary $Msubset f(M)implies M=f(M))$
metric-spaces compactness
metric-spaces compactness
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
asked Jan 21 at 15:45
John CataldoJohn Cataldo
1,1881316
1,1881316
1
$begingroup$
It shows $a=limlimits_{ktoinfty}f^{n_{k+1}-n_k}(a).$ Note that $f^{n_{k+1}-n_k}(a)in f(M)$ and $f(M)$ is closed.
$endgroup$
– Song
Jan 21 at 15:52
$begingroup$
@Song Why not an official answer?
$endgroup$
– Paul Frost
Jan 21 at 17:37
$begingroup$
@PaulFrost I saw someone had already answered it, but he/she deleted his/her answer now. Thank you for suggestion :)
$endgroup$
– Song
Jan 21 at 17:53
add a comment |
1
$begingroup$
It shows $a=limlimits_{ktoinfty}f^{n_{k+1}-n_k}(a).$ Note that $f^{n_{k+1}-n_k}(a)in f(M)$ and $f(M)$ is closed.
$endgroup$
– Song
Jan 21 at 15:52
$begingroup$
@Song Why not an official answer?
$endgroup$
– Paul Frost
Jan 21 at 17:37
$begingroup$
@PaulFrost I saw someone had already answered it, but he/she deleted his/her answer now. Thank you for suggestion :)
$endgroup$
– Song
Jan 21 at 17:53
1
1
$begingroup$
It shows $a=limlimits_{ktoinfty}f^{n_{k+1}-n_k}(a).$ Note that $f^{n_{k+1}-n_k}(a)in f(M)$ and $f(M)$ is closed.
$endgroup$
– Song
Jan 21 at 15:52
$begingroup$
It shows $a=limlimits_{ktoinfty}f^{n_{k+1}-n_k}(a).$ Note that $f^{n_{k+1}-n_k}(a)in f(M)$ and $f(M)$ is closed.
$endgroup$
– Song
Jan 21 at 15:52
$begingroup$
@Song Why not an official answer?
$endgroup$
– Paul Frost
Jan 21 at 17:37
$begingroup$
@Song Why not an official answer?
$endgroup$
– Paul Frost
Jan 21 at 17:37
$begingroup$
@PaulFrost I saw someone had already answered it, but he/she deleted his/her answer now. Thank you for suggestion :)
$endgroup$
– Song
Jan 21 at 17:53
$begingroup$
@PaulFrost I saw someone had already answered it, but he/she deleted his/her answer now. Thank you for suggestion :)
$endgroup$
– Song
Jan 21 at 17:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The argument shows $$lim_{ktoinfty}d(a,f^{n_{k+1}-n_k}(a))=0,$$ i.e. $limlimits_{ktoinfty} f^{n_{k+1}-n_k}(a) =a$. Note that since $n_{k+1}-n_kge 1$, $$
f^{n_{k+1}-n_k}(a)in f(M)
$$ and $f(M)$ is closed (as a continuous image of the compact set $M$). This shows
$$
ain f(M)
$$ as desired.
answered Jan 21 at 17:44
SongSong
16.7k11044
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The argument shows $$lim_{ktoinfty}d(a,f^{n_{k+1}-n_k}(a))=0,$$ i.e. $limlimits_{ktoinfty} f^{n_{k+1}-n_k}(a) =a$. Note that since $n_{k+1}-n_kge 1$, $$
f^{n_{k+1}-n_k}(a)in f(M)
$$ and $f(M)$ is closed (as a continuous image of the compact set $M$). This shows
$$
ain f(M)
$$ as desired.
answered Jan 21 at 17:44
SongSong
16.7k11044
$endgroup$
add a comment |
$begingroup$
The argument shows $$lim_{ktoinfty}d(a,f^{n_{k+1}-n_k}(a))=0,$$ i.e. $limlimits_{ktoinfty} f^{n_{k+1}-n_k}(a) =a$. Note that since $n_{k+1}-n_kge 1$, $$
f^{n_{k+1}-n_k}(a)in f(M)
$$ and $f(M)$ is closed (as a continuous image of the compact set $M$). This shows
$$
ain f(M)
$$ as desired.
answered Jan 21 at 17:44
SongSong
16.7k11044
$endgroup$
add a comment |
$begingroup$
The argument shows $$lim_{ktoinfty}d(a,f^{n_{k+1}-n_k}(a))=0,$$ i.e. $limlimits_{ktoinfty} f^{n_{k+1}-n_k}(a) =a$. Note that since $n_{k+1}-n_kge 1$, $$
f^{n_{k+1}-n_k}(a)in f(M)
$$ and $f(M)$ is closed (as a continuous image of the compact set $M$). This shows
$$
ain f(M)
$$ as desired.
answered Jan 21 at 17:44
SongSong
16.7k11044
$endgroup$
The argument shows $$lim_{ktoinfty}d(a,f^{n_{k+1}-n_k}(a))=0,$$ i.e. $limlimits_{ktoinfty} f^{n_{k+1}-n_k}(a) =a$. Note that since $n_{k+1}-n_kge 1$, $$
f^{n_{k+1}-n_k}(a)in f(M)
$$ and $f(M)$ is closed (as a continuous image of the compact set $M$). This shows
$$
ain f(M)
$$ as desired.
answered Jan 21 at 17:44
SongSong
16.7k11044
answered Jan 21 at 17:44
SongSong
16.7k11044
answered Jan 21 at 17:44
SongSong
16.7k11044
answered Jan 21 at 17:44
SongSong
16.7k11044
16.7k11044
add a comment |
add a comment |
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$begingroup$
It shows $a=limlimits_{ktoinfty}f^{n_{k+1}-n_k}(a).$ Note that $f^{n_{k+1}-n_k}(a)in f(M)$ and $f(M)$ is closed.
$endgroup$
– Song
Jan 21 at 15:52
$begingroup$
@Song Why not an official answer?
$endgroup$
– Paul Frost
Jan 21 at 17:37
$begingroup$
@PaulFrost I saw someone had already answered it, but he/she deleted his/her answer now. Thank you for suggestion :)
$endgroup$
– Song
Jan 21 at 17:53