Unique Completetion of a metric spaces: A question.
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i'm just trying to solidify my understanding of whats going on in the above case so i can visualise it and why it's important so please add or correct away.
Consider $(X,d)$ a metric space, then $exists~ text{ a complete metric space }(hat{X},hat{d}) $ such that $W subset_{dense} hat{X}$ which is isometric to X.
so in practice let $T:X rightarrow W$ be a distance preserving bijection, then $forall x,y in X, ~exists ~ T(x), T(y) in W $ and we have
$$hat{d}(Tx,Ty) = d(x,y)$$
and $bar{W} = hat{X}$
So; why is this important? i'm thinking the major effect of this is that for any sequence $x_n in_{cauchy} X$ which is divergent (by that i mean just doesnt converge), then $exists ~T(x_n) in_{cauchy} W$ which is convergent in $hat{X} = bar{W}$ since $hat{X}$ is complete.
i'm guessing this will be utilised with the contraction mapping theorem but this is thus far the only major revelation i could imagine as to why this is overwhelmingly important... am i missing anything?
Thanks for taking the time to read this, i appreciate it.
general-topology metric-spaces
asked Nov 16 at 3:41
Vaas
337213
|
show 2 more comments
up vote
0
down vote
favorite
i'm just trying to solidify my understanding of whats going on in the above case so i can visualise it and why it's important so please add or correct away.
Consider $(X,d)$ a metric space, then $exists~ text{ a complete metric space }(hat{X},hat{d}) $ such that $W subset_{dense} hat{X}$ which is isometric to X.
so in practice let $T:X rightarrow W$ be a distance preserving bijection, then $forall x,y in X, ~exists ~ T(x), T(y) in W $ and we have
$$hat{d}(Tx,Ty) = d(x,y)$$
and $bar{W} = hat{X}$
So; why is this important? i'm thinking the major effect of this is that for any sequence $x_n in_{cauchy} X$ which is divergent (by that i mean just doesnt converge), then $exists ~T(x_n) in_{cauchy} W$ which is convergent in $hat{X} = bar{W}$ since $hat{X}$ is complete.
i'm guessing this will be utilised with the contraction mapping theorem but this is thus far the only major revelation i could imagine as to why this is overwhelmingly important... am i missing anything?
Thanks for taking the time to read this, i appreciate it.
general-topology metric-spaces
asked Nov 16 at 3:41
Vaas
337213
1
The very concept of completion is to add new points (and associated distances) to make the original space complete. In other words, we extend the original space without changing the old distances. To do this in a minimal way, we don't want to add any points that aren't needed to make the old Cauchy sequences converge, hence we want the original space to be dense in the completion.
– quasi
Nov 16 at 3:57
so in essence in each case all we'll ever be doing is adding the limit points so that any divergent cauchy sequence in X converges...and doing so with the minimum amount of points needing to be added required?
– Vaas
Nov 16 at 4:03
1
Right. It's the same process used to get $mathbb{R}$ as the completion of $mathbb{Q}$.
– quasi
Nov 16 at 4:05
1
That's the basic idea. We want Cauchy sequences to converge, but if some don't, we can force it by adding "ideal" points for the missing limits.
– quasi
Nov 16 at 4:10
1
Well, you always have the option of passing to the completion, if it helps. The original space is still there, densely embedded.
– quasi
Nov 16 at 4:12
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
i'm just trying to solidify my understanding of whats going on in the above case so i can visualise it and why it's important so please add or correct away.
Consider $(X,d)$ a metric space, then $exists~ text{ a complete metric space }(hat{X},hat{d}) $ such that $W subset_{dense} hat{X}$ which is isometric to X.
so in practice let $T:X rightarrow W$ be a distance preserving bijection, then $forall x,y in X, ~exists ~ T(x), T(y) in W $ and we have
$$hat{d}(Tx,Ty) = d(x,y)$$
and $bar{W} = hat{X}$
So; why is this important? i'm thinking the major effect of this is that for any sequence $x_n in_{cauchy} X$ which is divergent (by that i mean just doesnt converge), then $exists ~T(x_n) in_{cauchy} W$ which is convergent in $hat{X} = bar{W}$ since $hat{X}$ is complete.
i'm guessing this will be utilised with the contraction mapping theorem but this is thus far the only major revelation i could imagine as to why this is overwhelmingly important... am i missing anything?
Thanks for taking the time to read this, i appreciate it.
general-topology metric-spaces
asked Nov 16 at 3:41
Vaas
337213
i'm just trying to solidify my understanding of whats going on in the above case so i can visualise it and why it's important so please add or correct away.
Consider $(X,d)$ a metric space, then $exists~ text{ a complete metric space }(hat{X},hat{d}) $ such that $W subset_{dense} hat{X}$ which is isometric to X.
so in practice let $T:X rightarrow W$ be a distance preserving bijection, then $forall x,y in X, ~exists ~ T(x), T(y) in W $ and we have
$$hat{d}(Tx,Ty) = d(x,y)$$
and $bar{W} = hat{X}$
So; why is this important? i'm thinking the major effect of this is that for any sequence $x_n in_{cauchy} X$ which is divergent (by that i mean just doesnt converge), then $exists ~T(x_n) in_{cauchy} W$ which is convergent in $hat{X} = bar{W}$ since $hat{X}$ is complete.
i'm guessing this will be utilised with the contraction mapping theorem but this is thus far the only major revelation i could imagine as to why this is overwhelmingly important... am i missing anything?
Thanks for taking the time to read this, i appreciate it.
general-topology metric-spaces
general-topology metric-spaces
asked Nov 16 at 3:41
Vaas
337213
asked Nov 16 at 3:41
Vaas
337213
asked Nov 16 at 3:41
Vaas
337213
asked Nov 16 at 3:41
Vaas
337213
asked Nov 16 at 3:41
Vaas
337213
337213
1
The very concept of completion is to add new points (and associated distances) to make the original space complete. In other words, we extend the original space without changing the old distances. To do this in a minimal way, we don't want to add any points that aren't needed to make the old Cauchy sequences converge, hence we want the original space to be dense in the completion.
– quasi
Nov 16 at 3:57
so in essence in each case all we'll ever be doing is adding the limit points so that any divergent cauchy sequence in X converges...and doing so with the minimum amount of points needing to be added required?
– Vaas
Nov 16 at 4:03
1
Right. It's the same process used to get $mathbb{R}$ as the completion of $mathbb{Q}$.
– quasi
Nov 16 at 4:05
1
That's the basic idea. We want Cauchy sequences to converge, but if some don't, we can force it by adding "ideal" points for the missing limits.
– quasi
Nov 16 at 4:10
1
Well, you always have the option of passing to the completion, if it helps. The original space is still there, densely embedded.
– quasi
Nov 16 at 4:12
|
show 2 more comments
1
The very concept of completion is to add new points (and associated distances) to make the original space complete. In other words, we extend the original space without changing the old distances. To do this in a minimal way, we don't want to add any points that aren't needed to make the old Cauchy sequences converge, hence we want the original space to be dense in the completion.
– quasi
Nov 16 at 3:57
so in essence in each case all we'll ever be doing is adding the limit points so that any divergent cauchy sequence in X converges...and doing so with the minimum amount of points needing to be added required?
– Vaas
Nov 16 at 4:03
1
Right. It's the same process used to get $mathbb{R}$ as the completion of $mathbb{Q}$.
– quasi
Nov 16 at 4:05
1
That's the basic idea. We want Cauchy sequences to converge, but if some don't, we can force it by adding "ideal" points for the missing limits.
– quasi
Nov 16 at 4:10
1
Well, you always have the option of passing to the completion, if it helps. The original space is still there, densely embedded.
– quasi
Nov 16 at 4:12
1
1
The very concept of completion is to add new points (and associated distances) to make the original space complete. In other words, we extend the original space without changing the old distances. To do this in a minimal way, we don't want to add any points that aren't needed to make the old Cauchy sequences converge, hence we want the original space to be dense in the completion.
– quasi
Nov 16 at 3:57
The very concept of completion is to add new points (and associated distances) to make the original space complete. In other words, we extend the original space without changing the old distances. To do this in a minimal way, we don't want to add any points that aren't needed to make the old Cauchy sequences converge, hence we want the original space to be dense in the completion.
– quasi
Nov 16 at 3:57
so in essence in each case all we'll ever be doing is adding the limit points so that any divergent cauchy sequence in X converges...and doing so with the minimum amount of points needing to be added required?
– Vaas
Nov 16 at 4:03
so in essence in each case all we'll ever be doing is adding the limit points so that any divergent cauchy sequence in X converges...and doing so with the minimum amount of points needing to be added required?
– Vaas
Nov 16 at 4:03
1
1
Right. It's the same process used to get $mathbb{R}$ as the completion of $mathbb{Q}$.
– quasi
Nov 16 at 4:05
Right. It's the same process used to get $mathbb{R}$ as the completion of $mathbb{Q}$.
– quasi
Nov 16 at 4:05
1
1
That's the basic idea. We want Cauchy sequences to converge, but if some don't, we can force it by adding "ideal" points for the missing limits.
– quasi
Nov 16 at 4:10
That's the basic idea. We want Cauchy sequences to converge, but if some don't, we can force it by adding "ideal" points for the missing limits.
– quasi
Nov 16 at 4:10
1
1
Well, you always have the option of passing to the completion, if it helps. The original space is still there, densely embedded.
– quasi
Nov 16 at 4:12
Well, you always have the option of passing to the completion, if it helps. The original space is still there, densely embedded.
– quasi
Nov 16 at 4:12
|
show 2 more comments
1 Answer
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We want to think of the completion of $X$ as a metric space $hat{X}$ which is the space $X$ plus extra points that we add to ensure that all the Cauchy sequences of $X$ that do not have a limit in $X$, get a limit in $hat{X}$.
We don't want to change $X$ in any way, so all points of $X$ keep their old distance. We only have to come up with distances $hat{d}(x,y)$ when at least one of them is a "new" point.
We don't want to add unnecessary points, so any new point in $hat{X}$ should in fact be a limit of some sequence from $X$. Note that this sequence, as seen in $X$ is in fact a Cauchy sequence (it must be Cauchy in $hat{d}$ (convergent implies Cauchy) and so in $d$ too as the distance is the same) that does not converge in $X$ (as limits in $hat{X}$ are unique, and we already have one among the new points.). This explains why $X$ must be dense in $hat{X}$.
But in practice $hat{X}$ will be a different object from the original $X$ (equivalence classes of Cauchy sequences, usually) so as long as we have an isometric dense copy of $X$ inside a space $hat{X}$ we identify $X$ with that copy via the isometry $T$. The points in $hat{X}setminus T[X]$ are then the "newly added" points.
It's more of a category theory view: the $X$ and $hat{X}$ are different objects in the category, with a "reflection morphism" (the isometric embedding) between them such that certain diagrams can be made commutative in a unique way. There isn't always a notion of "subobject" but morphisms we always have.
answered Nov 16 at 9:21
Henno Brandsma
101k344107
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We want to think of the completion of $X$ as a metric space $hat{X}$ which is the space $X$ plus extra points that we add to ensure that all the Cauchy sequences of $X$ that do not have a limit in $X$, get a limit in $hat{X}$.
We don't want to change $X$ in any way, so all points of $X$ keep their old distance. We only have to come up with distances $hat{d}(x,y)$ when at least one of them is a "new" point.
We don't want to add unnecessary points, so any new point in $hat{X}$ should in fact be a limit of some sequence from $X$. Note that this sequence, as seen in $X$ is in fact a Cauchy sequence (it must be Cauchy in $hat{d}$ (convergent implies Cauchy) and so in $d$ too as the distance is the same) that does not converge in $X$ (as limits in $hat{X}$ are unique, and we already have one among the new points.). This explains why $X$ must be dense in $hat{X}$.
But in practice $hat{X}$ will be a different object from the original $X$ (equivalence classes of Cauchy sequences, usually) so as long as we have an isometric dense copy of $X$ inside a space $hat{X}$ we identify $X$ with that copy via the isometry $T$. The points in $hat{X}setminus T[X]$ are then the "newly added" points.
It's more of a category theory view: the $X$ and $hat{X}$ are different objects in the category, with a "reflection morphism" (the isometric embedding) between them such that certain diagrams can be made commutative in a unique way. There isn't always a notion of "subobject" but morphisms we always have.
answered Nov 16 at 9:21
Henno Brandsma
101k344107
add a comment |
up vote
1
down vote
We want to think of the completion of $X$ as a metric space $hat{X}$ which is the space $X$ plus extra points that we add to ensure that all the Cauchy sequences of $X$ that do not have a limit in $X$, get a limit in $hat{X}$.
We don't want to change $X$ in any way, so all points of $X$ keep their old distance. We only have to come up with distances $hat{d}(x,y)$ when at least one of them is a "new" point.
We don't want to add unnecessary points, so any new point in $hat{X}$ should in fact be a limit of some sequence from $X$. Note that this sequence, as seen in $X$ is in fact a Cauchy sequence (it must be Cauchy in $hat{d}$ (convergent implies Cauchy) and so in $d$ too as the distance is the same) that does not converge in $X$ (as limits in $hat{X}$ are unique, and we already have one among the new points.). This explains why $X$ must be dense in $hat{X}$.
But in practice $hat{X}$ will be a different object from the original $X$ (equivalence classes of Cauchy sequences, usually) so as long as we have an isometric dense copy of $X$ inside a space $hat{X}$ we identify $X$ with that copy via the isometry $T$. The points in $hat{X}setminus T[X]$ are then the "newly added" points.
It's more of a category theory view: the $X$ and $hat{X}$ are different objects in the category, with a "reflection morphism" (the isometric embedding) between them such that certain diagrams can be made commutative in a unique way. There isn't always a notion of "subobject" but morphisms we always have.
answered Nov 16 at 9:21
Henno Brandsma
101k344107
add a comment |
up vote
1
down vote
up vote
1
down vote
We want to think of the completion of $X$ as a metric space $hat{X}$ which is the space $X$ plus extra points that we add to ensure that all the Cauchy sequences of $X$ that do not have a limit in $X$, get a limit in $hat{X}$.
We don't want to change $X$ in any way, so all points of $X$ keep their old distance. We only have to come up with distances $hat{d}(x,y)$ when at least one of them is a "new" point.
We don't want to add unnecessary points, so any new point in $hat{X}$ should in fact be a limit of some sequence from $X$. Note that this sequence, as seen in $X$ is in fact a Cauchy sequence (it must be Cauchy in $hat{d}$ (convergent implies Cauchy) and so in $d$ too as the distance is the same) that does not converge in $X$ (as limits in $hat{X}$ are unique, and we already have one among the new points.). This explains why $X$ must be dense in $hat{X}$.
But in practice $hat{X}$ will be a different object from the original $X$ (equivalence classes of Cauchy sequences, usually) so as long as we have an isometric dense copy of $X$ inside a space $hat{X}$ we identify $X$ with that copy via the isometry $T$. The points in $hat{X}setminus T[X]$ are then the "newly added" points.
It's more of a category theory view: the $X$ and $hat{X}$ are different objects in the category, with a "reflection morphism" (the isometric embedding) between them such that certain diagrams can be made commutative in a unique way. There isn't always a notion of "subobject" but morphisms we always have.
answered Nov 16 at 9:21
Henno Brandsma
101k344107
We want to think of the completion of $X$ as a metric space $hat{X}$ which is the space $X$ plus extra points that we add to ensure that all the Cauchy sequences of $X$ that do not have a limit in $X$, get a limit in $hat{X}$.
We don't want to change $X$ in any way, so all points of $X$ keep their old distance. We only have to come up with distances $hat{d}(x,y)$ when at least one of them is a "new" point.
We don't want to add unnecessary points, so any new point in $hat{X}$ should in fact be a limit of some sequence from $X$. Note that this sequence, as seen in $X$ is in fact a Cauchy sequence (it must be Cauchy in $hat{d}$ (convergent implies Cauchy) and so in $d$ too as the distance is the same) that does not converge in $X$ (as limits in $hat{X}$ are unique, and we already have one among the new points.). This explains why $X$ must be dense in $hat{X}$.
But in practice $hat{X}$ will be a different object from the original $X$ (equivalence classes of Cauchy sequences, usually) so as long as we have an isometric dense copy of $X$ inside a space $hat{X}$ we identify $X$ with that copy via the isometry $T$. The points in $hat{X}setminus T[X]$ are then the "newly added" points.
It's more of a category theory view: the $X$ and $hat{X}$ are different objects in the category, with a "reflection morphism" (the isometric embedding) between them such that certain diagrams can be made commutative in a unique way. There isn't always a notion of "subobject" but morphisms we always have.
answered Nov 16 at 9:21
Henno Brandsma
101k344107
answered Nov 16 at 9:21
Henno Brandsma
101k344107
answered Nov 16 at 9:21
Henno Brandsma
101k344107
answered Nov 16 at 9:21
Henno Brandsma
101k344107
101k344107
add a comment |
add a comment |
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The very concept of completion is to add new points (and associated distances) to make the original space complete. In other words, we extend the original space without changing the old distances. To do this in a minimal way, we don't want to add any points that aren't needed to make the old Cauchy sequences converge, hence we want the original space to be dense in the completion.
– quasi
Nov 16 at 3:57
so in essence in each case all we'll ever be doing is adding the limit points so that any divergent cauchy sequence in X converges...and doing so with the minimum amount of points needing to be added required?
– Vaas
Nov 16 at 4:03
1
Right. It's the same process used to get $mathbb{R}$ as the completion of $mathbb{Q}$.
– quasi
Nov 16 at 4:05
1
That's the basic idea. We want Cauchy sequences to converge, but if some don't, we can force it by adding "ideal" points for the missing limits.
– quasi
Nov 16 at 4:10
1
Well, you always have the option of passing to the completion, if it helps. The original space is still there, densely embedded.
– quasi
Nov 16 at 4:12