# Open Ball is Subset of Open Ball

## Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Let $x, y$ be points of $A$.

Then:

$\epsilon-\delta \geq d\left({x, y}\right) \implies B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$

where $B_\epsilon \left({x}\right)$ denotes the open $\epsilon$-ball in $M = \left({A, d}\right)$.

## Proof

Let $\epsilon - \delta \geq d \left({x, y}\right)$.

Then $\epsilon \geq d \left({x, y}\right) + \delta$.

If $z \in B_\delta \left({y}\right)$, then $d \left({y, z}\right) < \delta$.

So:

 $\displaystyle d \left({x, z}\right)$ $\le$ $\displaystyle d \left({x, y}\right) + d \left({y, z}\right)$ metric space axioms $\displaystyle$ $<$ $\displaystyle d \left({x, y}\right) + \delta$ definition of open ball $\displaystyle$ $\le$ $\displaystyle \epsilon$

Thus $z \in B_\epsilon \left({x}\right)$.

So $B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$.

$\blacksquare$