No metric on $\mathbb R^{\infty}$ whose projections are the absolute-value metric Let $\mathbb R^{\infty}$ be the space of all real sequences. I have the following conjecture: There is no metric on this space such that if two sequences differ in only one coordinate, then the distance between them is the absolute value of the difference between the respective coordinate values. More formally:

There exists no metric $\rho:\mathbb R^{\infty}\times\mathbb R^{\infty}\to[0,\infty)$ with the following property: if $\mathbf x,\mathbf y\in\mathbb R^{\infty}$ are such that there exists an $i\in\mathbb N$ such that $x_j=y_j$ for every $j\in\mathbb N\setminus\{i\}$, then $$\rho(\mathbf x,\mathbf y)=|x_i-y_i|.$$

For example, such a metric would yield $\rho((1,2,3,4,5,6,7,\ldots),(1,2,12,4,5,6,7,\ldots))=9$.
I tried showing that such a candidate metric would necessarily violate the triangle inequality, or it would yield infinite values, but to no avail. Any hints would be appreciated.
 A: Such a metric $\rho$ does exist.
Let $(e_n)_{n=1}^\infty$ be the canonical vectors in $\mathbb{R}^\infty$. Extend $(e_n)_{n=1}^\infty$ to a Hamel basis $B$ for $\mathbb{R}^\infty$ and define an inner product as:
$$\left\langle\sum_{b \in B} \alpha_b b, \sum_{b \in B} \beta_b b\right\rangle = \sum_{b \in B} \alpha_b\beta_b$$
The sums are finite because every vector in $\mathbb{R}^\infty$ is a finite linear combination of vectors in $B$. The inner product defines a norm $\|x\| = \sqrt{\langle x, x \rangle}$, and the norm defines a metric $\rho(x,y) = \|x-y\|$.
Indeed, $\rho$ satisfies the desired property. Let $x, y \in \mathbb{R}^\infty$ be two sequences such that $x_i = y_i$ for all $i \in \mathbb{N} \setminus \{j\}$.
If $$x = \sum_{b \in B\setminus\{e_j\}} \alpha_b b + \alpha e_j$$
then 
$$y = x + (y_j - x_j)e_j = \sum_{b \in B\setminus\{e_j\}} \alpha_b b + (\alpha  + y_j - x_j) e_j$$
necessarily, because the representation as a linear combination of basis elements is unique.
Therefore, $$\rho(x, y) = \|(y_j - x_j) e_j\| = |y_j - x_j|$$
