Proving that a function represents a metric space in $\mathbb{R^2}$

Question

How can one prove, that this function represents a metric space in $\mathbb{R^2}$?

I know, that certain conditions have to be proved to prove a metric:

(i) For all $x, y \in X$ the following has to hold $d(x,y) \geq 0$ and $d(x,y) = 0$ holds exactly when $x = y$. (non-negativity)

(ii) $\forall x,y \in X (d(x,y) = d(y, x)$ (symmetry)

(iii) $\forall x,y, z \in X (d(x,z) \leq d(x,y) + d(y,z))$ (triangle inequality)

So, one has to check the conditions (i), (ii), (iii) and here I don't know if it is correct what I did:

(i): For all $x, y \in \mathbb{R^2} $ the following has to hold: $d_{TGV}(x,y) \geq 0$ and $d_{TGV}(x,y) = 0$ $\iff$ $x = y$, because if $\left\lVert x - y \right\rVert_2 = 0$, $x = y = 0$ due to absolute homogenity in norms.

(ii): Symmetry follows because $\left\lVert - x \right\rVert$ = $\left\lVert x \right\rVert$ and therefore $\left\lVert x - y \right\rVert$ = $\left\lVert y - x \right\rVert$ and $\left\lVert x \right\rVert_2 + \left\lVert y \right\rVert_2$ = $\left\lVert y \right\rVert_2 + \left\lVert x \right\rVert_2$ because of commutativity.

(iii) Now $\forall x,y, z \in \mathbb{R^2} (d_{TGV}(x,z) \leq d_{TGV}(x,y) + d_{TGV}(y,z))$ (triangle inequality), because $\left\lVert x + y \right\rVert_2 \leq \left\lVert x \right\rVert + \left\lVert y \right\rVert$ (triangle inequality in norms)

Answer 1

Your proof of the triangle inequality is wrong, because you worked as if $d(x,y)=\|x-y\|_2$. You must consider each possibility:

1. $x$, $y$, and $z$ are linearly dependent;
2. $x$ and $y$ are linear dependent, but $z$ is independent of both;
3. $x$ and $z$ are linearly independent, but $y$ is independent of the other two;
4. $y$ and $z$ are linearly independent, but $x$ is independent of the other two.

Answer 2

Draw pictures. Think about what the formulae mean.

Two vectors in $\mathbb{R}^2$ are linearly independent iff they lie on the same line through the origin and their distance is then measured the standard way, which is along that common line of course, as standard Euclidean distance follows straight lines. So the lines through the origin are like rail lines (TGV?) where one can travel along these lines as a fast connection.

If however two vectors are independent, so they lie on different through $0$, we first have to travel to the origin (Paris?) and take the straight line to the other point. This gives us the sum of the two norms. So detours are necessary as the only straight line connections are along lines through the origin.

So this makes us believe this is indeed a valid distance that makes some "physical" sense. I can imagine a world where this is the shortest distance between points, but that's not yet a proof.

To check the axioms, one just has to distinguish cases. It's clear from the fact that $\|\cdot\|_2$ is a norm that the function only assumes positive values and in both subcases the symmetry is also clear. If $d(x,y)=0$ then $\|x-y\|_2 = 0$ which implies $x=y$ and otherwise $\|x\|_2 = \|y\|_2 = 0$ which cannot happen (as we only apply that formula in the independent case anyway).

The triangle equality requires some case checking, but we can always assume that $x,y,z$ in it are all distinct and that leaves us with a few subcases ($x,y$ dependent or not, $y,z$ ditto, $x,z$ likewise) which you should all check case by case.