Relation $\rho$ with given constraint on $\mathbb R^2$

Question

$\rho$ is a binary relation on $\mathbb R\times \mathbb R$ defined by: $$(a,b)\rho(c,d)\iff(a<c \lor (a=c \wedge b\leq d))$$ Is $\rho$ a partial order relation on $\mathbb R\times \mathbb R$? Is it a total order? Check for the partial order: reflexivity: $$(a,b)\rho(a,b)\iff (a<a\lor (a=a \wedge b\leq b)$$ yes.

antysimmetry - $\forall(a,b),(c,d)\in \mathbb R^2,(a,b)\rho(c,d)\wedge(c,d)\rho (a,b)\overset{?}\iff (a,b)=(c,d)$$\iff a=c\wedge b=d$ $$(c,d)\rho(a,b)\iff (c<a\lor(c=a\wedge d\leq b)$$ no.

transitivity: $\forall (a,b),(c,d),(e,f)\in \mathbb R^2\;\;\; (a,b)\rho(c,d) \wedge (c,d)\rho(e,f)\overset{?}\implies (a,b)\rho(e,f)$

$\rho$ is neither a partial order nor a total order relation. Can I just stop after non-antisimmetry? If it happened to be antisymmetric, would I have to write all the cases for transitivity condition?

Answer 1

Can I just stop after non-antisimmetry? If it happened to be antisymmetric, would I have to write all the cases for transitivity condition?

Sure. If you know a relation is not anti-symmetric, you know it's not a total order. Moreover, you don't even need to prove it in all generality as you did: if you wanted to use a specific pair of ordered pairs, you could do that since a single counterexample is enough to rule out antisymmetry.

This idea holds in general: if something has a list of conditions it needs in order to be the definition of $X$ thing, and it doesn't meet any one of those conditions, that alone is sufficient to say it is not $X$ thing.

Not that the practice in proving transitivity would be a bad thing!

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That said, I'm not convinced this relation isn't anti-symmetric.

Suppose $(a,b)\rho(c,d)$ and $(c,d)\rho(a,b)$. If it follows $(a,b)=(c,d)$, we have antisymmetry.

By $(a,b)\rho(c,d)$, we know

• $(1)$: $a<c$, or
• $(2)$: $a=c$ and $b \le d$

From $(c,d)\rho(a,b)$, we know

• $(3)$: $c>a$, or
• $(4)$: $c=a$ and $d \le b$

Since both are true simultaneously, we cannot have $a<c<a$, so $(1),(3)$ do not hold. Thus, $a=c$. If we have $b \le d \le b$, which must hold since $a=c$, then $b=d$. And thus $(a,b)=(c,d)$.

Thus, anti-symmetry, I believe, unless I overlooked something.

Answer 2

No, you wouldn't. Since both partial orders and total (lax) orders are antisymmetric, demonstrating that a relation is not antisymmetric is enough evidence to assert that the relation is neither a partial nor a total order.

That said, your argument doesn't do that; it only makes the claim "no" with no evidence. In order to demonstrate that $\rho$ is not antisymmetric, you need to give actual values for $a,b,c,d\in\mathbb R$ such that $(a,b)\rho(c,d)$ and $(c,d)\rho(a,b)$ are true but the two pairs are different. (If you find it hard to come up with such numbers, keep in mind that the relation may actually be antisymmetric.)