Ngô Quốc Anh

Tháng Ba 4, 2009

Show that the closure of a convex set also is convex

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 8 (MA5206), Linh Tinh — Ngô Quốc Anh @ 23:29

Today on the way, my Chinese friend and I have discussed the following question: Let denote a convex subset of a locally convex topological linear space . Show that the closure of also is convex.

I have suggested the following solution.

We define f: X \times X \times \mathbb{R} \to X as following f: \left( {x,y,\lambda } \right) \mapsto \lambda x + \left( {1 - \lambda } \right)y. Then is continuous and f\left( {K \times K \times \left[ {0,1} \right]} \right) \subset K since is convex. We then obtain that f\left( {\overline K \times \overline K \times \left[ {0,1} \right]} \right) \subset \overline K due to the continuity of , that is \lambda \overline K + \left( {1 - \lambda } \right)\overline K \in \overline K for every . Therefore is a convex set.

So what I am going to tell you is how correct the solution is? If no, what’s the problem, otherwise, what’s the main point? I will show you a little bit latter. I think I should go for sleep, it’s late now :(

Blog at WordPress.com.