Open Ball is Subset of Open Ball
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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
- Mizar article TOPGEN_5:22