Total order and the strict less-than

Statement

Verify that the relation $\le$ on $\mathbb{Z}$ is a total order and that $<$ is not an order relation (which property is missing?).

Solution

For $\le$ on $\mathbb{Z}$:

• Reflexive: $a\le a$ for every $a$.
• Antisymmetric: if $a\le b$ and $b\le a$ then $a=b$.
• Transitive: if $a\le b$ and $b\le c$ then $a\le c$.
• Total: given any two integers, either $a\le b$ or $b\le a$ holds.

Hence $\le$ is a total order.

The relation $<$ is not reflexive, because $a<a$ is false for every $a$. Thus reflexivity is missing: $<$ is not an order relation (in the sense adopted here, of a “weak” order). $\boxed{\text{Manca la riflessività}}$

Links

Topics: Set theory
Concepts: Total order · Order relation
Skills: Proving
Exercise type: Proof