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.










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    $A$ is just a cone, not a vector space.
    – Alan Muniz
    Sep 8 at 13:24














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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.










share|cite|improve this question

















  • 1




    $A$ is just a cone, not a vector space.
    – Alan Muniz
    Sep 8 at 13:24












up vote
0
down vote

favorite









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.










share|cite|improve this question













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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asked Sep 8 at 13:21









Diamir

165111




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  • 1




    $A$ is just a cone, not a vector space.
    – Alan Muniz
    Sep 8 at 13:24












  • 1




    $A$ is just a cone, not a vector space.
    – Alan Muniz
    Sep 8 at 13:24







1




1




$A$ is just a cone, not a vector space.
– Alan Muniz
Sep 8 at 13:24




$A$ is just a cone, not a vector space.
– Alan Muniz
Sep 8 at 13:24










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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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    up vote
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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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      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?






      share|cite|improve this answer






















        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?






        share|cite|improve this answer












        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?







        share|cite|improve this answer












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        answered Sep 8 at 13:26









        Martin Argerami

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