# Closed Ball is Closed/Normed Vector Space

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## 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 $M$.

$\blacksquare$

## Sources

- 2017: Amol Sasane:
*A Friendly Approach to Functional Analysis*... (previous) ... (next): $\S 1.3$: Normed and Banach spaces. Topology of normed spaces