# 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.

• If you had such a metric, then couldn't you use induction and triangle inequality to show that $\rho(0,(1,1,1,\dotsc))$ is greater than $n$ for every $n$? Commented Dec 30, 2017 at 15:11
• @ziggurism I did try precisely this, but I am stuck finding the right chain of triangle inequalities doing the trick. Commented Dec 30, 2017 at 15:22
• @ziggurism Also, if one restricts attention to the space of bounded real sequences $\ell^{\infty}$, then the sup metric clearly satisfies the condition indicated in my conjecture and the distance between $(0,0,\ldots)$ and $(1,1,\ldots)$ is finite. This observation makes me wonder whether the approach you suggested works. Commented Dec 30, 2017 at 15:47
• yes, I tried to work it out using my naive suggestion, and did not get anywhere. Sorry for the red herring. Commented Dec 30, 2017 at 16:08

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, $$\rho(x, y) = \|(y_j - x_j) e_j\| = |y_j - x_j|$$

• I guess the norm you have defined is not compatible with the topology of pointwise convergence (which I believe is non-metrizable)? Commented Dec 30, 2017 at 16:19
• @ziggurism The topology of pointwise convergence (aka the product topology) is metrisable. The metric constructed in this answer defines a different topology. Commented Dec 30, 2017 at 16:21
• @DanielFischer Right. I guess I am thinking of pointwise convergence on $\mathbb{R}^\mathbb{R}$ Commented Dec 30, 2017 at 16:23
• @ziggurism Indeed, the sequence $(e_n)_{n=1}^\infty$ converges to $0$ in the product topology, but does not converge with respect to $\rho$ because $\rho(e_m, e_n) = \sqrt{2}$ for all $m \ne n$. Commented Dec 30, 2017 at 16:53
• Wow, very nice answer. Thank you! Commented Dec 30, 2017 at 17:44