I try get it why relation divisibility is not relation partially ordered set.

$A=\{−2, 2, 4, 6, 8, 10\}$ with relation divisibility "|"

$R$ is relation divisibility | when $a,b,c \in Z : a = b \cdot c$

For relation partially ordered set must be relation:

  • reflexive (fulfil) - everery number could have divisible with yourself

  • antisymmetric (fulfil) - smaller number could have divisible with bigger, but not the otherway

  • transitive (not fulfil) - why please?


1 Answer 1


Divisibility is not antisymmetric on your $A$ because $-2\mid 2$ and $2\mid{-2}$.

  • $\begingroup$ Thank you. So antisymmetric is not fulfil and transitive is fulfil. Is that correct? $\endgroup$
    – user288083
    Nov 15, 2015 at 20:27
  • $\begingroup$ On notation: Some authors use a different def'n of poset (partially ordered set). In Forcing (a method in set theory) a poset relation $\leq$ is defined to be symmetric and transitive. And $x\leq y\leq x$ need not imply $y=x$ $\endgroup$ Nov 15, 2015 at 21:09
  • $\begingroup$ Yes it's transitive but not antisymmetric. @user254665 It's unfortunate, and confusing, to use the symbol "$\le$" for a relation that is not reflexive, as the lower line implies to every mathematician alive "or equal". Where have you seen such a (bad) definition of "poset"? $\endgroup$
    – BrianO
    Nov 15, 2015 at 21:44
  • $\begingroup$ @BrianO. in every article and book I've seen on forcing, including Kunen, Jech, Shelah, and in seminar & class at Univ. of Toronto. $\endgroup$ Nov 16, 2015 at 0:15
  • 1
    $\begingroup$ @user254665 Nonsense. Kunen, 1st Ed 5th printing can be found here [<-- link]. On p. 52 he defines a partial order, what many would call a preorder, to be a transitive and reflexive relation. I don't have Jech at hand but I know it very well and I don't believe you about his definition. Nobody defines a partial order as being symmetric, otherwise $<$ and $\le$ on our favorite number systems wouldn't be partial orders. $\endgroup$
    – BrianO
    Nov 16, 2015 at 0:27

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