# This convex hull is balanced?

Let be (X,S) a locally convex space, and $$B \subset X$$ a nonempty sequentially closed bounded and convex set such that $$\hat{0} \notin clB$$,(the closure of B). Define the set T:=s-clco$$\{B \cup-B \}$$ where s-clco denotes the sequential closure of the convex hull. This appeared in a paper that I was reading, and then the author assume that T is a balanced set. Why it is true? The correct definition of the convex hull of B is $$\left\{ \displaystyle\sum_{i = 1}^n t_i b_i: \sum_{i=1}^n t_i = 1, t_i \geq 0, b_i \in B \right\}$$ right?

Assuming that the scalar field is $$\mathbb R$$ note first that $$0=\frac {b+(-b)} 2 \in T$$. Hence, for any $$t \in T$$ and any $$c \in [0,1]$$, we have $$ct=ct+(1-c)0 \in T$$. Also $$T$$ is symmetric. Hence $$ct \in T$$ whenever $$t \in T$$ and $$|c| \leq 1$$.
• In general It is not true when the field is other than $\mathbb{R}$ right? – The Student Mar 7 at 4:48
• @TheStudent Yes, for the complex field we cannot say that $T$ is balanced. – Kavi Rama Murthy Mar 7 at 5:15