Let $(O, \leq)$ be a preorder or preference relation, that is a reflexive complete transitive not neccessarily antisymmetric relation over an abitary set, let's shorten $(O,\leq)$ with $O$. (Complete here means for any $x,y\in O$ holds $x\leq y \vee x=y \vee x\geq y$.) A finite lottery $L$ over $O$ is a probability distribution over $O$ with finite support, so $$\sum_{x\in\operatorname{support} L}L(x) = 1$$ Let's denote the set of all finite Lotteries over $O$ by $\operatorname{Lot}(O)$, we can then assign a preference relation $\leq_{\operatorname{Lot}(O)}$ to $\operatorname{Lot}(O)$ such that we have a new preorder. This process can be iterated and one can image a series of so-called compound lottery sets build upon each other. It is even possible to have one such that it is preserving the ordering, i.e. when the lottery assigning probability 1 to element $x$ from a preorder $A$ is denoted with $[x]$, we have that $x \leq y$ implies $[x] \leq_{\operatorname{Lot}(A)} [y]$. Now the compound level of $A$ in this series of $O$ is the number of steps needed to reach $A$ from $O$, so for example the compound level of $O$ in $O$ is $0$ and the compound level of $(\operatorname{Lot}(O), \leq_{\operatorname{Lot}(O)})$ in $O$ is $1$. Also the compound level of $(\operatorname{Lot}(O), \leq_{\operatorname{Lot}(O)})$ in $(\operatorname{Lot}(O), \leq_{\operatorname{Lot}(O)})$ is $0$

The above is a more formal description of the notions given in chapter 2 of "Game theory" in Maschler et al. Now my question is: Given $A$ but not $O$, is it possible to get an $O'$ such that for all preorders $O$ which can lead to $A$, the compound level of $A$ in $O$ is maximal for $O=O'$? In other words, exists a preorder $O'$ such that the compound level of $A$ in $O'$ is $n$ and for all preorders $O$ holds: $$\mbox{the compound level of $A$ in $O$} \geq n$$

This question seems trivial, because one may think "well, I can only possibly have reached $A$ from $O'$ in a finite number of steps, so surely I can get back to it?" But that is not true, the starting $O$ could already been a compound lottery so who knows? If $O$ only contains 1 element, so does $\operatorname{Lot}(O)$ how can we get back to the original $o$ that is not a Lottery of another object? When $O$ contains at least two elements, cardinality of $\operatorname{Lot}(O)$ is at least the one of the continuum (this can be seen by fixing two elements of $O$ and showing that the finite lotteries over these two elements correspond to the interval $[0,1]$).

Personally, I'm thinking there must be some result telling me that it cannot be decided if there is a minimal preorder or some way of showing that all preorders can be constructed (up to isomorphism) by the process above and the initial $O$ being any ordinal number. But that is by far not my area of expertise.

Help with tagging is appreciated.

**EDIT:** About assigning a preference relation to $\operatorname{Lot}(O)$: One can always find a reflexive complete transitive over $\operatorname{Lot}(O)$ that preserves the ordering of $O$ as described above, so existence is assured. I proved the preceding statement in MIZAR by introducing a lexicographical ordering, which is easy if $O$ is antisymmetric and a little bit sophisticated if not.

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