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

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

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle \map d {x, a} \le \delta\) | As $\epsilon \le \delta$ | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle x \in \map {B^-_\delta} a\) | Definition of closed ball |

By definition of subset:

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

$\blacksquare$