We are talking about the same thing here just at different levels of generality. The function you describe is the same as the one I’m describing, except on a much narrower domain (only a single binary lottery between A and B). Then you project the range to just a question about C.
In the specific function you are talking about, you must hold that this is true for all A, B, and C to get continuity. In the function I describe, the A, B, and C are generalized out, so the continuity property is equivalent to the continuity of the function.
Note, setting the limit to “no preference” does not resolve the discontinuity. But by intermediate value, there will exist at least one such point in any continuous approximation of the discontinuous function.
I meant that setting the limit to no preference for a given C doesn’t equate to a globally continuous function. But that when you adjust your preferences function to approximate the discontinuous function by a continuous one, the result will contain (at least one) no preference point between any two A < B.
Now perhaps there is a result which says that if you take the limit as you set all discontinuous C to no preference, that the resulting function is complete, consistent, transitive, and continuous, but I wouldn’t take that to be automatic.
Consider, for example, a step discontinuity, where an entire swatch of pA + (1-p)B are stuck on the same set of < and = mappings and then there is a sharp jump to a very large set of < and = mappings at a critical p’. If you map the ordinals to the real line, this is analogous to a y-coordinate jump. To remove this discontinuity you would need to do more than split the preferences at p’ around no preference, because all this does is add a single point to the mix. To fully resolve it, you need to add an entire continuous curve, which means a process of selecting new A, B, and C, and showing that the transfinite limit always converges to a valid result.
We are talking about the same thing here just at different levels of generality. The function you describe is the same as the one I’m describing, except on a much narrower domain (only a single binary lottery between A and B). Then you project the range to just a question about C.
In the specific function you are talking about, you must hold that this is true for all A, B, and C to get continuity. In the function I describe, the A, B, and C are generalized out, so the continuity property is equivalent to the continuity of the function.
So what did you mean by
I meant that setting the limit to no preference for a given C doesn’t equate to a globally continuous function. But that when you adjust your preferences function to approximate the discontinuous function by a continuous one, the result will contain (at least one) no preference point between any two A < B.
Now perhaps there is a result which says that if you take the limit as you set all discontinuous C to no preference, that the resulting function is complete, consistent, transitive, and continuous, but I wouldn’t take that to be automatic.
Consider, for example, a step discontinuity, where an entire swatch of pA + (1-p)B are stuck on the same set of < and = mappings and then there is a sharp jump to a very large set of < and = mappings at a critical p’. If you map the ordinals to the real line, this is analogous to a y-coordinate jump. To remove this discontinuity you would need to do more than split the preferences at p’ around no preference, because all this does is add a single point to the mix. To fully resolve it, you need to add an entire continuous curve, which means a process of selecting new A, B, and C, and showing that the transfinite limit always converges to a valid result.