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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 $\{x_n: \ n\in \mathbb{N}\}\cup\{0\}$ in $E$ is equal to the closure of the convex hull of the set $\{x_n: \ n\in \mathbb{N}\}$ in $E$, that is, is it true that $\overline{co(\{x_n: \ n\in \mathbb{N}\}\cup\{0\})}=\overline{co(\{x_n: \ n\in \mathbb{N}\})}$ ? (Here we assume that for all $n\in \mathbb{N}$, $x_n\neq 0$.)

However, in may books, instead of writing $\overline{co(\{x_n: \ n\in \mathbb{N}\})}$, I see that it is written $\overline{co(\{x_n: \ n\in \mathbb{N}\}\cup\{0\})}$. 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 $A\subset E$, namely, $co(A)=\{\sum_{n=1}^N\lambda_nx_n: \ x_n\in A, \lambda_n\geq0, \sum_{n=1}^N\lambda_n=1, N\in\mathbb{N}\}$.

Am I right?

Thanks for any comment/answer.

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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 $\{x_n:n \in \mathbb N\}$.

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  • $\begingroup$ 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. $\endgroup$ – serenus Nov 19 '18 at 7:58

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