# Geometric explanation of the product metric

Can someone describe to me the geometric intuition behind using a mapping $$((x_1,y_1),(x_2,y_2)) \mapsto \frac{d_1(x_1,y_1)}{1+d_1(x_1,y_1)} + \frac{1}{2} \frac{d_2(x_2,y_2)}{1+d_2(x_2,y_2)}$$ to define a metric on the product of the metric spaces $(X,d_1),(Y,d_2)$ ?

Of course I can check the axioms, but that doesn't give me any insight; so why was it defined like this and not differently (especially, why the $\frac{1}{2}$)? Why not use $((x_1,y_1),(x_2,y_2))$ $\mapsto d_1(x_1,y_1)$ $+ d_2(x_2,y_2)$ ?

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Why not indeed? What is the source of this mapping? – Harald Hanche-Olsen Feb 24 '12 at 14:35
Where did you encounter this? I don't think it's a standard definition. – Brad Feb 24 '12 at 14:37
It could be viewed as an artificially constructed metric $d^*$ which, firstly, makes $(X \times Y,d^*)$ into a bounded metric space (any two points having distance $d^*<\frac32$) and, secondly, makes $Y$ appear smaller by deflating its metric $d_2$. Might be a good counterexample for something, at least... – bgins Feb 24 '12 at 14:49

Jim's answer inspired a thought: Perhaps the author of the suggestion is building up to a metric for a countably infinite product: $$d(x,y)=\sum_{n=0}^\infty 2^{-n}\frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}.$$ In this case, the normalization and the factors $2^{-n}$ are required to make the series converge.
Given a metric $d$, the metric $d(x,y)/(1+d(x,y))$ is equivalent, but bounded. In a number of contexts, it is useful to have a bounded metric. A simpler way of creating a bounded metric is by defining it to be $\min\{d(x,y),1\}$. As for the $1/2$ factor on the second summand, it's not clear what purpose that serves. In any event, it is true that the metric you are asking about generates the same topology as $d_1(x_1,y_1)+d_2(x_2,y_2)$.