If we’re using the Independence II as an axiom, you should be a little more precise, when you introduced it above, you referred to the base four axioms, including continuity.
Now, I only noticed consistency needed to convert between the two Independence formulations, which would make your statement correct. But on the face of things, it looks like you are trying to show a money pump theorem under discontinuous preferences by calling upon the continuity axiom.
For all A, B, C, D and p, if A ≤ B and C ≤ D, then pA + (1-p)C ≤ pB + (1-p)D.
If we now take C and D to be the same lottery, we get independence, as long as C ≤ C. Now, given completeness, C ≤ C is always true (because at least one of C=C, CC must be true, and thus we can always get C ≤ C, -- switching C with C if needed!).
So we don’t need consistency, we need a weak form of completeness, in which every lottery can be at least compared with itself.
If we’re using the Independence II as an axiom, you should be a little more precise, when you introduced it above, you referred to the base four axioms, including continuity.
Now, I only noticed consistency needed to convert between the two Independence formulations, which would make your statement correct. But on the face of things, it looks like you are trying to show a money pump theorem under discontinuous preferences by calling upon the continuity axiom.
Mathematically:
Independence + other 3 axioms ⇒ Independence II
Independence II ⇒ Independence
Hence: ~Independence ⇒ ~Independence II
My theorem implies: ~Independence II ⇒ You can be money pumped
Hence: ~Independence ⇒ You can be money pumped
Note, Independence II does not imply Independence, without using at least the consistency axiom.
The contrapositive of independence II is:
For all A, B, C, D and p, if A ≤ B and C ≤ D, then pA + (1-p)C ≤ pB + (1-p)D.
If we now take C and D to be the same lottery, we get independence, as long as C ≤ C. Now, given completeness, C ≤ C is always true (because at least one of C=C, CC must be true, and thus we can always get C ≤ C, -- switching C with C if needed!).
So we don’t need consistency, we need a weak form of completeness, in which every lottery can be at least compared with itself.
Transitivity and Continuity are unnecessary, however.
That is my reading of it too. I know Stuart is putting forward analytic results here, I was concerned that this one was not correctly represented.