Exercise 1167Logic, sets and proofs › RelationsL1Added 3 Nov 2026

Total orders

#total order#comparability#real numbers

Statement

Show that $(\mathbb{R},\le)$ is a total order.

Hint

Check reflexivity, antisymmetry, transitivity, totality.

Show solutionHide solution

Reflexive. $x\le x$ .

Antisymmetric. $x\le y$ and $y\le x\Rightarrow x=y$ .

Transitive. $x\le y\le z\Rightarrow x\le z$ .

Total. For all $x,y$ , either $x\le y$ or $y\le x$ .

Conclusion. $(\mathbb{R},\le)$ is a total order.