Fix positive integers $k$ and $n$.
For which finite posets $(X,\lesssim)$ with $\#X=k$ does there exist an order embedding $\phi\colon(X,\lesssim)\to (\mathbb{R}^n,\le)$, where $\le$ is the standard partial order (product order) on $\mathbb{R}^n$?
Are there quantitative bounds on the minimal such $n$ for specific subclasses of these posets?
If $(P,\le)$ is a poset, the least $n$ such that $(P,\le)$ embeds into a product of $n$ total orders (or equivalently, such that $\le$ is the intersection of $n$ total orders on $P$) is known as the dimension of $(P,\le)$. If $P$ is finite, this is equivalent to embedding into the product of $n$ copies of any fixed infinite total order, such as $(\mathbb R,\le)$. Thus, the question is exactly about which posets of size $k$ have dimension at most $n$.
As Sam Hopkins notes in a comment, now banished to the netherworld, classifying posets of a given dimension is likely very hard. But for some basic bounds, all posets of size $k$ have dimension at most $k$, as $(P,\le)$ embeds to $(\mathcal P(P),\subseteq)\simeq\{0,1\}^k$ via $a\mapsto\{x\in P:x\le a\}$; on the other hand, the Wikipedia article gives an example of a poset of dimension $k/2$.
