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 $[0,1]$
  • $\begingroup$ Independence and Archimedean together imply (mixture) continuity. So you can't have the first two without the third. See Kreps (1988) for a proof. $\endgroup$
    – Herr K.
    Nov 16, 2018 at 2:47
  • $\begingroup$ @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. $\endgroup$
    – yurnero
    Nov 16, 2018 at 3:24
  • 1
    $\begingroup$ It's Lemma 5.6(b) $\endgroup$
    – Herr K.
    Nov 16, 2018 at 5:09


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