Absolute value of an ordered pair

Question

On the following answer the user takes $\mathbb{R}^2$ as a metric space with the metric defined as: $$\delta(A,B)=\begin{cases} |A|+|B|, & \text{if $A\neq \lambda B$} \\ |A-B|, & \text{if $A= \lambda B$ for some $\lambda$} \end{cases}$$

I am confused with this definition. If we are in $\mathbb{R}^2$ then $A$ and $B$ are ordered pairs. What does $|A| + |B|$ mean in this case? Metric should return a real number and this is summarizing ordered pairs? Also he mentions that this second case of the function $|A - B|$ is the usual Euclidean distance. It would be in $\mathbb{R}^1$, but not in $\mathbb{R}^2$.

Can somebody please clear this up?

This looks like something that is often called the French railway metric. The idea is that Paris is at the origin, and to go between any two other points in the plane, you need to go via Paris (unless those two points are colinear with the origin). Does that help? If $A \in \mathbb{R}^2$ is a point $A = (a_1, a_2)$, then $|A|$ is the usual Euclidean norm $|A| = \sqrt{a_1^2 + a_2^2}$. — Joppy, Mar 07 '20 at 06:45

score 1 · Accepted Answer · answered Mar 07 '20 at 12:57

1

This is indeed the so-called French railroad metric, mentioned in the comment. For a point (or vector) $A=(a_1,a_2)$ in the plane this author denotes by $|A|$ its usual "norm" $\sqrt{a_1^2 + a_2^2}$. We imagine the points as being "stations" on a railroad system where the "railroad lines" are the lines through $O$, the origin (Paris, as the name suggests),so all lines go there. If two stations lie on the same railroad line, i.e. $B = \lambda A$ for some real $\lambda$, then to travel from $A$ to $B$ (or vice versa), we can just take their usual distance $|B-A|$ (as the straight line is then the shortest route). If this is not the case, you first go to Paris ($O$), which is Euclidean distance $|A|$ along the railroad and then switch in Paris to the line to $B$, and to get there then is $|B|$ as well, so in total your travel takes $|A|+|B|$. If you'd pay per distance of track you travelled, $\delta(A,B)$ would be a measure of the time or cost the route from $A$ to $B$ would take.

So it sort of makes sense if you look at it this way.

So only balls around $O$ look like the normal Euclidean ones, for other points they can be small intervals along a straight line (to the origin), e.g.

answered Mar 07 '20 at 12:57

Henno Brandsma

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How was I supposed to know that those vertical lines represent a norm and not absolute value? – Michael Munta Mar 08 '20 at 10:02
@MichaelMunta it does occur. It’s experience – Henno Brandsma Mar 08 '20 at 10:56
What about case $A \neq \lambda B$? Cant we always find a real scalar that will make the result equal to $A$? – Michael Munta Mar 08 '20 at 12:53
@MichaelMunta no, because it says that $A$ and $B$ are linearly dependent; i.e. lie on the same line through $O$. E.g. for $A=(1,2)$ and $B=(1,3)$ you cannot find such a $\lambda$. – Henno Brandsma Mar 08 '20 at 12:56
Can you please work it out? How to scale points? – Michael Munta Mar 08 '20 at 13:13
@MichaelMunta it's just a vector! $\lambda A = (\lambda, 2\lambda)$ in the previous example. Standard scalar multiplication of a real with a vector in $\Bbb R^2$. – Henno Brandsma Mar 08 '20 at 13:15

Absolute value of an ordered pair

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