# Archimedean but not mixture continuous

In the context of preferences on a set of lotteries on a finite set $$X$$, what is an example of a preference that is independent, Archimedean but not mixture continuous?

I know the mixture continuous implies Archimedean but cannot think of an example where the reverse implication fails.

For completeness, I provide here the definitions of the terms above. Suppose $$X=\{x_1,\ldots,x_n\}$$ is finite. Let $$\Delta X=\{\pi\in\mathbb{R}^n:\sum_{i=1}^n\pi_i=1, \pi_i\geq 0\}$$. A preference $$\succeq$$ on the elements of $$\Delta X$$ is:

• Independent: if $$\forall$$ lotteries $$\pi,\rho,\sigma$$ and $$\forall \alpha\in(0,1)$$: $$\pi\succ\rho\iff\alpha\pi+(1-\alpha)\sigma\succ \alpha\rho+(1-\alpha)\sigma,\\ \pi\sim\rho\iff\alpha\pi+(1-\alpha)\sigma\sim \alpha\rho+(1-\alpha)\sigma;$$
• Archimedean: if $$\forall$$ lotteries $$\pi,\rho,\sigma$$, $$\pi\succ\rho\succ\sigma\implies\exists\alpha,\beta\in(0,1): \alpha \pi+(1-\alpha)\sigma\succ\rho, \beta \pi+(1-\beta)\sigma\prec\rho$$
• Mixture continuous: if for all $$\pi,\rho,\sigma$$, the sets $$\{\alpha\in[0,1]:\alpha\pi+(1-\alpha)\rho\succeq\sigma\}\quad\text{and}\quad \{\alpha\in[0,1]:\alpha\pi+(1-\alpha)\rho\preceq\sigma\}$$ are closed in $$[0,1]$$
• Independence and Archimedean together imply (mixture) continuity. So you can't have the first two without the third. See Kreps (1988) for a proof. – Herr K. Nov 16 '18 at 2:47
• @HerrK. I haven't used that book in the past but on your comment, I've looked at its Chapter 5. But I cannot find the result you refer to. – yurnero Nov 16 '18 at 3:24
• It's Lemma 5.6(b) – Herr K. Nov 16 '18 at 5:09