Closed set in a finite-dimensional normed space.
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A finite-dimensional normed space X is complete. Every complete space is also closed. If I take linearly independent $x_1,dots,x_n in X$, then the set $sum_i=1^n a_i x_i : a_i > 0=:A$ is finite-dimensional. Hence this set should be closed. But obviously $0 notin A$. But I can choose every scalar $a_i$ smaller and smaller hence I can obtain a sequence in $A$ converging to $0$. Therefore it shouldn't be closed in my opinion.
analysis convergence normed-spaces
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A finite-dimensional normed space X is complete. Every complete space is also closed. If I take linearly independent $x_1,dots,x_n in X$, then the set $sum_i=1^n a_i x_i : a_i > 0=:A$ is finite-dimensional. Hence this set should be closed. But obviously $0 notin A$. But I can choose every scalar $a_i$ smaller and smaller hence I can obtain a sequence in $A$ converging to $0$. Therefore it shouldn't be closed in my opinion.
analysis convergence normed-spaces
1
$A$ is just a cone, not a vector space.
â Alan Muniz
Sep 8 at 13:24
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up vote
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down vote
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up vote
0
down vote
favorite
A finite-dimensional normed space X is complete. Every complete space is also closed. If I take linearly independent $x_1,dots,x_n in X$, then the set $sum_i=1^n a_i x_i : a_i > 0=:A$ is finite-dimensional. Hence this set should be closed. But obviously $0 notin A$. But I can choose every scalar $a_i$ smaller and smaller hence I can obtain a sequence in $A$ converging to $0$. Therefore it shouldn't be closed in my opinion.
analysis convergence normed-spaces
A finite-dimensional normed space X is complete. Every complete space is also closed. If I take linearly independent $x_1,dots,x_n in X$, then the set $sum_i=1^n a_i x_i : a_i > 0=:A$ is finite-dimensional. Hence this set should be closed. But obviously $0 notin A$. But I can choose every scalar $a_i$ smaller and smaller hence I can obtain a sequence in $A$ converging to $0$. Therefore it shouldn't be closed in my opinion.
analysis convergence normed-spaces
analysis convergence normed-spaces
asked Sep 8 at 13:21
Diamir
165111
165111
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$A$ is just a cone, not a vector space.
â Alan Muniz
Sep 8 at 13:24
add a comment |Â
1
$A$ is just a cone, not a vector space.
â Alan Muniz
Sep 8 at 13:24
1
1
$A$ is just a cone, not a vector space.
â Alan Muniz
Sep 8 at 13:24
$A$ is just a cone, not a vector space.
â Alan Muniz
Sep 8 at 13:24
add a comment |Â
1 Answer
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Your set $A$ is not a subspace. Actually, take $n=1$, $X=mathbb R$, $x_1=1$, and now you are saying that $(0,infty)$ should be closed. Why would that be?
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your set $A$ is not a subspace. Actually, take $n=1$, $X=mathbb R$, $x_1=1$, and now you are saying that $(0,infty)$ should be closed. Why would that be?
add a comment |Â
up vote
2
down vote
Your set $A$ is not a subspace. Actually, take $n=1$, $X=mathbb R$, $x_1=1$, and now you are saying that $(0,infty)$ should be closed. Why would that be?
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your set $A$ is not a subspace. Actually, take $n=1$, $X=mathbb R$, $x_1=1$, and now you are saying that $(0,infty)$ should be closed. Why would that be?
Your set $A$ is not a subspace. Actually, take $n=1$, $X=mathbb R$, $x_1=1$, and now you are saying that $(0,infty)$ should be closed. Why would that be?
answered Sep 8 at 13:26
Martin Argerami
117k1071165
117k1071165
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1
$A$ is just a cone, not a vector space.
â Alan Muniz
Sep 8 at 13:24