Strongest topology that makes vector space locally convexDefinition of locally convex topological vector spaceLocally convex space characterization in terms of dualityTwo different opinions on whether a topological vector space is a uniform spaceOn Some Locally Convex Topologies of a Vector SpaceHow can we ensure that a space is a subset of locally convex topological space?Locally convex topological vector space using semi normsInitial topology coincides with the locally convex topologyLocally convex spaces - is any space that contains a locally convex space as a subspace, also locally convex?Locally Convex Topology on $C_b(Omega)$Finest locally convex topology - Conway, ex 20, sec 1, chapter 4
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Strongest topology that makes vector space locally convex
Definition of locally convex topological vector spaceLocally convex space characterization in terms of dualityTwo different opinions on whether a topological vector space is a uniform spaceOn Some Locally Convex Topologies of a Vector SpaceHow can we ensure that a space is a subset of locally convex topological space?Locally convex topological vector space using semi normsInitial topology coincides with the locally convex topologyLocally convex spaces - is any space that contains a locally convex space as a subspace, also locally convex?Locally Convex Topology on $C_b(Omega)$Finest locally convex topology - Conway, ex 20, sec 1, chapter 4
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
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Here is an exercise from Barvinok's "A Course in Convexity" (ex. III.3.3.3, p.119):
Prove that the strongest topology that makes a vector space $V$ a locally convex topological vector space is the topology where $U subseteq V$ is open if and only if it is a union of convex algebraically open sets.
Isn't the discrete topology (all sets are open) also turning $V$ into a locally convex TVS? Indeed, every singleton set $x$ is convex and open, the operations are continuous, and every singleton set is also closed.
Am I missing something or is there a problem with the exercise? If the statement is wrong, then any clues as to what should be the correct statement?
functional-analysis convex-analysis topological-vector-spaces
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
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add a comment |
$begingroup$
Here is an exercise from Barvinok's "A Course in Convexity" (ex. III.3.3.3, p.119):
Prove that the strongest topology that makes a vector space $V$ a locally convex topological vector space is the topology where $U subseteq V$ is open if and only if it is a union of convex algebraically open sets.
Isn't the discrete topology (all sets are open) also turning $V$ into a locally convex TVS? Indeed, every singleton set $x$ is convex and open, the operations are continuous, and every singleton set is also closed.
Am I missing something or is there a problem with the exercise? If the statement is wrong, then any clues as to what should be the correct statement?
functional-analysis convex-analysis topological-vector-spaces
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
$endgroup$
add a comment |
$begingroup$
Here is an exercise from Barvinok's "A Course in Convexity" (ex. III.3.3.3, p.119):
Prove that the strongest topology that makes a vector space $V$ a locally convex topological vector space is the topology where $U subseteq V$ is open if and only if it is a union of convex algebraically open sets.
Isn't the discrete topology (all sets are open) also turning $V$ into a locally convex TVS? Indeed, every singleton set $x$ is convex and open, the operations are continuous, and every singleton set is also closed.
Am I missing something or is there a problem with the exercise? If the statement is wrong, then any clues as to what should be the correct statement?
functional-analysis convex-analysis topological-vector-spaces
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
$endgroup$
Here is an exercise from Barvinok's "A Course in Convexity" (ex. III.3.3.3, p.119):
Prove that the strongest topology that makes a vector space $V$ a locally convex topological vector space is the topology where $U subseteq V$ is open if and only if it is a union of convex algebraically open sets.
Isn't the discrete topology (all sets are open) also turning $V$ into a locally convex TVS? Indeed, every singleton set $x$ is convex and open, the operations are continuous, and every singleton set is also closed.
Am I missing something or is there a problem with the exercise? If the statement is wrong, then any clues as to what should be the correct statement?
functional-analysis convex-analysis topological-vector-spaces
functional-analysis convex-analysis topological-vector-spaces
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
edited 7 hours ago
Abdelmalek Abdesselam
1,0214 silver badges12 bronze badges
1,0214 silver badges12 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
asked 9 hours ago
FernandoFernando
524 bronze badges
524 bronze badges
add a comment |
add a comment |
1 Answer
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No! The scalar multiplication $mathbbFtimes Vto V$ ($mathbbF=mathbbR,mathbbC$) ceases to be continuous if you put the discrete topology on a nontrivial $V$. Indeed, fixing $vneq 0$, the inverse image of the open set $v$ intersecting the open $mathbbFtimesvsubsetmathbbFtimes V$ is a singleton $(1,v)$, which is not open in $mathbbFtimesv$.
This should also hint at how to recover the "algebraically open".
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
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Very useful! Thanks!
$endgroup$
– Fernando
7 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No! The scalar multiplication $mathbbFtimes Vto V$ ($mathbbF=mathbbR,mathbbC$) ceases to be continuous if you put the discrete topology on a nontrivial $V$. Indeed, fixing $vneq 0$, the inverse image of the open set $v$ intersecting the open $mathbbFtimesvsubsetmathbbFtimes V$ is a singleton $(1,v)$, which is not open in $mathbbFtimesv$.
This should also hint at how to recover the "algebraically open".
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
$endgroup$
$begingroup$
Very useful! Thanks!
$endgroup$
– Fernando
7 hours ago
add a comment |
$begingroup$
No! The scalar multiplication $mathbbFtimes Vto V$ ($mathbbF=mathbbR,mathbbC$) ceases to be continuous if you put the discrete topology on a nontrivial $V$. Indeed, fixing $vneq 0$, the inverse image of the open set $v$ intersecting the open $mathbbFtimesvsubsetmathbbFtimes V$ is a singleton $(1,v)$, which is not open in $mathbbFtimesv$.
This should also hint at how to recover the "algebraically open".
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
$endgroup$
$begingroup$
Very useful! Thanks!
$endgroup$
– Fernando
7 hours ago
add a comment |
$begingroup$
No! The scalar multiplication $mathbbFtimes Vto V$ ($mathbbF=mathbbR,mathbbC$) ceases to be continuous if you put the discrete topology on a nontrivial $V$. Indeed, fixing $vneq 0$, the inverse image of the open set $v$ intersecting the open $mathbbFtimesvsubsetmathbbFtimes V$ is a singleton $(1,v)$, which is not open in $mathbbFtimesv$.
This should also hint at how to recover the "algebraically open".
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
$endgroup$
No! The scalar multiplication $mathbbFtimes Vto V$ ($mathbbF=mathbbR,mathbbC$) ceases to be continuous if you put the discrete topology on a nontrivial $V$. Indeed, fixing $vneq 0$, the inverse image of the open set $v$ intersecting the open $mathbbFtimesvsubsetmathbbFtimes V$ is a singleton $(1,v)$, which is not open in $mathbbFtimesv$.
This should also hint at how to recover the "algebraically open".
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
edited 9 hours ago
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
answered 9 hours ago
user10354138user10354138
17.3k2 gold badges12 silver badges32 bronze badges
17.3k2 gold badges12 silver badges32 bronze badges
$begingroup$
Very useful! Thanks!
$endgroup$
– Fernando
7 hours ago
add a comment |
$begingroup$
Very useful! Thanks!
$endgroup$
– Fernando
7 hours ago
$begingroup$
Very useful! Thanks!
$endgroup$
– Fernando
7 hours ago
$begingroup$
Very useful! Thanks!
$endgroup$
– Fernando
7 hours ago
add a comment |
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