Can the null-set underly a relation structure? Can it underly a well-order? If so, would such a well-order have order-type $0$? (Can one even have an order-type of $0$?)
My motivation for this question is distantly but completely related to an earlier posting I made re a problem that's been bugging me this whole day: I've been trying to show that for any set $W$ of well-orders, there is a well-order $(A, X)$ such that for all $(B, Y) \in W, \exists a \in A$ such that $(B, Y) \cong I_a$, where $I_a$ is the initial segment of $(A, X)$ up to $a$.
A wise user on here showed me that one could form the set $W'/R$, where $W'$ is the set of all (in his qualifying words, "not necessarily proper") initial segments in $W$, and $R$ is order-isomorphism. But my main fear is that very qualification: I know that $W'/R$ is itself well-ordered by order-type comparison. I'm just worried that $W'/R$ is only just as "large" as the largest order-type in $W'$, because $W'/R$ must be "larger" than $W'$ in order for everything in $W$ to be isomorphic to a proper initial segment of $W'/R$. If $0$ is indeed an order-type, then $W'/R$ would be "+1 larger" than $W$, which is what we need, I'm pretty sure.