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 $\epsilon, \delta \in \R_{\gt 0}$ 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

\(\displaystyle x \in \map {B_\epsilon} a\) \(\leadstoandfrom\) \(\displaystyle \map d {x, a} < \epsilon\) 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
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle x \in \map {B^-_\delta} a\) Definition of closed ball

By definition of subset:

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

$\blacksquare$