# Closed Ball is Closed/Normed Vector Space

## Theorem

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

Let $x \in X$.

Let $\epsilon \in \R_{> 0}$.

Let $\map {B_\epsilon^-} x$ be the closed $\epsilon$-ball of $x$ in $M$.

Then $\map {B_\epsilon^-} x$ is a closed set of $M$.

## Proof

We show that the complement $X \setminus \map {B_\epsilon^-} x$ is open in $M$.

Let $y \in X \setminus \map {B_\epsilon^-} x$.

Then by definition of closed ball:

$\norm {x - y} > \epsilon$

Put:

$\delta := \norm {x - y} - \epsilon > 0$

Then:

$\norm {x - y} - \delta = \epsilon$

Let $z \in \map {B_\delta} y$.

Then:

 $\ds \norm {x - z}$ $\ge$ $\ds \norm {x - y} - \norm {y - z}$ Reverse Triangle Inequality $\ds$ $>$ $\ds \norm {x - y} - \delta$ $\ds$ $=$ $\ds \epsilon$

and so:

$z \notin \map {B_\epsilon^-} x$

Then:

$\map {B_\delta} y \subseteq X \setminus \map {B_\epsilon^-} x$

so $X \setminus \map {B_\epsilon^-} x$ is open in $M$.

Hence, by definition of closed set:

$\map {B_\epsilon^-} x$ is closed in $M$.

$\blacksquare$