Given a cardinal $\kappa\geq \aleph\_0$, is there a poset $(P,\leq)$ with $|P| = \kappa$ such that every poset of cardinality $\kappa$ can be order-embedded into $(P,\leq)$?



Universal poset for cardinals $\kappa \geq \aleph_0$
The easy case occurs when $\kappa^{<\kappa}=\kappa$. Under GCH, this includes every regular cardinal. As Emil mentions in the comments, the result for this case follows from general model-theoretic considerations of saturated models.


But for definiteness, let me describe a concrete construction. We construct a universal partial order in a process of length $\kappa$. Start with nothing $P\_0=\emptyset$. At each stage $\alpha$, consider all possible subsets $A,B,C\subseteq P\_\alpha$ each of size less than $\kappa$, such that every element of $A$ is strictly below every element of $B$, and no element of $C$ is below an element of $A$ or above an element of $B$. Add a new point to $P\_{\alpha+1}$ that is above every element of $A$, below every element of $B$, and incomparable with every element of $C$. That is, for each possible way of relating to a given fewer than $\kappa$ many elements we have so far, we add such a point realizing that possibility. Since $\kappa^{<\kappa}=\kappa$, there are at most $\kappa$ many $A,B,C$, and so we have added at most $\kappa$ many points at each stage, and so inductively every $P\_\alpha$ has size at most $\kappa$. At limit stages, we take the union of everything we've added up to that stage.


Consider the ultimate order $P\_\kappa$ that we've built by stage $\kappa$. This order has size $\kappa$, and I claim it is universal for all orders of size $\kappa$. To see this, suppose that you have an order $\langle Q,\leq\rangle$ of size $\kappa$. Enumerate it in a $\kappa$ sequence $q\_\alpha$ for $\alpha<\kappa$, and build an embedding into $P\_\kappa$ in stages. We will ensure that each $q\_\alpha$ finds its image in $P\_{\alpha+1}$.


At each stage, $q\_\alpha$ relates to the previous elements $q\_\beta$ for $\beta<\alpha$ in a certain way — above some, below others, and incomparable with the rest. But at stage $\alpha$ in the universal construction, we added a point of exactly that type relative to the images of those points. So there is a point in $P\_{\alpha+1}$ to which to map $q\_\alpha$ so as to extend the embedding one more step. Thus, $Q$ embeds into $P\_\kappa$, and so our partial order is universal for all orders of size continuum.


The model we've constructed is not only universal, but also homogeneous, since any given embedding of a less-than-size-$\kappa$ order can be extended. But if CH fails, then it is easy to see that there can be no such order of size $\aleph\_1$, since for a given copy of $\mathbb{Q}$ it would have to realize all the cuts, but there are too many. More generally, whenever $2^\gamma>\kappa$ for some $\gamma<\kappa$, then there would have to be an antichain of size $\gamma$, and every subset of it would have to have a point above it and not the other points in the antichain, but there are too many subsets. It is much more subtle, however, to rule out merely universal orders of these sizes.


To get an idea of the difficulties, consider the following paper of Kojman and Shelah concerning the corresponding case of universal linear orders.


* *Kojman, Menachem; Shelah, Saharon*, [**Nonexistence of universal orders in many cardinals**](https://doi.org/10.2307/2275437), J. Symb. Log. 57, No. 3, 875-891 (1992). [ZBL0790.03036](https://zbmath.org/?q=an:0790.03036).


They prove, for example (theorem 3.6), that if $\aleph\_1<\lambda=\text{cof}(\lambda)<2^{\aleph\_0}$, then there is no $\lambda$-universal linear order of size $\lambda$.


I am unsure of the extent to which those results apply to partial orders.


