ah, I think I am starting to follow. It is a bit ambigious whether it is supposed to be two instances of one arbitraliy small finite or two (perhaps different) arbitrarily small finites. If it is only one then the tails are again relevant.
I don’t see what you mean. It doesn’t make a different whether you have only one or two small-but-finite quantities (though usually you do have on for each quantity you are dealing with), as long as they are in general position. For instance, in my modification to the example you gave, I only used one: While it’s true that the tail becomes relevant in (0, 1)(1, 0) vs (1, 0)(0, 1) because 0*1=1*0, it is not true that the tail becomes relevant in the slightly modified (eps, 1)(1, 0) vs (1, 0)(0, 1) for any real eps != 0, as eps*1 != 1*0.
So the only case where infinitesimal preferences are relevant are in astronomically unlikely situations where the head of the comparison exactly cancels out.
I thought we were comparing (eps, 1)(1, 0) and (1, 0)(eps, 1). if eps=eps strict equality. if it was (a,1)(1,0) and (1,0)(b,1) and its possible that a!=b it is unsure whether there is equality.
We were comparing epsilon to no-epsilon (what I had in mind with my post).
Anyway, the point is that strict equality would require astronomical consequences, and so only be measure 0. So outside of toy examples it would be a waste to consider lexicographic preferences or probabilities.
I don’t see what you mean. It doesn’t make a different whether you have only one or two small-but-finite quantities (though usually you do have on for each quantity you are dealing with), as long as they are in general position. For instance, in my modification to the example you gave, I only used one: While it’s true that the tail becomes relevant in (0, 1)(1, 0) vs (1, 0)(0, 1) because 0*1=1*0, it is not true that the tail becomes relevant in the slightly modified (eps, 1)(1, 0) vs (1, 0)(0, 1) for any real eps != 0, as eps*1 != 1*0.
So the only case where infinitesimal preferences are relevant are in astronomically unlikely situations where the head of the comparison exactly cancels out.
I thought we were comparing (eps, 1)(1, 0) and (1, 0)(eps, 1). if eps=eps strict equality. if it was (a,1)(1,0) and (1,0)(b,1) and its possible that a!=b it is unsure whether there is equality.
yeah (eps,0) behaves differently from (0,1)
We were comparing epsilon to no-epsilon (what I had in mind with my post).
Anyway, the point is that strict equality would require astronomical consequences, and so only be measure 0. So outside of toy examples it would be a waste to consider lexicographic preferences or probabilities.