Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed.
Still, I would like to know:
- if it is sound, because absolute value always creates me some problem, and
- if there is a shorter (neater) way to get the result (maybe without using contradiction).
Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$.
Proof:
We proceed by cases.
Case 1: $d(x,y)-d(y,z) \geq 0$.
Assume by contradiction that $d(x,y)-d(y,z) > d(x,z)$. Hence, $d(x,y)> d(y,z) + d(x,z)$, contradicting the triangle inequality.Case 2: $d(x,y)-d(y,z) < 0$.
Assume by contradiction that $ – d(x,y) + d(y,z) > d(x,z)$. Hence, $d(x,y) < d(y,z) – d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$
As always, any feedback is welcome.
Thank you for your time.
Best Answer
Your work seems right (although you could get the easier contradiction in the second case by adding $d(x,y)$), but there is an easier proof: Let $x,y,z$ be points in a metric space. Then
$$d(x,y)\leq d(x,z)+d(z,y).$$
Then,
$$d(x,y)-d(y,z)\leq d(x,z).$$
On the other hand,
$$d(y,z)\leq d(x,z)+d(x,y).$$
Then,
$$d(y,z)-d(x,y)\leq d(x,z).$$
Therefore,
$$|d(y,z)-d(x,y)|\leq d(x,z).$$