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For example,a relation,R defined on integers. $R = \{(a,b)\mid a^3=b^3\}$

Could this be partial order?

amWhy
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Manuel
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  • Do you know the definition of a partial order? Can you use the definition to prove that the given relation is a partial order? (Hint: there are three things to prove.) – Code-Guru Apr 05 '13 at 00:54
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    A symmetric partial order $R$ is not very interesting: it is a diagonal. Indeed, for every comparable $x,y$, you must have $x R y$ and $y R x$, hence $x=y$. – Julien Apr 05 '13 at 00:57
  • @julien, as discussed in this Q, Wikipd. entry for equality relation states that it is the only relation that is an equivalence and an order. However the definition is impredicative and isomorphism also seems to satisfy that proposition when correctly substituted. Maybe you can shed some light? "Diagonal" seems ok as a descriptor however. – alancalvitti Apr 06 '13 at 19:17
  • @alancalvitti What light? I just said that a partial order which can't compare any pair of distinct elements is not a very interesting partial order. What else can I say? – Julien Apr 06 '13 at 19:20
  • @julien, I'm sayin, do you think the Wikip. article is correct that equality is the only such relation, or is isomorphism also both an equivalence and an order? I maintain that isomorphism would have to be substituted for equality also in the definition of order and so also satisfies that "equation". – alancalvitti Apr 06 '13 at 19:27
  • @alancalvitti Yes, $xRy$ if $x=y$ is the only relation on a given set which is simultaneously an equivalence relation and a partial order. This is plain and simple. I have no idea why any kind of "isomorphism" would have to be considered after this has been observed. – Julien Apr 06 '13 at 19:33
  • @julien, keeping in mind that order has "=" built into its definition, so defining equality in terms of a relation that already defines equality is impredicative. Isomorphism "~" should be substituted. So it's not that simple. – alancalvitti Apr 06 '13 at 19:41

1 Answers1

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A partial order can be symmetric, but that tells us nothing respect to being a partial order. (Antisymmetry is not the antonym of symmetry. A relation can be both.)

To show that a relation is a partial order:

  • You need to show that the relation is reflexive

    • For all $x \in \mathbb Z$, is it true that $x^3 = x^3$?

  • You need to show that the relation is antisymmetric

    • For all $x, y \in \mathbb Z$, is it true that IF $\;x^3 = y^3\;$ AND $y^3 = x^3$, then $x = y$?

  • You need to show that the relation is transitive:

    • For all $x, y, z \in \mathbb Z$, is it true that IF $x^3 = y^3,\;$ AND $\;y^3 = z^3,\;$ then $x^3 = z^3\;$?

(Note: in this case, the relation is both symmetric and antisymmetric.)

amWhy
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