Open Ball in Euclidean Plus Metric is Subset of Equivalent Ball in Euclidean Metric
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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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $5$