Right- let’s just call the points A, B, C, -B, and -A. C is going to be the middle of that set, even if you remove (B and -B) and/or (A and -A). When you remove (-B and -A), you’ve broken the symmetry of the set, even though the set has new symmetry by virtue of there being a new middle point.
If you’re postulating a (much much) weaker version of IIA saying something like “if you remove symmetric irrelevant points from a symmetric set, the outcome doesn’t change” then you’d be right. But IIA does not require that symmetry be preserved.
Yep, I agree that strong IIA of “if x is chosen from T, and S is a subset of T, then x is chosen from S” doesn’t apply if preferences are based on the relative merits of x rather than the individual merits of x. That statement seems obviously true on its own, and so I think the picture proof of this particular example detracts more than it adds, because there is a natural weak IIA here.
Right- let’s just call the points A, B, C, -B, and -A. C is going to be the middle of that set, even if you remove (B and -B) and/or (A and -A). When you remove (-B and -A), you’ve broken the symmetry of the set, even though the set has new symmetry by virtue of there being a new middle point.
If you’re postulating a (much much) weaker version of IIA saying something like “if you remove symmetric irrelevant points from a symmetric set, the outcome doesn’t change” then you’d be right. But IIA does not require that symmetry be preserved.
Yep, I agree that strong IIA of “if x is chosen from T, and S is a subset of T, then x is chosen from S” doesn’t apply if preferences are based on the relative merits of x rather than the individual merits of x. That statement seems obviously true on its own, and so I think the picture proof of this particular example detracts more than it adds, because there is a natural weak IIA here.