A partial order $\mathbb{B}$ is *universal* for a class $\cal{P}$ of partial orders if every order in $\cal{P}$ embeds
order-preservingly into $\mathbb{B}$.

For example, every partial order $\langle\mathbb{P},\lt\rangle$ maps order-preservingly into its power set by the map $$p\mapsto\{q\in\mathbb{P}\mid q\leq p\}$$ that sends each element $p$ to its lower cone.

Thus, the power set order $\langle P(\{1,2,\ldots,n\}),{\subseteq}\rangle$ is universal for the class of partial orders of size $n$. This provides an order of size $2^n$ that is universal for orders of size $n$.

**Question.** What is the minimal size of a partial
order that is universal for orders of size $n$?

In particular, is there a polynomial upper bound?

One can make at least slight improvements to the $2^n$
upper bound, by observing that the emptyset was not needed,
as it never arises as a lower cone, and we don't need *all*
the atoms, since if they are needed, then one can use the
co-atoms instead. I suspect that there is a lot of waste in
the power set order, but the best upper bound I know is
still exponential.

For a lower bound, my current knowledge is weak and far from exponential. Any order that is universal for orders of size $n$ will contain a chain and an antichain, making it have size at least $2n-1$. (That bound is exact for $n\leq 3$.) A student in my intro logic course extended this to $n\log(n)$ by considering $k$ chains (and antichains) of size $n/k$.

Can one find better lower bounds?

Interestingly, the same student observed that we cannot in
general expect to find *unique* smallest universal orders,
since he found several orders of size 5 that are
universal for orders of size 3 and which are minimal with
that property. So in general, we cannot expect a unique
optimal universal order. Does this phenomenon occur for
every $n$? (He also found minimal universal orders of size larger than the minimal size universal order.)