TITLE: Is closure of convex subset of $X$ is again a convex subset of $X$?
QUESTION [21 upvotes]: Today I was going through my text book exercise and got hold of the following question.

Let $X$ be a normed linear space with norm $\lVert\cdot\rVert$ and $A$ is a non empty convex subset of $X$ then prove that $\operatorname{Closure}(A)$ is also a convex subset of $X$.
  I gave it a try but could not succeed.

REPLY [23 votes]: In fact, the result is true in any topological vector space $X$. 
Let $C$ be a nonempty convex subset of $X$. For $x,y \in \overline{C}$ and $\lambda \in [0,1]$, we prove that any neighborhood of $z= \lambda x+ (1-\lambda)y$ intersects $C$. So let $W$ be a neighborhood of $0$. Because $(u,v) \mapsto \lambda u+(1-\lambda)v$ is continuous, there exist open subsets $U$ and $V$ such that $\lambda U+(1-\lambda)V \subset W$; $x+U$ (resp. $y+V$) is an open neighborhood of $x$ (resp. of $y$) so there exists $x_1 \in (x+U) \cap C$ (resp. $y_1 \in (y+V) \cap C$). Therefore, $z_1= \lambda x_1+(1-\lambda)y_1 \in C$ because $C$ is convex and $z_1 \in \lambda (x+U)+(1-\lambda)(y+V) \subset z+W$, ie. $(z+W) \cap C \neq \emptyset$.