Open Ball contains Smaller Open Ball

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 open $\delta$-ball on $a$.

Then:

$\map {B_\epsilon} a \subseteq \map {B_\delta} a$

Proof

\(\ds x \in \map {B_\epsilon} a\)

\(\leadstoandfrom\)

\(\ds \map d {x, a} < \epsilon\)

Definition of Open Ball of Metric Space

\(\ds \)

\(\leadsto\)

\(\ds \map d {x, a} < \delta\)

As $\epsilon \le \delta$

\(\ds \)

\(\leadstoandfrom\)

\(\ds x \in \map {B_\delta} a\)

Definition of Open Ball of Metric Space

By definition of subset:

$\map {B_\epsilon} a \subseteq \map {B_\delta} a$

$\blacksquare$