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

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?

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