Closed Ball contains Smaller Open 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 open $\epsilon$-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

 $\displaystyle x \in \map {B_\epsilon} a$ $\leadsto$ $\displaystyle \map d {x, a} < \epsilon$ Definition of Open Ball $\displaystyle$ $\leadsto$ $\displaystyle \map d {x, a} < \delta$ As $\epsilon \le \delta$ $\displaystyle$ $\leadsto$ $\displaystyle \map d {x, a} \le \delta$ Definition of Total Ordering $\displaystyle$ $\leadsto$ $\displaystyle x \in \map {B^-_\delta} a$ Definition of Closed Ball

By definition of subset:

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

$\blacksquare$