# Is there a term for an $L^1$-combination of discrete metrics?

Let X be some set and $X^r$ be the space of $r$-tuples of elements of $X$. The discrete metric on $X$ is $d(x,y) = 1$ if $x \neq y$ and $d(x,y) = 0$ if $x=y$. Now, suppose I combine discrete metrics on all of the $r$ dimensions of my tuple, using $L^1$, i.e. $d(\vec{x}, \vec{y}) = \sum_{i=1}^{d} d(x_i,y_i)$, i.e. the distance between two tuples is the number of coordinates on which they differ. This is a metric on $X^r$.

My question: Is there a commonly-used name or term for this metric? I'm sure there must be and it's just on the tip of my tongue.

Note: If $d$ had been the $L^1$ norm (i.e. $d(x,y) = \left|x-y\right|$) then the tuple metric would have been the Taxicab Metric, a.k.a. the Manhattan Distance.

• the $1$-product metric en.wikipedia.org/wiki/Product_metric ? If you meant $\sum_i d(x_i , y_i)$. – Calvin Khor Feb 12 '18 at 14:40
• @CalvinKhor: I did, but the 1-product metric is relative to any basic metric. So "the 1-product of discrete metrics"? – einpoklum Feb 12 '18 at 15:21
• Perhaps you are thinking of the "taxicab metric"? I don't think that phrase is used much outside of the contex of the Euclidean plane, but I see no reason that one could not talk about the "taxicab product" of two metric spaces (although in that context there's nothing special about discrete metrics). – Lee Mosher Feb 12 '18 at 15:29
• Sure. The wikipedia page defines it for a product of metric spaces $(X_i,d_i)$ for which $d_i$ is already chosen beforehand, so no ambiguity should arise – Calvin Khor Feb 12 '18 at 15:30
• @CalvinKhor: I was hoping for more of a "non-compound term", like what LeeMosher suggests. But thanks. – einpoklum Feb 12 '18 at 15:34

## 1 Answer

This is called the $\ell_0$ distance (or $L_0$ distance), e.g., second page of these notes or this blog post.

The idea is that "the $\ell_0$ norm" of a vector is the size of its support, i.e., the number of nonzero coordinates it has. Accordingly, $\ell_0$ metric is the number of coordinates in which two points differ.

There is a terminological conflict because "$L_0$ norm" is often understood as the limit of "$L^p$ norms" as $p\to 0$, which isn't the same. So one needs to clarify what the term means.

• For finite dimension, why are these not the same? – einpoklum Feb 12 '18 at 19:22
• Maybe because people want consistent definition for finite and infinite case, and so they make modifications that affect the finite case as well... like weight $2^{-n}$ for $n$th variable. – user357151 Feb 12 '18 at 19:43