Open Ball is Subset of Open Ball

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

$\blacksquare$


Sources