# Having difficulties showing the triangle inequality of metric in the plane

Let $P \in \mathbb{R}^2$ and define

$$d(x,y) = \left\{ \begin{array}{lr} ||x-y|| & \text{if} \; \; x,y,P \; \; \text{are collinear,}\\ || x - P|| + ||y-P||& \;\;\;\; \text{otherwise.} \end{array} \right.$$

How can I show this metric satisfies the triangle inequality? I am trying to visualize how the open balls look but I cannot see to find out.

• My first instinct would be to break it into 4 cases, with your x,y,z, and P....case 1 no pair is colinear with P, case 2 both pairs are (those should be trivial), case 3 and 4 one pair is colinear, then attack with triangle inequalities on the underlying norms
– Alan
Oct 20, 2014 at 6:15
• Alan's comment is right - although cases 3 and 4 are the same, by symmetry - althoughalthough the proof will need to consider which of $x$ and $y$ is closer to $P$. Oct 20, 2014 at 6:36
• By the way, since you asked about the open balls: if $r\le\|x-P\|$, then the open ball of radius $r$ around $x$ is just an open line segment of length $2r$ whose midpoint is $x$ and that lies on the line joining $x$ to $P$. If $r>\|x-P\|$, then the open ball of radius $r$ around $x$ includes the aforementioned open line segment, but also includes the open Eucliden ball of radius $r-\|x-P\|$ around $P$. See if you believe those statements - that should help get a handle on this metric. Oct 20, 2014 at 6:38
• @achillehui this is wrong, since eg. $x,y,P$ are not on the same line or in other words: $x,y$ are not on a line through the origin. The LHS is therefore $d(x,P)+d(P,y) +d(y,P)+d(P,z)$ which is $\geq$ RHS Oct 22, 2014 at 6:10

In German it is called the french railway metric . The reason for this is that in the earlier France, the distance to get from one city to the other by train was the same as getting to Paris and then to the target city (the railway system being centralized in Paris). This explanation also yields the intuition for your metric. (Also called the British Rail metric by substituting Paris with London)

The proof is as easy as it gets: On every line through $P$ it is clearly a distance. Then go through each case whether you have two points on a line through $P$ and one not or three points which are not on any common line through $P$. Keep in mind the intuition from above and you will get there (you essentially do nothing but saying: to go from one city over Paris to another city over Paris to a third city is longer than going from one city over Paris to another - this amounts to the triangle inequality) (so another hint to express this in mathematical language: $d(x,y) = d(x,P) + d(P,y)$ (i.e. go from $x$ to $x$ over $P$) and now use triangle inequalities for each summand i.e. triangle inequalities for the triples $x,P,z$ and $z,P,y$ and see what happens).

If you would like me to give a detailed solution, I will be happy to do that. But I think I gave you some good hints, which will lead you to the solution, and you will learn a lot from this, hence the shortness.

Cheers

• For the $\epsilon$-balls: Go to whatever city to like and now think: where can you get by train with covered distance less than $\epsilon$. Answer: you could get to the near stations on your line, and if you even get to $P$aris, then you can get to the whole $\delta$-ball around $P$aris where $\delta$ is the distance left from $\epsilon$, i.e. the $\epsilon -$ the distance to the city you started. Oct 22, 2014 at 6:18

(By request of the OP.)

If $x$ is a point different from $P$, and $\epsilon\le\|x-P\|$, the $\epsilon$-ball centred at $x$ is an open interval of length $2\epsilon$ centred at $x$ on the line $\overline{Px}$:

If $\epsilon>\|x-P\|$, the $\epsilon$-ball at $x$ consists of the segment $\overline{Px}$, the open ray of length $\epsilon$ extending from $x$ away from $P$ on the line $\overline{Px}$, and the ordinary open ball of radius $\epsilon-\|x-P\|$ centred at $P$:

In other words, in every case it contains all of the points that are less than $\epsilon$ away from $x$ provided that you can move only along lines passing through $P$. You must always be moving directly towards or directly away from $P$.

Finally, the $\epsilon$-ball at $P$ is just the ordinary Euclidean $\epsilon$-ball at $P$.

For the triangle inequality, let $x,y$, and $z$ be any three points; we wish to show that $$d(x,y)\le d(x,z)+d(z,y)\;.$$

Consider first the case in which $x,y$, and $P$ are collinear. If $z$ is also on this line, $d$ reduces to the usual Euclidean metric, so assume that this is not the case. Then $d(x,z)=\|x-P\|+\|P-z\|$, $d(z,y)=\|z-P\|+\|P-y\|$, and $d(x,y)\le\|x-P\|+\|P-y\|$ by the triangle inequality for the Euclidean metric, so

\begin{align*} d(x,y)&\le\|x-P\|+\|P-y\|\\ &\le\big(\|x-P\|+\|P-z\|\big)+\big(\|z-P\|+\|P-y\|\big)\\ &=d(x,z)+d(z,y)\;. \end{align*}

Now assume that $x,y$, and $P$ are not collinear, so that $d(x,y)=\|x-P\|+\|P-y\|$. If $z$ is not on either of the lines $\overline{Px}$ and $\overline{Py}$, then

\begin{align*} d(x,y)&=\|x-P\|+\|P-y\|\\ &<\big(\|x-P\|+\|P-z\|\big)+\big(\|z-P\|+\|P-y\|\big)\\ &=d(x,z)+d(z,y)\;. \end{align*}

If $z$ is on one of those lines, we may assume without loss of generality that it’s on $\overline{Px}$. Then $\|x-P\|\le\|x-z\|+\|z-P\|$ by the triangle inequality for the Euclidean metric, so

\begin{align*} d(x,y)&=\|x-P\|+\|P-y\|\\ &\le\big(\|x-z\|+\|z-P\|\big)+\|P-y\|\\ &=\|x-z\|+\big(\|z-P\|+\|P-y\|\big)\\ &=d(x,z)+d(z,y)\;. \end{align*}

• Thank you very much for your time Professor. I really appreciate it. Oct 29, 2014 at 0:17
• @FromCuba: You’re very welcome. Oct 29, 2014 at 0:19