Is closure of convex subset of $X$ is again a convex subset of $X$?

Solution 1:

Pick two points in the closure of the set. Say $x,y \in \bar{X}$. So there exist some sequences $\{x_i\}$ and $\{y_i\}$ in $X$ which tend to $x$ and $y$ respectively.

But for all $i$ and $0 < t < 1$ we have that: $$ tx_{i} + (1-t)y_{i} \in X$$

Now taking limits gives the correct result.

Solution 2:

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. $(z+W) \cap C \neq \emptyset$.