Open Ball in Euclidean Plus Metric is Subset of Equivalent Ball in Euclidean Metric

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\R$ be the set of real numbers.

Let $d: \R \times \R \to \R$ be the Euclidean plus metric:

$\map d {x, y} := \size {x - y} + \ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$

Let $d': \R \times \R \to \R$ be the Euclidean metric.


Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.

Let $\map {B_\epsilon} {p; d}$ be an open $\epsilon$-ball of $p$ in $\R$ on $d$.

Let $\map {B'_\epsilon} {p; d'}$ be an open $\epsilon$-ball of $p$ in $\R$ on $d'$.


Then:

$\map {B_\epsilon} {p; d} \subseteq \map {B'_\epsilon} {p; d'}$


Proof

Let $p \in \R$.

Let $\epsilon \in \R_{>0}$.

Let $x \in \map {B_\epsilon} {p; d}$.


Then:

\(\ds \map d {x, p}\) \(<\) \(\ds \epsilon\)
\(\ds \leadsto \ \ \) \(\ds \epsilon\) \(>\) \(\ds \size {x - y} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }\)
\(\ds \) \(\ge\) \(\ds \size {x - y}\)
\(\ds \) \(=\) \(\ds \map {d'} {x, p}\)

Thus:

$x \in \map {B_\epsilon} {p; d} \implies x \in \map {B_\epsilon} {p; d'}$

and the result follows by Definition of Subset.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Open_Ball_in_Euclidean_Plus_Metric_is_Subset_of_Equivalent_Ball_in_Euclidean_Metric&oldid=539973"