On the closure of the convex hull of a sequence in normed spaces











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Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)



However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.



Am I right?



Thanks for any comment/answer.






functional-analysis normed-spaces







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    up vote
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    down vote

    favorite












    Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)



    However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.



    Am I right?



    Thanks for any comment/answer.






    functional-analysis normed-spaces







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)



      However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.



      Am I right?



      Thanks for any comment/answer.






      functional-analysis normed-spaces







      share|cite|improve this question













      Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)



      However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.



      Am I right?



      Thanks for any comment/answer.






      functional-analysis normed-spaces




      functional-analysis normed-spaces






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      serenus

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          Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.






          share|cite|improve this answer





















          • Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
            – serenus
            5 hours ago













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          Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.






          share|cite|improve this answer





















          • Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
            – serenus
            5 hours ago

















          up vote
          0
          down vote













          Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.






          share|cite|improve this answer





















          • Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
            – serenus
            5 hours ago















          up vote
          0
          down vote










          up vote
          0
          down vote









          Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.






          share|cite|improve this answer












          Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          Kavi Rama Murthy

          39.9k31750




          39.9k31750












          • Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
            – serenus
            5 hours ago




















          • Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
            – serenus
            5 hours ago


















          Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
          – serenus
          5 hours ago






          Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
          – serenus
          5 hours ago




















           

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