Translation-invariant total preorder of sets of integers

Question

Does there exist a total preorder $\leq$ on $P(\mathbb{Z})$, the set all sets of integers, with both of the following properties?

(a) If $X \subsetneq Y$, then $X \ngeq Y$.

(b) If $X \leq Y$ then $X + z \leq Y + z$ for any integer $z$ (where $X + z = \{ x + z ∣ x ∈ X \}$).

EDIT. Since there's no answer yet, maybe it will help if I add some things I know about related questions.

1. Suppose we drop condition (a). Then the answer is yes. (Consider any total ordering of the equivalence classes of $P\mathbb{Z}$ under translation. The Axiom of Choice guarantees that any set can be totally ordered.)

2. Suppose we drop condition (b). Then the answer is yes. (We can use a straightforward Zorn's Lemma argument to faithfully extend the subset ordering to a total preorder.)

3. Suppose we strengthen condition (b) to this: $X \sim X + z$ for any integer $z$. Then the answer is no. (Let $X$ be the set of positive integers. Then $X$ is a strict subset of its translate $X - 1$.)

4. Suppose we strengthen condition (b) to this: if $X \leq Y$ then $f(X) \leq f(Y)$ for any permutation $f : \mathbb{Z} \to \mathbb{Z}$. Then the answer is no. (Let $X$ be the set of positive integers, let $f$ be the reflection of $\mathbb{Z}$ around zero, and let $g$ be the reflection around one. We can use the fact that $f^2$ and $g^2$ are the identity to show that, if $\leq$ is a total preorder, then $g(X) \leq X \leq f(X)$ (and also conversely); but $f(X) \subsetneq g(X)$.)

Also, here is a more general question of which my question is a special case:

Let $G$ be a group acting on a set $A$. Suppose $\leq$ is a preorder on $A$ which is invariant under $G$. Under what conditions does there exist a $G$-invariant total preorder $\leq^*$ on $A$ which faithfully extends $\leq$, in the sense that if $x < y$ then $x <^* y$?

Answer 1

Yes, I think there does exist such a relation.

By compactness it suffices to prove that for any finite set $\mathcal S\subset \mathcal P(\mathbb Z)$ there is a total preorder on $\mathcal S$ obeying your conditions (a) and (b). (So (b) holds whenever $X,Y,X+z,Y+z\in\mathcal S.$)

Define an equivalence relation $\sim$ where $A\sim B$ means $A+m\subseteq B\subseteq A+n$ for some integers $n,m.$

For each equivalence class $\mathcal C\subseteq \mathcal S$ that is not a singleton, pick $n\neq 0$ such that $A\subsetneq A+n$ for all $A\in\mathcal C.$ (Or pick $|n|$ minimal.) For $A,B\in\mathcal C$ define $A\preceq B$ to mean $n\sum_{k\in (A+n)\setminus A}k\leq n\sum_{k\in(B+n)\setminus B}k.$ This makes sense because there are at most $n$ integers in each sum. This defines a translation-invariant total preorder on (all translates of) sets in $\mathcal C.$ And it ensures that $A\prec B$ if $A\subsetneq B.$

For $A,B\in \mathcal S$ define $A\preceq' B$ to mean $A+n\subseteq B$ for some $n.$ I think transitivity is clear. If $A\preceq' B\preceq' A$ then $A\sim A.$ So $\preceq'$ defines a partial order on $\mathcal S/\sim,$ which can be extended to a total order.

Combining this total order on $\mathcal S/\sim$ with the total preorder on each equivalence class we get a total preorder on $\mathcal S$ satisfying those conditions (a) and (b).