Recall that $\set {r_j}_{j \mathop \in \N}$ is an enumeration of the rational numbers $\Q$. Also, we note that: {{begin-eqn}} {{eqn | o =        | r = \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 {x - r_j} } } }       | c =  }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^\infty 2^{-i} 1       | c = {{Defof|Infimum of Subset of Real Numbers}} }} {{eqn | r = 1       | c = Sum of Infinite Geometric Sequence: Corollary 1 with $z = \dfrac 1 2$ }} {{end-eqn}} meaning that this is a convergent series and so the definition is meaningful. Next the axioms for a metric are checked in turn. === Proof of $\text M 1$ === {{begin-eqn}} {{eqn | l = \map d {x, x}       | r = \size {x - x} + \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 {x - r_j} } } } }} {{eqn | r = 0 + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, 0} }} {{eqn | r = 0 }} {{end-eqn}} {{qed|lemma}} === Proof of $\text M 2$ === Let $i \in \N$ be fixed. Define: :$\map {f_i} x := \displaystyle \max_{j \mathop \le i} \frac 1 {\size {x - r_j} }$ Then: {{begin-eqn}} {{eqn | o =        | r = \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } + \inf \set {1, \size {\map {f_i} y - \map {f_i} z} }       | c =  }} {{eqn | r = \inf \set {1 + 1, 1 + \size {\map {f_i} y - \map {f_i} z}, \size {\map {f_i} x - \map {f_i} y} + 1, \size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z} }       | c = Sum of Infima is Infimum of Sums }} {{end-eqn}} Now since $\inf \set {1, \size {\map {f_i} x - \map {f_i} z} } \le 1$, it follows that: {{begin-eqn}} {{eqn | l = \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }       | o = \le       | r = 1 + 1 }} {{eqn | l = \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }       | o = \le       | r = 1 + \size {\map {f_i} y - \map {f_i} z} }} {{eqn | l = \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }       | o = \le       | r = \size {\map {f_i} x - \map {f_i} y} + 1 }} {{end-eqn}} Suppose now that $\size {\map {f_i} x - \map {f_i} z} \le 1$. Then: {{begin-eqn}} {{eqn | l = \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }       | r = \size {\map {f_i} x - \map {f_i} z} }} {{eqn | o = \le       | r = \size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z}       | c = Triangle Inequality for Real Numbers }} {{end-eqn}} On the other hand, if: :$\size {\map {f_i} x - \map {f_i} z} > 1$ then also: :$\size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z} > 1$ and: {{begin-eqn}} {{eqn | l = \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }       | r = 1 }} {{eqn | o = <       | r = \size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z} }} {{end-eqn}} Combining both cases with the estimates above, we conclude: :$\displaystyle \inf \set {1, \size {\map {f_i} x - \map {f_i} z} } \le \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } + \inf \set {1, \size {\map {f_i} y - \map {f_i} z} }$ Finally, now, we have: {{begin-eqn}} {{eqn | l = \map d {x, z}       | r = \size {x - z} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\map {f_i} x - \map {f_i} z} } }} {{eqn | o = \le       | r = \paren {\size {x - y} + \size {y - z} } + \sum_{i \mathop = 1}^\infty 2^{-i} \le \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } + \inf \set {1, \size {\map {f_i} y - \map {f_i} z} } }} {{eqn | r = \paren {\size {x - y} + \sum_{i \mathop = 1}^\infty 2^{-i} \le \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } } + \paren {\size {y - z} + \sum_{i \mathop = 1}^\infty 2^{-i} \le \inf \set {1, \size {\map {f_i} y - \map {f_i} z} } } }} {{eqn | r = \map d {x, y} + \map d {y, z} }} {{end-eqn}} {{qed|lemma}} === Proof of $\text M 3$ === We have that: {{begin-eqn}} {{eqn | l = \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} } } }       | r = \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {-\paren {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } } }       | c = Absolute Value of Negative }} {{eqn | r = \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {y - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {x - r_j} } } } }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = \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} } } } }} {{eqn | r = \size {y - x} + \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} } } }       | c = $\size {\, \cdot \,}$ is a metric }} {{eqn | r = \size {y - x} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {y - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {x - r_j} } } }       | c = from above }} {{eqn | r = \map d {y, x} }} {{end-eqn}} {{qed|lemma}} === Proof of $\text M 4$ === Suppose that $x \ne y$. Then: {{begin-eqn}} {{eqn | l = \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} } } }       | o = \ge       | r = \size {x - y} }} {{eqn | o = >       | r = 0       | c = $\size {\, \cdot \,}$ is a metric }} {{end-eqn}} {{qed}}