Topology induced by the Euclidean metric
up vote
1
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Let $V$ be a finite dimensional real vector space, then $V$ is isomorphic to $mathbb{R}^n$.
So, is the Euclidean metric defined by $d(x,y)= ||(lambda_i)-(gamma_i) ||$ where $lambda_i$ and $gamma_i$ are the coefficients?
In this case, if $(V,tau)$ is a topological vector space then the linear map $id: (V,tau_d) rightarrow (V,tau)$ is continuous because $(V,tau_d)$ is Hausdorff and finite dimensional and finally the topology induced by the Euclidean metris $ tau_d$ is at least as fine as $tau$ ?
Am I right?
Thanks!
general-topology functional-analysis
edited Nov 22 at 14:49
caverac
12.7k21028
asked Nov 22 at 14:44
Diego Cornejo Koc
63
add a comment |
up vote
1
down vote
favorite
Let $V$ be a finite dimensional real vector space, then $V$ is isomorphic to $mathbb{R}^n$.
So, is the Euclidean metric defined by $d(x,y)= ||(lambda_i)-(gamma_i) ||$ where $lambda_i$ and $gamma_i$ are the coefficients?
In this case, if $(V,tau)$ is a topological vector space then the linear map $id: (V,tau_d) rightarrow (V,tau)$ is continuous because $(V,tau_d)$ is Hausdorff and finite dimensional and finally the topology induced by the Euclidean metris $ tau_d$ is at least as fine as $tau$ ?
Am I right?
Thanks!
general-topology functional-analysis
edited Nov 22 at 14:49
caverac
12.7k21028
asked Nov 22 at 14:44
Diego Cornejo Koc
63
No, the coefficient stuff used to define a metric is wrong.
– William Elliot
Nov 23 at 0:49
How is defined the Euclidean metric? I am confused.
– Diego Cornejo Koc
Nov 23 at 2:06
For vectors v,w, d(v,w) = ||v - w||.
– William Elliot
Nov 23 at 5:33
Yes, but $v$ and $w$ are in $V$. So $d(v,w)$ is the norm of their coefficients.
– Diego Cornejo Koc
Nov 23 at 5:46
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $V$ be a finite dimensional real vector space, then $V$ is isomorphic to $mathbb{R}^n$.
So, is the Euclidean metric defined by $d(x,y)= ||(lambda_i)-(gamma_i) ||$ where $lambda_i$ and $gamma_i$ are the coefficients?
In this case, if $(V,tau)$ is a topological vector space then the linear map $id: (V,tau_d) rightarrow (V,tau)$ is continuous because $(V,tau_d)$ is Hausdorff and finite dimensional and finally the topology induced by the Euclidean metris $ tau_d$ is at least as fine as $tau$ ?
Am I right?
Thanks!
general-topology functional-analysis
edited Nov 22 at 14:49
caverac
12.7k21028
asked Nov 22 at 14:44
Diego Cornejo Koc
63
Let $V$ be a finite dimensional real vector space, then $V$ is isomorphic to $mathbb{R}^n$.
So, is the Euclidean metric defined by $d(x,y)= ||(lambda_i)-(gamma_i) ||$ where $lambda_i$ and $gamma_i$ are the coefficients?
In this case, if $(V,tau)$ is a topological vector space then the linear map $id: (V,tau_d) rightarrow (V,tau)$ is continuous because $(V,tau_d)$ is Hausdorff and finite dimensional and finally the topology induced by the Euclidean metris $ tau_d$ is at least as fine as $tau$ ?
Am I right?
Thanks!
general-topology functional-analysis
general-topology functional-analysis
edited Nov 22 at 14:49
caverac
12.7k21028
asked Nov 22 at 14:44
Diego Cornejo Koc
63
edited Nov 22 at 14:49
caverac
12.7k21028
asked Nov 22 at 14:44
Diego Cornejo Koc
63
edited Nov 22 at 14:49
caverac
12.7k21028
edited Nov 22 at 14:49
caverac
12.7k21028
edited Nov 22 at 14:49
caverac
12.7k21028
12.7k21028
asked Nov 22 at 14:44
Diego Cornejo Koc
63
asked Nov 22 at 14:44
Diego Cornejo Koc
63
asked Nov 22 at 14:44
Diego Cornejo Koc
63
63
No, the coefficient stuff used to define a metric is wrong.
– William Elliot
Nov 23 at 0:49
How is defined the Euclidean metric? I am confused.
– Diego Cornejo Koc
Nov 23 at 2:06
For vectors v,w, d(v,w) = ||v - w||.
– William Elliot
Nov 23 at 5:33
Yes, but $v$ and $w$ are in $V$. So $d(v,w)$ is the norm of their coefficients.
– Diego Cornejo Koc
Nov 23 at 5:46
add a comment |
No, the coefficient stuff used to define a metric is wrong.
– William Elliot
Nov 23 at 0:49
How is defined the Euclidean metric? I am confused.
– Diego Cornejo Koc
Nov 23 at 2:06
For vectors v,w, d(v,w) = ||v - w||.
– William Elliot
Nov 23 at 5:33
Yes, but $v$ and $w$ are in $V$. So $d(v,w)$ is the norm of their coefficients.
– Diego Cornejo Koc
Nov 23 at 5:46
No, the coefficient stuff used to define a metric is wrong.
– William Elliot
Nov 23 at 0:49
No, the coefficient stuff used to define a metric is wrong.
– William Elliot
Nov 23 at 0:49
How is defined the Euclidean metric? I am confused.
– Diego Cornejo Koc
Nov 23 at 2:06
How is defined the Euclidean metric? I am confused.
– Diego Cornejo Koc
Nov 23 at 2:06
For vectors v,w, d(v,w) = ||v - w||.
– William Elliot
Nov 23 at 5:33
For vectors v,w, d(v,w) = ||v - w||.
– William Elliot
Nov 23 at 5:33
Yes, but $v$ and $w$ are in $V$. So $d(v,w)$ is the norm of their coefficients.
– Diego Cornejo Koc
Nov 23 at 5:46
Yes, but $v$ and $w$ are in $V$. So $d(v,w)$ is the norm of their coefficients.
– Diego Cornejo Koc
Nov 23 at 5:46
add a comment |
1 Answer
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up vote
0
down vote
Let V be a normed vector space with norm n:V -> R.
The norm induces a metric for V, d(u,v) = n(u - v).
The topology Td, induced by the norm metric cannot be
compared to other topologies making V a TVS.
That is because V with the discrete topology
Tdis and V with the indiscrete are both TVS's.
Notice that the map id:(V,Td) -> (V,Tdis) is not continuous.
In the event, as is common, a TVS is defined to also
be Hausdorff, then for the TVS V with topology T
if id:(V,T) -> (V,Td) is continuous, then T is finer than Td.
answered Nov 23 at 11:45
William Elliot
7,0222518
It should be $d(f(x),f(y)) = || x-y ||$. But what about the topologies?
– Diego Cornejo Koc
Nov 23 at 13:49
Then use f$^{-1}$. Metrics generate topologies.
– William Elliot
Nov 24 at 0:37
What about the topologies?
– Diego Cornejo Koc
Nov 27 at 22:35
It is the topology generated by the metric or the norm.
– William Elliot
Nov 28 at 4:17
I mean the second question. Is the topology generated by the metric at least as fine as another topology $tau$ , where $(V,tau)$ is a t.v.s. ?
– Diego Cornejo Koc
Nov 28 at 13:57
|
show 4 more comments
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
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up vote
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down vote
Let V be a normed vector space with norm n:V -> R.
The norm induces a metric for V, d(u,v) = n(u - v).
The topology Td, induced by the norm metric cannot be
compared to other topologies making V a TVS.
That is because V with the discrete topology
Tdis and V with the indiscrete are both TVS's.
Notice that the map id:(V,Td) -> (V,Tdis) is not continuous.
In the event, as is common, a TVS is defined to also
be Hausdorff, then for the TVS V with topology T
if id:(V,T) -> (V,Td) is continuous, then T is finer than Td.
answered Nov 23 at 11:45
William Elliot
7,0222518
It should be $d(f(x),f(y)) = || x-y ||$. But what about the topologies?
– Diego Cornejo Koc
Nov 23 at 13:49
Then use f$^{-1}$. Metrics generate topologies.
– William Elliot
Nov 24 at 0:37
What about the topologies?
– Diego Cornejo Koc
Nov 27 at 22:35
It is the topology generated by the metric or the norm.
– William Elliot
Nov 28 at 4:17
I mean the second question. Is the topology generated by the metric at least as fine as another topology $tau$ , where $(V,tau)$ is a t.v.s. ?
– Diego Cornejo Koc
Nov 28 at 13:57
|
show 4 more comments
up vote
0
down vote
Let V be a normed vector space with norm n:V -> R.
The norm induces a metric for V, d(u,v) = n(u - v).
The topology Td, induced by the norm metric cannot be
compared to other topologies making V a TVS.
That is because V with the discrete topology
Tdis and V with the indiscrete are both TVS's.
Notice that the map id:(V,Td) -> (V,Tdis) is not continuous.
In the event, as is common, a TVS is defined to also
be Hausdorff, then for the TVS V with topology T
if id:(V,T) -> (V,Td) is continuous, then T is finer than Td.
answered Nov 23 at 11:45
William Elliot
7,0222518
It should be $d(f(x),f(y)) = || x-y ||$. But what about the topologies?
– Diego Cornejo Koc
Nov 23 at 13:49
Then use f$^{-1}$. Metrics generate topologies.
– William Elliot
Nov 24 at 0:37
What about the topologies?
– Diego Cornejo Koc
Nov 27 at 22:35
It is the topology generated by the metric or the norm.
– William Elliot
Nov 28 at 4:17
I mean the second question. Is the topology generated by the metric at least as fine as another topology $tau$ , where $(V,tau)$ is a t.v.s. ?
– Diego Cornejo Koc
Nov 28 at 13:57
|
show 4 more comments
up vote
0
down vote
up vote
0
down vote
Let V be a normed vector space with norm n:V -> R.
The norm induces a metric for V, d(u,v) = n(u - v).
The topology Td, induced by the norm metric cannot be
compared to other topologies making V a TVS.
That is because V with the discrete topology
Tdis and V with the indiscrete are both TVS's.
Notice that the map id:(V,Td) -> (V,Tdis) is not continuous.
In the event, as is common, a TVS is defined to also
be Hausdorff, then for the TVS V with topology T
if id:(V,T) -> (V,Td) is continuous, then T is finer than Td.
answered Nov 23 at 11:45
William Elliot
7,0222518
Let V be a normed vector space with norm n:V -> R.
The norm induces a metric for V, d(u,v) = n(u - v).
The topology Td, induced by the norm metric cannot be
compared to other topologies making V a TVS.
That is because V with the discrete topology
Tdis and V with the indiscrete are both TVS's.
Notice that the map id:(V,Td) -> (V,Tdis) is not continuous.
In the event, as is common, a TVS is defined to also
be Hausdorff, then for the TVS V with topology T
if id:(V,T) -> (V,Td) is continuous, then T is finer than Td.
answered Nov 23 at 11:45
William Elliot
7,0222518
edited Nov 29 at 10:30
answered Nov 23 at 11:45
William Elliot
7,0222518
answered Nov 23 at 11:45
William Elliot
7,0222518
answered Nov 23 at 11:45
William Elliot
7,0222518
7,0222518
It should be $d(f(x),f(y)) = || x-y ||$. But what about the topologies?
– Diego Cornejo Koc
Nov 23 at 13:49
Then use f$^{-1}$. Metrics generate topologies.
– William Elliot
Nov 24 at 0:37
What about the topologies?
– Diego Cornejo Koc
Nov 27 at 22:35
It is the topology generated by the metric or the norm.
– William Elliot
Nov 28 at 4:17
I mean the second question. Is the topology generated by the metric at least as fine as another topology $tau$ , where $(V,tau)$ is a t.v.s. ?
– Diego Cornejo Koc
Nov 28 at 13:57
|
show 4 more comments
It should be $d(f(x),f(y)) = || x-y ||$. But what about the topologies?
– Diego Cornejo Koc
Nov 23 at 13:49
Then use f$^{-1}$. Metrics generate topologies.
– William Elliot
Nov 24 at 0:37
What about the topologies?
– Diego Cornejo Koc
Nov 27 at 22:35
It is the topology generated by the metric or the norm.
– William Elliot
Nov 28 at 4:17
I mean the second question. Is the topology generated by the metric at least as fine as another topology $tau$ , where $(V,tau)$ is a t.v.s. ?
– Diego Cornejo Koc
Nov 28 at 13:57
It should be $d(f(x),f(y)) = || x-y ||$. But what about the topologies?
– Diego Cornejo Koc
Nov 23 at 13:49
It should be $d(f(x),f(y)) = || x-y ||$. But what about the topologies?
– Diego Cornejo Koc
Nov 23 at 13:49
Then use f$^{-1}$. Metrics generate topologies.
– William Elliot
Nov 24 at 0:37
Then use f$^{-1}$. Metrics generate topologies.
– William Elliot
Nov 24 at 0:37
What about the topologies?
– Diego Cornejo Koc
Nov 27 at 22:35
What about the topologies?
– Diego Cornejo Koc
Nov 27 at 22:35
It is the topology generated by the metric or the norm.
– William Elliot
Nov 28 at 4:17
It is the topology generated by the metric or the norm.
– William Elliot
Nov 28 at 4:17
I mean the second question. Is the topology generated by the metric at least as fine as another topology $tau$ , where $(V,tau)$ is a t.v.s. ?
– Diego Cornejo Koc
Nov 28 at 13:57
I mean the second question. Is the topology generated by the metric at least as fine as another topology $tau$ , where $(V,tau)$ is a t.v.s. ?
– Diego Cornejo Koc
Nov 28 at 13:57
|
show 4 more comments
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No, the coefficient stuff used to define a metric is wrong.
– William Elliot
Nov 23 at 0:49
How is defined the Euclidean metric? I am confused.
– Diego Cornejo Koc
Nov 23 at 2:06
For vectors v,w, d(v,w) = ||v - w||.
– William Elliot
Nov 23 at 5:33
Yes, but $v$ and $w$ are in $V$. So $d(v,w)$ is the norm of their coefficients.
– Diego Cornejo Koc
Nov 23 at 5:46