Making proof rigorous that isometries of $mathbb{R}^n$ are bijective using ballsCan you prove this three-way...
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Making proof rigorous that isometries of $mathbb{R}^n$ are bijective using balls
Can you prove this three-way linear map composition?Isometries of the realsClassification of isometries in euclidean planeAbout isometries $(mathbb{R}^3,left|{cdot}right|_{infty})to (mathbb{R}^n,left|{cdot}right|_2)$Does a group of isometries uniquely characterize a metric?Every non-abelian group of orientation-preserving plane isometries contains a non-trivial translationProof that a function is bijective if and only if it is both surjective and injectiveIsometries of complex spaceQuick question on the definition of isometriesHow many “elementary” isometries are needed to generate any isometry of an n-dimensional Euclidean space?
$begingroup$
I was wondering how the proof of the fact that isometries of $mathbb{R}^n$ are bijective could be made rigorous using the following argument. I understand that there is a more general (topological) argument involving isometries of compact metric spaces, but I'm doing this as a practice question for a class on groups (we prove if $f : mathbb{R}^n rightarrow mathbb{R}^n$ is an isometry it is bijective, and thus that the set of all such isometries forms a group under composition).
The argument (for surjectivity, injectivity is fine) goes as follows:
Let $x in mathbb{R}^n$, and consider $B_{delta}(x) = {y in mathbb{R}^n : d(x,y) < delta } subset mathbb{R}^n$ the open ball of radius $delta$ around $x$ in $mathbb{R}^n$. Now since $f$ is an isometry, $f(B_{delta}(x)) = B_{delta}(x+a)$, where $a in mathbb{R}^n$. That is, $f$ is a translation of $B_{delta}(x)$ since $f$ is distance preserving. Now if we let $delta rightarrow infty$, any $pin mathbb{R}^n$ will eventually be in some $B_{delta}(x)$ as this ball will fill $mathbb{R}^n$. Thus $exists q in mathbb{R}^n$ such that $f(q) = p$ and hence $f$ is surjective. Then as $f$ is injective and surjective, $f$ is a bijection.
I'm wondering how this argument could be made more rigorous. The proof for injectivity and that this does form a group under composition is fine, just that this part is a little trickier and I was wondering if anyone had any advice for this.
proof-verification proof-writing isometry
asked yesterday
jamesmbcnjamesmbcn
224
$endgroup$
|
show 2 more comments
$begingroup$
I was wondering how the proof of the fact that isometries of $mathbb{R}^n$ are bijective could be made rigorous using the following argument. I understand that there is a more general (topological) argument involving isometries of compact metric spaces, but I'm doing this as a practice question for a class on groups (we prove if $f : mathbb{R}^n rightarrow mathbb{R}^n$ is an isometry it is bijective, and thus that the set of all such isometries forms a group under composition).
The argument (for surjectivity, injectivity is fine) goes as follows:
Let $x in mathbb{R}^n$, and consider $B_{delta}(x) = {y in mathbb{R}^n : d(x,y) < delta } subset mathbb{R}^n$ the open ball of radius $delta$ around $x$ in $mathbb{R}^n$. Now since $f$ is an isometry, $f(B_{delta}(x)) = B_{delta}(x+a)$, where $a in mathbb{R}^n$. That is, $f$ is a translation of $B_{delta}(x)$ since $f$ is distance preserving. Now if we let $delta rightarrow infty$, any $pin mathbb{R}^n$ will eventually be in some $B_{delta}(x)$ as this ball will fill $mathbb{R}^n$. Thus $exists q in mathbb{R}^n$ such that $f(q) = p$ and hence $f$ is surjective. Then as $f$ is injective and surjective, $f$ is a bijection.
I'm wondering how this argument could be made more rigorous. The proof for injectivity and that this does form a group under composition is fine, just that this part is a little trickier and I was wondering if anyone had any advice for this.
proof-verification proof-writing isometry
asked yesterday
jamesmbcnjamesmbcn
224
$endgroup$
2
$begingroup$
Unfortunately, $f$ being an isometry tells you only that $f(B_delta(x)) subseteq B_delta(f(x))$. You fail to obtain $supseteq$ in infinite-dimensions, for example. If this can be shown through topology, it will take an argument that uses properties peculiar to finite dimensions (e.g. compactness of the closed unit ball?).
$endgroup$
– Theo Bendit
yesterday
1
$begingroup$
It seems like this argument asserts that every isometry is a translation. (There is a possible typo of “if” for “of,” so I may be mistaken.) This is certainly not true, since reflections are isometries.
$endgroup$
– Steve Kass
yesterday
$begingroup$
@TheoBendit right, but since I am considering only the case in which $f$ is a map from $mathbb{R}^n$ to itself we are working only in finite dimensions, no?
$endgroup$
– jamesmbcn
yesterday
$begingroup$
@SteveKass Right of course, so I should fix this by possibly having some orthogonal operation on $B_{delta}(x)$ also? As I understand it, it turns out isometries of $mathbb{R}^n$ turn out to be the affine maps, but I was trying to use a different argument.
$endgroup$
– jamesmbcn
yesterday
2
$begingroup$
Of course; I'm just saying that you may need to go into the anatomy of $Bbb{R}^n$, or the linear structure of finite-dimensional vector spaces, if you want your proof to work. At some point in your proof, you've got to actually use the assumption that you're in finite-dimensional space, otherwise the infinite-dimensional spaces will be a counterexample to your arguments!
$endgroup$
– Theo Bendit
yesterday
|
show 2 more comments
$begingroup$
I was wondering how the proof of the fact that isometries of $mathbb{R}^n$ are bijective could be made rigorous using the following argument. I understand that there is a more general (topological) argument involving isometries of compact metric spaces, but I'm doing this as a practice question for a class on groups (we prove if $f : mathbb{R}^n rightarrow mathbb{R}^n$ is an isometry it is bijective, and thus that the set of all such isometries forms a group under composition).
The argument (for surjectivity, injectivity is fine) goes as follows:
Let $x in mathbb{R}^n$, and consider $B_{delta}(x) = {y in mathbb{R}^n : d(x,y) < delta } subset mathbb{R}^n$ the open ball of radius $delta$ around $x$ in $mathbb{R}^n$. Now since $f$ is an isometry, $f(B_{delta}(x)) = B_{delta}(x+a)$, where $a in mathbb{R}^n$. That is, $f$ is a translation of $B_{delta}(x)$ since $f$ is distance preserving. Now if we let $delta rightarrow infty$, any $pin mathbb{R}^n$ will eventually be in some $B_{delta}(x)$ as this ball will fill $mathbb{R}^n$. Thus $exists q in mathbb{R}^n$ such that $f(q) = p$ and hence $f$ is surjective. Then as $f$ is injective and surjective, $f$ is a bijection.
I'm wondering how this argument could be made more rigorous. The proof for injectivity and that this does form a group under composition is fine, just that this part is a little trickier and I was wondering if anyone had any advice for this.
proof-verification proof-writing isometry
asked yesterday
jamesmbcnjamesmbcn
224
$endgroup$
I was wondering how the proof of the fact that isometries of $mathbb{R}^n$ are bijective could be made rigorous using the following argument. I understand that there is a more general (topological) argument involving isometries of compact metric spaces, but I'm doing this as a practice question for a class on groups (we prove if $f : mathbb{R}^n rightarrow mathbb{R}^n$ is an isometry it is bijective, and thus that the set of all such isometries forms a group under composition).
The argument (for surjectivity, injectivity is fine) goes as follows:
Let $x in mathbb{R}^n$, and consider $B_{delta}(x) = {y in mathbb{R}^n : d(x,y) < delta } subset mathbb{R}^n$ the open ball of radius $delta$ around $x$ in $mathbb{R}^n$. Now since $f$ is an isometry, $f(B_{delta}(x)) = B_{delta}(x+a)$, where $a in mathbb{R}^n$. That is, $f$ is a translation of $B_{delta}(x)$ since $f$ is distance preserving. Now if we let $delta rightarrow infty$, any $pin mathbb{R}^n$ will eventually be in some $B_{delta}(x)$ as this ball will fill $mathbb{R}^n$. Thus $exists q in mathbb{R}^n$ such that $f(q) = p$ and hence $f$ is surjective. Then as $f$ is injective and surjective, $f$ is a bijection.
I'm wondering how this argument could be made more rigorous. The proof for injectivity and that this does form a group under composition is fine, just that this part is a little trickier and I was wondering if anyone had any advice for this.
proof-verification proof-writing isometry
proof-verification proof-writing isometry
asked yesterday
jamesmbcnjamesmbcn
224
asked yesterday
jamesmbcnjamesmbcn
224
edited yesterday
jamesmbcn
asked yesterday
jamesmbcnjamesmbcn
224
asked yesterday
jamesmbcnjamesmbcn
224
asked yesterday
jamesmbcnjamesmbcn
224
224
2
$begingroup$
Unfortunately, $f$ being an isometry tells you only that $f(B_delta(x)) subseteq B_delta(f(x))$. You fail to obtain $supseteq$ in infinite-dimensions, for example. If this can be shown through topology, it will take an argument that uses properties peculiar to finite dimensions (e.g. compactness of the closed unit ball?).
$endgroup$
– Theo Bendit
yesterday
1
$begingroup$
It seems like this argument asserts that every isometry is a translation. (There is a possible typo of “if” for “of,” so I may be mistaken.) This is certainly not true, since reflections are isometries.
$endgroup$
– Steve Kass
yesterday
$begingroup$
@TheoBendit right, but since I am considering only the case in which $f$ is a map from $mathbb{R}^n$ to itself we are working only in finite dimensions, no?
$endgroup$
– jamesmbcn
yesterday
$begingroup$
@SteveKass Right of course, so I should fix this by possibly having some orthogonal operation on $B_{delta}(x)$ also? As I understand it, it turns out isometries of $mathbb{R}^n$ turn out to be the affine maps, but I was trying to use a different argument.
$endgroup$
– jamesmbcn
yesterday
2
$begingroup$
Of course; I'm just saying that you may need to go into the anatomy of $Bbb{R}^n$, or the linear structure of finite-dimensional vector spaces, if you want your proof to work. At some point in your proof, you've got to actually use the assumption that you're in finite-dimensional space, otherwise the infinite-dimensional spaces will be a counterexample to your arguments!
$endgroup$
– Theo Bendit
yesterday
|
show 2 more comments
2
$begingroup$
Unfortunately, $f$ being an isometry tells you only that $f(B_delta(x)) subseteq B_delta(f(x))$. You fail to obtain $supseteq$ in infinite-dimensions, for example. If this can be shown through topology, it will take an argument that uses properties peculiar to finite dimensions (e.g. compactness of the closed unit ball?).
$endgroup$
– Theo Bendit
yesterday
1
$begingroup$
It seems like this argument asserts that every isometry is a translation. (There is a possible typo of “if” for “of,” so I may be mistaken.) This is certainly not true, since reflections are isometries.
$endgroup$
– Steve Kass
yesterday
$begingroup$
@TheoBendit right, but since I am considering only the case in which $f$ is a map from $mathbb{R}^n$ to itself we are working only in finite dimensions, no?
$endgroup$
– jamesmbcn
yesterday
$begingroup$
@SteveKass Right of course, so I should fix this by possibly having some orthogonal operation on $B_{delta}(x)$ also? As I understand it, it turns out isometries of $mathbb{R}^n$ turn out to be the affine maps, but I was trying to use a different argument.
$endgroup$
– jamesmbcn
yesterday
2
$begingroup$
Of course; I'm just saying that you may need to go into the anatomy of $Bbb{R}^n$, or the linear structure of finite-dimensional vector spaces, if you want your proof to work. At some point in your proof, you've got to actually use the assumption that you're in finite-dimensional space, otherwise the infinite-dimensional spaces will be a counterexample to your arguments!
$endgroup$
– Theo Bendit
yesterday
2
2
$begingroup$
Unfortunately, $f$ being an isometry tells you only that $f(B_delta(x)) subseteq B_delta(f(x))$. You fail to obtain $supseteq$ in infinite-dimensions, for example. If this can be shown through topology, it will take an argument that uses properties peculiar to finite dimensions (e.g. compactness of the closed unit ball?).
$endgroup$
– Theo Bendit
yesterday
$begingroup$
Unfortunately, $f$ being an isometry tells you only that $f(B_delta(x)) subseteq B_delta(f(x))$. You fail to obtain $supseteq$ in infinite-dimensions, for example. If this can be shown through topology, it will take an argument that uses properties peculiar to finite dimensions (e.g. compactness of the closed unit ball?).
$endgroup$
– Theo Bendit
yesterday
1
1
$begingroup$
It seems like this argument asserts that every isometry is a translation. (There is a possible typo of “if” for “of,” so I may be mistaken.) This is certainly not true, since reflections are isometries.
$endgroup$
– Steve Kass
yesterday
$begingroup$
It seems like this argument asserts that every isometry is a translation. (There is a possible typo of “if” for “of,” so I may be mistaken.) This is certainly not true, since reflections are isometries.
$endgroup$
– Steve Kass
yesterday
$begingroup$
@TheoBendit right, but since I am considering only the case in which $f$ is a map from $mathbb{R}^n$ to itself we are working only in finite dimensions, no?
$endgroup$
– jamesmbcn
yesterday
$begingroup$
@TheoBendit right, but since I am considering only the case in which $f$ is a map from $mathbb{R}^n$ to itself we are working only in finite dimensions, no?
$endgroup$
– jamesmbcn
yesterday
$begingroup$
@SteveKass Right of course, so I should fix this by possibly having some orthogonal operation on $B_{delta}(x)$ also? As I understand it, it turns out isometries of $mathbb{R}^n$ turn out to be the affine maps, but I was trying to use a different argument.
$endgroup$
– jamesmbcn
yesterday
$begingroup$
@SteveKass Right of course, so I should fix this by possibly having some orthogonal operation on $B_{delta}(x)$ also? As I understand it, it turns out isometries of $mathbb{R}^n$ turn out to be the affine maps, but I was trying to use a different argument.
$endgroup$
– jamesmbcn
yesterday
2
2
$begingroup$
Of course; I'm just saying that you may need to go into the anatomy of $Bbb{R}^n$, or the linear structure of finite-dimensional vector spaces, if you want your proof to work. At some point in your proof, you've got to actually use the assumption that you're in finite-dimensional space, otherwise the infinite-dimensional spaces will be a counterexample to your arguments!
$endgroup$
– Theo Bendit
yesterday
$begingroup$
Of course; I'm just saying that you may need to go into the anatomy of $Bbb{R}^n$, or the linear structure of finite-dimensional vector spaces, if you want your proof to work. At some point in your proof, you've got to actually use the assumption that you're in finite-dimensional space, otherwise the infinite-dimensional spaces will be a counterexample to your arguments!
$endgroup$
– Theo Bendit
yesterday
|
show 2 more comments
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$begingroup$
Unfortunately, $f$ being an isometry tells you only that $f(B_delta(x)) subseteq B_delta(f(x))$. You fail to obtain $supseteq$ in infinite-dimensions, for example. If this can be shown through topology, it will take an argument that uses properties peculiar to finite dimensions (e.g. compactness of the closed unit ball?).
$endgroup$
– Theo Bendit
yesterday
1
$begingroup$
It seems like this argument asserts that every isometry is a translation. (There is a possible typo of “if” for “of,” so I may be mistaken.) This is certainly not true, since reflections are isometries.
$endgroup$
– Steve Kass
yesterday
$begingroup$
@TheoBendit right, but since I am considering only the case in which $f$ is a map from $mathbb{R}^n$ to itself we are working only in finite dimensions, no?
$endgroup$
– jamesmbcn
yesterday
$begingroup$
@SteveKass Right of course, so I should fix this by possibly having some orthogonal operation on $B_{delta}(x)$ also? As I understand it, it turns out isometries of $mathbb{R}^n$ turn out to be the affine maps, but I was trying to use a different argument.
$endgroup$
– jamesmbcn
yesterday
2
$begingroup$
Of course; I'm just saying that you may need to go into the anatomy of $Bbb{R}^n$, or the linear structure of finite-dimensional vector spaces, if you want your proof to work. At some point in your proof, you've got to actually use the assumption that you're in finite-dimensional space, otherwise the infinite-dimensional spaces will be a counterexample to your arguments!
$endgroup$
– Theo Bendit
yesterday