**Exercise 2: Relation on $\mathbb{R}^3$** Given the binary relation $\mathcal{R}$ on $\mathbb{R}^3$ defined by: $$(x,y,z) \mathcal{R} (a,b,c) \iff (|x - a| \leq b - y \text{ and } z = c).$$ 1. **Show that $\mathcal{R}$ is an order relation on $\mathbb{R}^3$.** *An order relation must be reflexive, antisymmetric, and transitive.* 1. **Reflexivity:** For any $(x,y,z) \in \mathbb{R}^3$, check if $(x,y,z) \mathcal{R} (x,y,z)$. We have $|x - x| = 0 \leq y - y = 0$ and $z = z$, so the relation holds. Thus, $\mathcal{R}$ is reflexive. 2. **Antisymmetry:** Suppose $(x,y,z) \mathcal{R} (a,b,c)$ and $(a,b,c) \mathcal{R} (x,y,z)$. From the definition: $$|x - a| \leq b - y \text{ and } z = c,$$ $$|a - x| \leq y - b \text{ and } c = z.$$ Since $|x - a| = |a - x|$ and $z = c$, we get: $$|x - a| \leq b - y \quad \text{and} \quad |x - a| \leq y - b.$$ For both inequalities to hold, it must be that $b - y \geq 0$ and $y - b \geq 0$, implying: $$b - y = 0,$$ hence $y = b$. Thus, $|x - a| \leq 0$ implies $x = a$. Since also $z = c$, antisymmetry holds: if both $(x,y,z) \mathcal{R} (a,b,c)$ and $(a,b,c) \mathcal{R} (x,y,z)$ then $(x,y,z) = (a,b,c)$. 3. **Transitivity:** Suppose $(x,y,z) \mathcal{R} (a,b,c)$ and $(a,b,c) \mathcal{R} (p,q,r)$. We have: $$|x - a| \leq b - y,\quad z = c,$$ $$|a - p| \leq q - b,\quad c = r.$$ From $z = c$ and $c = r$, we get $z = r$. Using the triangle inequality: $$|x - p| \leq |x - a| + |a - p| \leq (b - y) + (q - b) = q - y.$$ Hence: $$(x,y,z) \mathcal{R} (p,q,r),$$ proving transitivity. **Conclusion:** $\mathcal{R}$ is reflexive, antisymmetric, and transitive, so it is an order relation on $\mathbb{R}^3$. 2. **Determine whether this order is total on $\mathbb{R}^3$.** *An order is total if for all $(x,y,z)$ and $(a,b,c)$ in $\mathbb{R}^3$, either $(x,y,z) \mathcal{R} (a,b,c)$ or $(a,b,c) \mathcal{R} (x,y,z)$ holds.* Consider $(x,y,z)$ and $(a,b,c)$ where $z \neq c$. Since $\mathcal{R}$ requires $z = c$ for the relation to hold, neither $(x,y,z) \mathcal{R} (a,b,c)$ nor $(a,b,c) \mathcal{R} (x,y,z)$ can hold. Therefore, $\mathcal{R}$ is **not a total order** on $\mathbb{R}^3$. **Final answers:** $$\boxed{\text{$\mathcal{R}$ is an order relation but not a total order on $\mathbb{R}^3$.}}$$