Topology induced by the Euclidean metric











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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!










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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















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!










share|cite|improve this question
























  • 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













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!










share|cite|improve this question
















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






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share|cite|improve this question













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share|cite|improve this question








edited Nov 22 at 14:49









caverac

11.8k21027




11.8k21027










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


















  • 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










1 Answer
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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.






share|cite|improve this answer























  • 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













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1 Answer
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1 Answer
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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.






share|cite|improve this answer























  • 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

















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.






share|cite|improve this answer























  • 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















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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered Nov 23 at 11:45









William Elliot

6,9052518




6,9052518












  • 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










  • 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




















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