Say an antichain of a poset $P$ is a set of pairwise incomparable elements of $P.$ Typically, the width of a partial order is defined to be the supremum of the cardinalities of antichains of $P.$ When Choice fails, however, it seems as if this definition may be invalid (or perhaps simply unsatisfactory).
For example, if Choice fails, then there are two disjoint sets $A$ and $B$ with incomparable cardinalities. Let $P=A\cup B,$ and given $a,b\in P,$ say $a\leq b$ iff $a\in A$ and $b\in B.$ Then $\leq$ makes $P$ a poset, and $A$ and $B$ are precisely the maximal antichains of $P$.
However, do the cardinalities necessarily form a lattice if Choice fails? If not, then it seems there need not be a supremum of $\bigl\{|A|,|B|\bigr\},$ so that this definition need not apply to this $P,$ and we need another one. If so, then what is the supremum of $\bigl\{|A|,|B|\bigr\}$? Clearly, $|A\cup B|=|A|+|B|$ is an upper bound, but must it be the least? If so, why? If not, then what could it be?
Indeed, the cardinalities do not need to form a lattice without the axiom of choice.
Suppose $A$ and $B$ are infinite Dedekind-finite sets of incomparable cardinality, and suppose $A, B$ both embed via $f, g$ into $C$, with $C=ran(f)\cup ran(g)$. Then it's easy to see that $C$ is also Dedekind-finite, so if we remove an element from $C$ we get a set of strictly smaller cardinality. On the other hand, since $A$ and $B$ are incomparable, we must have $X:=ran(f)\setminus ran(g)$ and $Y:=ran(g)\setminus ran(f)$ each nonempty, so both $A$ and $B$ embed into $C$ minus one element.
This shows that not only do such $A$ and $B$ not have a supremum, but given any set into which they both embed, there is a strictly smaller set into which they both embed.
Answering a question in the comments: there can indeed be posets with no maximal antichains. We can have a poset which is the disjoint union of $\omega$-many pieces $A_i$ which are each ordered as $\mathbb{Z}$, and elements of different pieces are incomparable, but so that the family $\{A_i: i\in\omega\}$ does not admit a choice function.