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:

$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$

$\blacksquare$