Are total orders also partial orders

Question

So I'm trying to understand total orders and partial orders. In order to prove a relation is a partial order, I need to prove that the relation is

• Transitive
• Anty-symmetric
• Reflexive

Can I say that a total order is a relation that is not an antisymmetric relation?

For example: if we have the relation where A is related to B iff A is less than or equal to B. This obviously looks like a total ordered relation, however, it is antisymmetric e.g. A = 5 and B = 5 (5, 5) is totally in the relation (they are equal) and A = B so the relation is antisymmetric,, how is this possible? Are all total orders relation also partial orders?

Thanks

Answer 1

Of course total orders are partial orders. A total order is a partial order in which any two elements are comparable. That is given any two elements $a,b$, either $a\leq b$ or $b\leq a$.

For an example of a partial order that isn't a total order, just look at the powerset on three elements ordered by containment. It is obviously not totally ordered. A second example: positive integers with the relation $a\leq b$ iff $a$ divides $b$.

Can I say that a total order is a relation that is not an antisymmetric relation?

I have no idea why you would say that. A total order satisfies more properties than a partial order, not less.

A partial order with the antisymmetric axiom taken away is called a preorder. For example, $x\leq y$ iff $|x|\leq |y|$ is a preorder of the real numbers which isn't a partial order.

Answer 2

A total order is a relation $R$ that is

• Transitive
• Anti-Symmetric
• Total

Total means that for any $a$ and $b$: either $aRb$ or $bRa$. But that means that for any $a$: $aRa$ or $aRa$, i.e. $aRa$. Hence, any total order is automaticaly reflexive. And thus, together with transitivity and anti-symmetry, a total order is automatically a partial order.