What is a motivation or geometric intuition of constructing a metric on a product space as an infinite sum? In this notes, the author stated the following lemma. 

Lemma $3.1$ Let $X_0,X_1,...$ be a countable sequence of Polish spaces. Then the product space $\prod X_n$ is also a Polish space. 

For all $n\in\mathbb{N},$ let $d_n$ be a complete metric on $X_n$ and assume that $d_n(x,y)<1$ for all $x,y\in X_n.$
To show that the product space $\prod X_n$ is completely metrizable, the author defined the following metric on $\prod X_n$ given by 
$$\hat{d}(f,g) =\sum_{n=0}^\infty \frac{1}{2^{n+1}} d_n(f(n),g(n))$$
where $f,g\in\prod X_n.$

Question: What is a motivation or geometric intuition of the metric $\hat{d}?$ 
  Also, how do we know that the metric $\hat{d}$ is a complete metric on $\prod X_n$ intuitively?

When I tried to prove the lemma myself, I did not think of the metric $\hat{d}.$
Of course, once I know that I have to consider $\hat{d},$ what remains to do is verification.
 A: This is not a  geometrically motivated construction; e.g., if you tried to use this metric on a product like $\mathbb{R}^3$, you'd find it does not have  geometrically appealing features. The idea is algebraic, we use addition of "pseudometrics" to separate the points of the product space. 
Each individual metric $d_n$ can be considered as  a pseudometric on the product space, via
$$
\rho_n(f, g) = d_n(f(n), g(n)) 
$$
A pseudometric is different from a metric in that it is allowed to be zero when $f, g$ are not equal. So now we have a bunch of pseudometrics $\{\rho_n\}$ and want to create a metric $\rho$ from them. The issue is how to make sure that $\rho(f,g)>0$ when $f, g$ are different. 
If $f\ne g$ then there exists some index $n$ such that $f(n)\ne g(n)$. So we have a pseudometric $\rho_n$ such that $\rho_n(f, g)>0$. Question is, what formula should we use to create $\rho$ from $\rho_n$, so that


*

*If at least one $\rho_n$ is positive, then $\rho$ is positive

*Triangle inequality holds for $\rho$, as it holds for $\rho_n$.

*If each $d_n$ is complete, $\rho$ is complete.

*The metric $\rho$ induces the product topology


For example, $\rho = \sup_n \rho_n$ works for (1-3) but fails (4); the topology it induces (uniform convergence) is stronger than the product topology (componentwise convergence). 
We could use a weighted supremum: $\rho = \sup_n c_n\rho_n$ where $c_n>0$ and $c_n\to 0$ (suggested by Daniel Fischer). Then all (1-4) hold. The fact that $c_n\to 0$ makes $\rho$ small when sufficiently many of the components $\rho_n$ are small, not necessarily all of them.
Or, we could use a sum instead of supremum. $\sum_n \rho_n$ doesn't work since it need not converge. So pick a convergent series $\sum c_n$ with positive terms, and let $\rho = \sum_n c_n \rho_n$. Again, properties (1) and (2) are easy to check. And now (3) is easier to prove. Indeed, a sequence $f_k$ being Cauchy with respect to $\rho$ implies that for every $n$, the  $n$-th component sequence $\{f_k(n)\}$ is Cauchy with respect to the corresponding $d_n$, and thus converges. So we have a coordinate-wise limit $f_k\to f$. Unlike the construction with supremum, coordinate-wise convergence implies convergence in the metric $\rho = \sum_n c_n \rho_n$. Indeed, for any $\epsilon>0$ there is $N$ such that $\sum_{n>N} c_n<\epsilon/2$, so the tail  of the sequence is negligible. And for large enough $k$ we will have $\rho_n(f_k, f)<\epsilon/(2N)$, $n=1,\dots, N$. Combine the two to get $\rho(f_k, f)<\epsilon$. 
Observations


*

*Most of this is not really about products; this is just a construction of a metric from countably many bounded pseudometrics. (Except for property 4 which is indeed specific to products.)

*There is nothing special about $1/2^{n}$, except it happens to be the world's favorite convergent series.  Any convergent series with positive terms could be used.
