Question

$(X,d)$ metrik uzay olmak üzere her $x,y,z\in X$ için $$|d(x,z)-d(z,y)|\leq d(x,y)$$ olduğunu gösteriniz.

Answer 1

$\left.\begin{array}{rr} x,y,z\in X \\ \\ (X,d) \text{ metrik uzay}\Rightarrow d, X\text{'de metrik}\end{array} \right\}\Rightarrow d(z,y)\leq d(z,x)+d(x,y)$

$\Rightarrow -d(x,y)\leq d(x,z)-d(z,y)\ldots (1)$

$\left.\begin{array}{rr} x,y,z\in X \\ \\ (X,d) \text{ metrik uzay}\Rightarrow d, X\text{'de metrik}\end{array} \right\}\Rightarrow d(x,z)\leq d(x,y)+d(y,z)$

$\Rightarrow d(x,z)-d(z,y)\leq d(x,y)\ldots (2)$

$$(1),(2)$$ $$\Rightarrow$$ $$-d(x,y)\leq d(x,z)-d(z,y)\leq d(x,y)$$ $$\Rightarrow$$ $$|d(x,z)-d(z,y)|\leq d(x,y).$$