4 ms·
I keep getting bounced out of the flow by the author describing the relation A<=B as "A is bigger than B". I get that the order relation is arbitrary, but that
by ajarmst 6y ago
I keep getting bounced out of the flow by the author describing the relation A<=B as "A is bigger than B". I get that the order relation is arbitrary, but that's just irritating. Perhaps using "precedes" rather than "is bigger" would be helpful. I also don't think the assertion that anti-symmetry excludes equality ("no ties are permitted"). The usual definition of antisymmetry specifically requires equality: a<=b && b<=a -> a==b (i.e. a<=b -> !(b<=a) is a stronger restriction than mere antisymmetry.)
- BoiledCabbage 6y agoI believe a tie is different from equality. The reason the above holds is because a and b must both be the same element. A "tie" in this case assume they are different elements. Ie the author is a different person from his grandmother. If that's the case, then it's false that author == grandmother. Ie two things can't be "tied" unless they are actually just one thing.
- ajarmst 6y agoOk. But I still don't agree that the property of antisymmetry implies or requires that restriction.
- ajarmst 6y agoIf we agree that the axiom of extensionality applies, then the restriction of no ties is implied. I don't think antisymmetry is necessary or sufficient for it.
- BoiledCabbage 6y agoOk, how are you defining a tie? I believe the author is defining a tie as the following: (a <= b) && (b <= a) && (a != b) Then a and b are "tied". Where "!=" means a and b are different.
- ajarmst 6y agoI'm fine with that restriction. But that isn't the axiom of antisymmetry. That's the axiom of antisymmetry plus a rule that holds that equality implies identity (which would be typically described in terms of the axiom of extensionality). My problem is with the implied claim that antisymmetry alone gives you that restriction, which is incorrect. Antisymmetry is entirely consistent with collections that contain equal but distinct elements.