Closed Ball contains Smaller Closed Ball
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{> 0}$ such that $\epsilon \le \delta$.
Let $\map {B^-_\epsilon} a$ be the closed $\epsilon$-ball on $a$.
Let $\map {B^-_\delta} a$ be the closed $\delta$-ball on $a$.
Then:
- $\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$
Proof
\(\ds x \in \map {B^-_\epsilon} a\) | \(\leadsto\) | \(\ds \map d {x, a} \le \epsilon\) | Definition of Closed Ball | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map d {x, a} \le \delta\) | As $\epsilon \le \delta$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds x \in \map {B^-_\delta} a\) | Definition of Closed Ball |
By definition of subset:
- $\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$
$\blacksquare$