Open Ball contains Strictly Smaller Closed Ball

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $a \in A$.

Let $\epsilon, \delta \in \R_{>0}$ such that $\epsilon < \delta$.

Let $\map {B^-_\epsilon} a$ be the closed $\epsilon$-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

\(\ds x\)

\(\in\)

\(\ds \map {B^-_\epsilon} a\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds \map \cl {\map {B_\epsilon} a}\)

Definition of Closed Ball of Metric Space

\(\ds \leadsto \ \ \)

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

\(\le\)

\(\ds \epsilon\)

Closure of Open Ball in Metric Space

\(\ds \leadsto \ \ \)

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

\(<\)

\(\ds \delta\)

As $\epsilon < \delta$

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

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

Definition of Open Ball of Metric Space

By definition of subset:

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

$\blacksquare$