• Philarete
    9
    Hi all,

    (1) I read today in a paper that if one gives up the asymmetry of a relation, one will also have to give up "at least one of irreflexivity or transitivity". But how can we prove this ? I am especially concerned by the move from symmetry to non-transitivity.

    (2) Some, in the literature on metaphysical grounding, contend that an irreflexive, asymmetric, but non-transitive relation might form loops. Instances of the relation, it is said, would form chains "double-backing" of themselves. I have a lot of trouble to grasp what this is supposed to mean, or how we could illustrate this principle. The best I can think of is something of the like : aRb, bRc, cRd, dRa. But here, my intuition is rather that a loop is introduced by transitivity (and not by non-transitivity). Isn't it because R is transitive that we may go all the way from a back to a ?

    I should stress that I have no background in set theory, which is why I had a hard time to answer these question by looking up online.

    Thanks in advance for your help
    Philarete
  • MathematicalPhysicist
    45
    Regarding question (1):
    A relation can have any properties we wish it to have.

    Perhaps you meant for an equivalence relation which is a binary relation which is transitive, reflexive and symmetric.

    You should give us a link to the paper.
  • sime
    1k
    (t) transitivity: aRb & bRc => aRc
    (s) symmetry: aRb <=> bRa
    (r) reflexivity: aRa

    (s) & (t) => (r) via substitution of (s) into (t)

    Therefore:

    (s) => { (t) => (r) } (giving up irreflexivity if transitivity is true)
    (s) & Not (r) => Not (t) (giving up transitivity if irreflexivity is true)
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