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

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

Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
 * $d \left({x, y}\right) := \left\vert{x - y}\right\vert + \displaystyle \sum_{i \mathop = 1}^\infty 2^\left({-i}\right) \inf \left({1, \left\vert{ \max_{j \mathop \le i} \frac 1 {\left\vert{x - r_j}\right\vert} - \max_{j \mathop \le i} \frac 1 {\left\vert{y - r_j}\right\vert} }\right\vert }\right)$

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

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

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

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

Then:
 * $B_\epsilon \left({p; d}\right) \subseteq B'_\epsilon \left({p; d'}\right)$

Proof
Let $p \in \R$.

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

Let $x \in B_\epsilon \left({p; d}\right)$.

Then:

Thus:
 * $x \in B_\epsilon \left({p; d}\right) \implies x \in B_\epsilon \left({p; d'}\right)$

and the result follows by.