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In ZFC, all cardinals are totally ordered under the relation $\mathfrak{a} \leq \mathfrak{b}$ if and only if there exist injective functions from sets with cardinality $\mathfrak{a}$ to sets with cardinality $\mathfrak{b}$. In fact, they are well-ordered, and therefore any set of cardinals is not dense. (Here, by "dense" I mean that for any $x,y$ such that $x < y$, there exists $z$ so that $x < z < y$.)

But if we exclude the Axiom of Choice, we can only guarantee that cardinals are partially ordered. However, we might still ask whether there exists any set $X$ of cardinals such that if $x,y \in X$ and $x < y$, then there exists $z \in X$ so that $x < z < y$.

My understanding is that very little is known about the structure of cardinals in models of ZF without Choice. Is this particular question open?

Arseniy Sheydvasser
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    It is consistent with ZF that the cardinals is a universal partial order. Namely, every partially ordered set can be realised as cardinals in that model. The proof is not trivial; but a simpler proof is that given any partial order, there is an extension of the universe in which that partial order embeds into the cardinals. – Asaf Karagila Dec 01 '22 at 15:35
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    I will add that there are a lot of open questions regarding the cardinal structure, but we also know that it can be "as wild as it gets". It is, in general, harder to break something with finesse, rather than just smash it completely, as hard as possible. – Asaf Karagila Dec 01 '22 at 15:41

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