What is the proof, that $d(x,y) = \sum_{n=0}^{\infty} \frac{1}{2^n}\frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$ is a metric? Where $x=(x_1,x_2,\cdots)$ , $y=(y_1,y_2,\cdots)$ and $d_n$ is a metric for $X_n$. How does one prove the triangle inequality? https://en.wikipedia.org/wiki/Metric_space#Product_metric_spaces
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1Try showing $f(x)=x/(1+x)$ is decreasing in $x$ for $x\geq 0$. – Alex R. Jul 30 '16 at 17:15
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I proved this once, but now have forgotten how the proof goes. – Jul 30 '16 at 17:15
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Basically like in this older post. – Jul 30 '16 at 17:16
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Ok, thanks. I will delete the question, since it is a duplicate! – Jul 30 '16 at 17:18
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As far as I'm concerned, you can leave it. Even more so that you have received an answer! – Jul 30 '16 at 17:19
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Apologies -- I didn't notice it was a duplicate before answering... – Clement C. Jul 30 '16 at 17:21
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You can do that component (summand) per component, relying on the fact that each $d_n$ does satisfy the triangle inequality.
In more detail: $$\begin{align} \frac{d_n(a,b)}{1+d_n(a,b)} + \frac{d_n(b,c)}{1+d_n(b,c)} &= \frac{d_n(a,b)+2d_n(a,b)d_n(b,c)+d_n(b,c)}{(1+d_n(a,b))(1+d_n(b,c))}\\ &= \frac{d_n(a,b)+2d_n(a,b)d_n(b,c)+d_n(b,c)}{1+d_n(a,b)+d_n(b,c)+d_n(a,b)d_n(b,c)} \\&= \frac{d_n(a,b)+d_n(b,c)+2u}{1+d_n(a,b)+d_n(b,c)+u} \\&\geq \frac{d_n(a,b)+d_n(b,c)+u}{1+d_n(a,b)+d_n(b,c)+u} \\&\geq \frac{d_n(a,b)+d_n(b,c)}{1+d_n(a,b)+d_n(b,c)} \end{align}$$ setting $u\stackrel{\rm def}{=}d_n(a,b)d_n(b,c)\geq 0$, and using the (easy) fact that if $\beta\geq \alpha > 0$ and $u\geq 0$, then $\frac{\alpha+u}{\beta+u} \geq \frac{\alpha}{\beta}$.
From there, we can rely on the monotonicity of the function $f\colon x\geq 0\mapsto \frac{x}{1+x}$: $$ \frac{d_n(a,b)+d_n(b,c)}{1+d_n(a,b)+d_n(b,c)} = f(d_n(a,b)+d_n(b,c)) \geq f(d_n(a,c)) = \frac{d_n(a,c)}{1+d_n(a,c)} $$ since by the triangle inequality for $d_n$, we know that $d_n(a,b)+d_n(b,c)\geq d_n(a,c)$.
Clement C.
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Since $d_i$ is metric, inequality $$d_i(x_i,z_i)\leq d_i(x_i,y_i)+d_i(y_i,z_i)$$ holds in i-th metric space from the product space.
Also, metrics $d(x,y)$ and $\frac{d(x,y)}{1+d(x,y)}$ are equivalent.
(If $d$ is a metric, then $d/(1+d)$ is also a metric).
Therefore, we have $$\frac{d_i(x_i,z_i)}{1+d_i(x_i,z_i)}\leq\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}+\frac{d_i(y_i,z_i)}{1+d_i(y_i,z_i)},$$
for every index $i$ and
$$\frac{1}{2^i}\frac{d_i(x_i,z_i)}{1+d_i(x_i,z_i)}\leq\frac{1}{2^i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}+\frac{1}{2^i}\frac{d_i(y_i,z_i)}{1+d_i(y_i,z_i)},$$
for every $i$. By summing, we get $$d(x,z)\leq d(x,y)+d(y,z).$$
