Let $(X,\succeq)$ be a poset. I have the following two questions:


1. Is it true that there exists a finite complemented distributive lattice (a Boolean lattice) $(S, \succeq^\*)$ such that $X\subseteq S$ and $\succeq^\*$ agrees with $\succeq$ over $X$? If not, are there conditions on $(X,\succeq)$ that guarantee this?
2. If the answer to question 1 is true, then is it true that $(S, \succeq^\*)$ is "unique" conditional on the cardinality of $S$? That is, if $(S\_1, \succeq^\*\_{1})$ and $(S\_2, \succeq^\*\_{2})$ satisfy the conditions described in question 1 above and $\lvert S\_1\rvert=\lvert S\_2\rvert$, then $(S\_1, \succeq^\*\_{1})$ is isomorphic to $(S\_2, \succeq^\*\_{2})$? If not, are there any uniqueness results along these lines?


I'm not a mathematician, but this question has come up in my research. Sorry if this is obvious, and thank you for your help!



Is every finite poset a subset of a finite complemented distributive lattice?
As Sam mentioned in the comments, the answer to question 1 is yes. One can map every condition $p$ in the partial order to the lower cone, the set $S\_p=\{q\in P\mid q\leq p\}$ of conditions below $p$. It is clear that $q\leq p\iff S\_q\subset S\_p$, and so this maps $P$ into the powerset algebra, which is a Boolean algebra. This idea works for any poset, whether finite or infinite.


Meanwhile, the answer to question 2 is no for infinite cardinalities. The particular Boolean algebra that we had used was atomic, but we may find a larger Boolean algebra of the same infinite cardinality by embedding that Boolean algebra inside an atomless algebra. Simply view every atom as sitting atop a copy of the countable atomless algebra, and then generate the corresponding Boolean algebra. This will have the same infinite cardinality, while still accepting an embedded copy of $P$, but this larger Boolean algebra, being atomless, is not isomorphic to the earlier one, which is.


For finite cardinalities, however, of course the answer to question 2 is yes, since every finite Boolean algebra is isomorphic to a finite powerset algebra, and so if two of them have the same size (must be $2^n$), then they are isomorphic.


