# Closure of Convex Subset in Normed Vector Space is Convex

## Theorem

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Let $C \subseteq X$ be a convex subset of $X$.

Let $C^-$ be the closure of $C$.

Then $C^- \subseteq X$ is also a convex subset of $X$.

## Proof

Let $x, y \in C^-$.

Suppose $x, y$ are limit points.

Then there are sequences $\sequence {x_n}_{n \mathop \in \N}, \sequence {x_n}_{n \mathop \in \N}$ in $C$, such that:

$\ds \lim_{n \mathop \to \infty} x_n = x$
$\ds \lim_{n \mathop \to \infty} x_y = y$

Let $\alpha \in \closedint 0 1$.

Then:

 $\ds \paren {1 - \alpha} x + \alpha y$ $=$ $\ds \paren {1 - \alpha} \lim_{n \mathop \to \infty} x_n + \alpha \lim_{n \mathop \to \infty} y_n$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \paren {\paren {1 - \alpha} x_n + \alpha y_n}$

Since $C$ is convex:

$\ds \forall n \in \N : \paren {1 - \alpha} x_n + \alpha y_n \in C \subseteq C^-$

$C^-$ is closed and contains its limit points.

Hence:

$\paren {1 - \alpha} x + \alpha y \in C^-$

$\blacksquare$