Infinite product metrics and their projections onto the factor spaces

Question

In this thread, I formulated a conjecture about the non-existence of a metric $\rho$ on $\mathbb R^{\infty}$ in such a way that for any $(x_1,x_2,x_3,\ldots)\in\mathbb R^{\infty}$, the restriction of $\rho$ to subspaces of the form \begin{align*} &\;\mathbb R\times\{x_2\}\times\{x_3\}\times\cdots,\\ &\;\{x_1\}\times\mathbb R\times\{x_3\}\times\cdots,\\ &\;\{x_1\}\times\{x_2\}\times\mathbb R\times\cdots, \end{align*} and so forth, essentially coincides with the absolute-value metric.

In that thread, @mechanodroid refuted this conjecture, ingenuously demonstrating that such a metric does exist on $\mathbb R^{\infty}$.

In the current thread, I am asking input toward establishing that the existence result in that other thread is more like the exception than the norm—that is, non-existence persists in general infinite product spaces.

I am looking for a counterexample such that

• for each $n\in\mathbb N$, $(X_n,d_n)$ is a (non-empty) metric space;
• there exists no metric $\rho$ on the product space $X\equiv \prod_{n\in\mathbb N} X_n$ with the following property: if $x,y\in X$ is such that there exists an $i\in\mathbb N$ such that $x_j=y_j$ for all $j\in\mathbb N\setminus\{i\}$, then $\rho(x,y)=d_i(x_i,y_i)$.

Intuitively, this means that general infinite product spaces may not be metrized in such a manner that the “marginal metric” coincides with the respective original metric on each factor space.

Many thanks in advance for any kind of contribution.

Answer 1

Such a metric always exists when the metrics are uniformly bounded:

$$\rho(x,y)=\sup_{i\in\mathbb{N}} d_i(x_i,y_i).$$

Addendum: It is generally not possible to metrize the topology of pointwise convergence this way. Let $X=\{0,1\}^\infty$ with each factor having the metric satisfying $d_i(0,1)=1$and let $x_n$ be the sequence with entry $1$ in the $n$-th place and zeros everywhere else. Let $x$ be the sequence consisting of zeros only. Then $\langle x_n\rangle$ converges pointwise to $x$ but $d(x,x_n)=d_n(0,1)=1$ for all $n$.