Open Ball is Subset of Open Ball

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Then:

$\epsilon-\delta \ge \map d {x, y} \implies \map {B_\delta} y \subseteq \map {B_\epsilon} x$

where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball in $M = \struct {A, d}$.


Proof

Let $\epsilon - \delta \ge \map d {x, y}$.

Then $\epsilon \ge \map d {x, y} + \delta$.

If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$.

So:

\(\ds \map d {x, z}\) \(\le\) \(\ds \map d {x, y} + \map d {y, z}\) Metric Space Axiom $(\text M 2)$: Triangle Inequality
\(\ds \) \(<\) \(\ds \map d {x, y} + \delta\) Definition of Open Ball of Metric Space
\(\ds \) \(\le\) \(\ds \epsilon\)

Thus $z \in \map {B_\epsilon} x$.

So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.

$\blacksquare$


Sources