It is to be demonstrated that $d$ satisfies all the metric space axioms. Recall from the definition of the '''Fréchet space''' that the distance function $d: \R^\omega \times \R^\omega \to \R$ is defined on $\R^\omega$ as: :$\forall x, y \in \R^\omega: \map d {x, y} = \displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }$ where $x := \sequence {x_i}_{i \mathop \in \N} = \tuple {x_0, x_1, x_2, \ldots,}$ and $y := \sequence {y_i}_{i \mathop \in \N} = \tuple {y_0, y_1, y_2, \ldots,}$ denote arbitrary elements of $\R^\omega$. First it is confirmed that Fréchet Product Metric is Absolutely Convergent on arbitrary $x$ and $y$, as follows. First note that: {{begin-eqn}} {{eqn | l = \dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} }       | r = 1 - \dfrac 1 {1 + \size {x_i - y_i} }       | c =  }} {{end-eqn}} implies that we have: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} } < 1       | c =  }} {{end-eqn}} By the Ratio Test we have that the series is absolutely convergent. An important property of absolutely convergent series is that any rearrangement of the terms of the series is also convergent and equal to the original series. This will be used in the proof of axiom $M2$. === Proof of $M1$ === {{begin-eqn}} {{eqn | l = \map d {x, x}       | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} }       | c = Definition of $d$ }} {{eqn | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} \size 0} {1 + \size 0}       | c =  }} {{eqn | r = \sum_{i \mathop \in \N} \dfrac 0 1       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }       | c = Definition of $d$ }} {{eqn | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {y_i - x_i} } {1 + \size {y_i - x_i} }       | c = {{Defof|Absolute Value}} }} {{eqn | r = \map d {y, x}       | c = Definition of $d$ }} {{end-eqn}} So axiom $M3$ holds for $d$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | lll=\leadsto       | ll= \exists k \in \N:       | l = x_k       | o = \ne       | r = y_k       | c =  }} {{end-eqn}} Thus for such a $k \in \N$: {{begin-eqn}} {{eqn | l = x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | lll=\leadsto       | l = \size {x_k - y_k}       | o = >       | r = 0       | c = {{Defof|Absolute Value}} }} {{eqn | lll=\leadsto       | l = 2^{-i} \size {x_k - y_k}       | o = >       | r = 0       | c =  }} {{eqn | ll= \land       | l = 1 + \size {x_k - y_k}       | o = >       | r = 0       | c =  }} {{eqn | lll=\leadsto       | l = \dfrac {2^{-k} \size {x_k - x_k} } {1 + \size {x_k - x_k} }       | o = >       | r = 0       | c =  }} {{end-eqn}} It has been established during the course of demonstrating compliance with axiom $M1$ (and is in any case trivially obvious) that: :$x_j = y_j \implies \dfrac {2^{-j} \size {x_j - x_j} } {1 + \size {x_j - x_j} } = 0$ Thus $\displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} }$ contains at least one term: :$\dfrac {2^{-k} \size {x_k - x_k} } {1 + \size {x_k - x_k} } > 0$ and any number of other terms: :$\dfrac {2^{-j} \size {x_j - x_j} } {1 + \size {x_j - x_j} } = 0$ Hence: :$\displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} } > 0$ for $x \ne y$. So axiom $M4$ holds for $d$. {{qed|lemma}} === Proof of $M2$ === Consider the function $f (a) = a / (1+a)$, $a \geq 0$. We want to show it satisfies :$f (a) \leq f(a') \quad \text{ for } \; a \geq 0 \; \text{ and } \; a' \geq a$. This very first inequality follows since $f (a) = 1 - 1/ (1+a)$ with $1/(1+a)$ decreasing as $a$ increases. We also want to show it satisfies  :$f (a) \leq f(a+b) \leq f(a) + f(b) \quad \text{ for } \; a,b >0$. The first inequality here follows from the above inequality. For the second inequality here we note $f(a) / a = 1/(1+a)$ gives :$f (a)/a \geq f(a+b)/(a+b), \quad f(b)/b \geq f(a+b)/(a+b) \text{ for } \; a,b >0$ from which we obtain $f(a) + f(b) \geq f(a+b) (a+b) / (a+b) = f (a+b)$ (the inequality is immediately verified in cases where $a$ or $b$ is $0$). We now use these inequalities to prove axiom $M2$ {{begin-eqn}} {{eqn | l = \map d {x, z}       | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - z_i} } {1 + \size {x_i - z_i} }       | c = Definition of $d$ }} {{eqn | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {(x_i - y_i) + (y_i - z_i)} } {1 + \size {(x_i - y_i) + (y_i - z_i)} }       | c =  }} {{eqn | o = \le       | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} (\size {x_i - y_i} + \size {y_i - z_i} } {1 + \size {x_i - y_i} + \size {y_i - z_i} }       | c = as $\size {(x_i - y_i) + (y_i - z_i)} \le \size {x_i - y_i} + \size {y_i - z_i}$ and $f(a) \le f(a')$ for $a' \ge a$ }} {{eqn | o = \le       | r = \sum_{i \mathop \in \N} \left({\dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} } + \dfrac {2^{-i} \size {y_i - z_i} } {1 + \size {y_i - z_i} } }\right)       | c = as $f((x_i - y_i) + (y_i - z_i)) \leq f(x_i - y_i) + f(y_i - z_i)$ }} {{eqn | r = \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} } + \displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {y_i - z_i} } {1 + \size {y_i - z_i} }       | c = Fréchet Product Metric is Absolutely Convergent }} {{eqn | r = \map d {x, y} + \map d {y, z}       | c = Definition of $d$ }} {{end-eqn}} So axiom $M2$ holds for $d$. {{qed|lemma}} Thus $d$ satisfies all the metric space axioms and so is a metric. {{qed}} Category:Fréchet Product Metric eqa8lrwt5g67sae7u91p8f6wvkjnj5d