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