Cfrgtkky

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.

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 ${x_n:n in mathbb N}$.