0

I understand it, but I don't know how I would find my way to it by myself, I've made some sketches to try to understand. But still i feel like missing something. Hope you understand me. $|d(x,y)-d(z,t)|\leq d(x,z)+d(y,t)$

I understand that the distance between 2 of those points less another distance will be less than the sum of other 2 distances, feels like obvious that a substracion deals a number smaller than a sum

José Osorio
  • 734
  • 1
    The relevancy of this inequality is the following: It shows that the metric $d$ is continuous as a function $d:>X\times X\to{\mathbb R}_{\geq0}$. – Christian Blatter Oct 28 '16 at 07:57

2 Answers2

2

This is a double application of the triangle inequality. $$d(x,y) \leq d(x,z)+d(z,y) \leq d(x,z)+d(z,t)+d(t,y),$$ so $$d(x,y)-d(z,t) \leq d(x,z)+d(y,t).$$ Similarly $$d(z,t)-d(x,y) \leq d(x,z)+d(y,t),$$ and the result follows. The proof should tell you how to sketch the situation convincingly; going directly from $x$ to $y$ is "faster" than going from $x$ to $z$, then from $z$ to $t$, then from $t$ to $y$.

JKRT
  • 492
2

Suppose (without loss of generality) that $d(x,y)\geq d(z,t)$. Then this inequality says $d(x,y)-d(z,t)\leq d(x,z)+d(y,t)$, or equivalently $$d(x,y)\leq d(x,z)+d(z,t)+d(t,y).$$ But this is essentially just the triangle inequality (or perhaps more accurately, the "quadrilateral inequality"): it says that if you go from $x$ to $z$ to $t$ to $y$ and add up the distances of each step, this can't be smaller than the distance just going directly from $x$ to $y$. You can formally prove this by using the triangle inequality twice: $$d(x,y)\leq d(x,z)+d(z,y)\leq d(x,z)+(d(z,t)+d(t,y)).$$

Eric Wofsey
  • 330,363