Closed Ball is Closed/Metric Space
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $x \in A$.
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 $A \setminus B_\epsilon^- \left({x}\right)$ is open in $M$.
Let $a \in A \setminus \map {B_\epsilon^-} x$.
Then by definition of closed ball:
- $\map d {x, a} > \epsilon$
Put:
- $\delta := \map d {x, a} - \epsilon > 0$
Then:
- $\map d {x, a} - \delta = \epsilon$
Let $b \in \map {B_\delta} a$.
Then:
\(\ds \map d {x, b}\) | \(\ge\) | \(\ds \map d {x, a} - \map d {a, b}\) | Reverse Triangle Inequality | |||||||||||
\(\ds \) | \(>\) | \(\ds \map d {x, a} - \delta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
and so:
- $b \notin \map {B_\epsilon^-} x$
Then:
- $\map {B_\delta} a \subseteq A \setminus \map {B_\epsilon^-} x$
so $A \setminus \map {B_\epsilon^-} x$ is open in $M$.
Hence, by definition of closed set:
- $\map {B_\epsilon^-} x$ is closed in $M$.
$\blacksquare$