There does not exist an order-preserving surjection from a poset to its down-set lattice? As in title. By "down-set lattice," I am specifically referring to the set of lower sets of the poset in question, ordered by set inclusion.
I'm not even sure how to start. It feels true, because it looks like a function from a set to a subset of its power set, and I'd be very surprised if that were surjective. But I'm not sure where to take that hunch. I'd be very grateful for any help proving this claim.
Thanks in advance!
 A: My purpose in posting this answer is to present a slightly streamlined version of the proof in Eric Wofsey's answer and a slightly cleaner formulation of the result, and to put it in its historical context by giving references.
Note that, if the ordering of $P$ is trivial (no two elements are comparable), then we are saying that there is no surjection from the set $P$ to its power set $\mathcal P(P).$ In other words, this result is a generalization of Cantor's theorem, although the proof is quite different from the familiar diagonal proof of that theorem. This generalized Cantor theorem was originally proved by Gleason and Dilworth, and further generalized by Davies, Hayes, and Rousseau:

A. M. Gleason and R. P. Dilworth, A generalized Cantor theorem, Proc. Amer. Math. Soc. 13 (1962), 704–705.
Roy O. Davies, Allan Hayes and George Rousseau, Complete lattices and the generalized Cantor theorem, Proc. Amer. Math. Soc. 27 (1971), 253–258.

Terminology. A binary relation $\le$ is a quasi-ordering (or "preordering") if it is reflexive and transitive. A mapping $f$ between quasi-ordered sets is isotone (or "order-preserving") if $x\le y\implies f(x)\le f(y).$ A subset $D$ of a quasi-ordered set $Q$ is an lower set (or "down-set") if, for all $x,y\in Q,\ x\le y\in D \implies x\in D.$
The question asks for a proof of (essentially) the following:
Theorem. If $Q$ is a quasi-ordered set and $f:Q\to\mathcal P(Q)$ an isotone mapping, then the range of $f$ can't contain all the lower sets of $Q.$
Note that the result can be restated in a way that doesn't refer explicitly to the ordering of $Q$:
Theorem. Given a set $Q$ and a mapping $f:Q\to\mathcal P(Q),$ we can find a set $D\in\mathcal P(Q)\setminus\operatorname{range}(f)$ such that for all $x,y\in Q$ we have: $f(x)\subseteq f(y),\ y\in D\implies x\in D.$
Proof. For each ordinal $\alpha$ define
$$D_\alpha=\{x\in Q:f(x)\subseteq\bigcup_{\xi\lt\alpha}D_\xi\}.$$
Clearly, each $D_\alpha$ satisfies the condition: $f(x)\subseteq f(y),\ y\in D_\alpha\implies x\in D_\alpha.$ All we have to do now is find some $D_\alpha$ which is not in the range of $f.$ Clearly, $\beta\lt\alpha\implies D_\beta\subseteq D_\alpha,$ so we always have $\bigcup_{\xi\lt\alpha}D_\xi\subseteq D_\alpha.$ If the proper inclusion $\bigcup_{\xi\lt\alpha}D_\xi\subsetneqq D_\alpha$ held for every $\alpha,$ then $\alpha\mapsto D_\alpha$ would be an injection from the class of all ordinals into $\mathcal P(Q),$ contradicting Hartogs' theorem. Thus, there is an ordinal $\alpha$ such that
$$D_\alpha\subseteq\bigcup_{\xi\lt\alpha}D_\xi.$$Considering the least such $\alpha,$ I claim that $D_\alpha$ is not in the range of $f.$
Assume for a contradiction that $D_\alpha=f(x)$ for some $x\in Q.$ Then $f(x)\subseteq\bigcup_{\xi\lt\alpha}D_\xi,$ so $x\in D_\alpha,$ so $x\in D_\beta$ for some $\beta\lt\alpha,$ and $f(x)\subseteq\bigcup_{\xi\lt\beta}D_\xi.$ But now we have
$$D_\beta\subseteq D_\alpha=f(x)\subseteq\bigcup_{\xi\lt\beta}D_\xi,$$
so $D_\beta\subseteq\bigcup_{\xi\lt\beta}D_\gamma$ and $\beta\lt\alpha,$ contradicting the minimality of $\alpha.$
A: Here's one way to do it.  Suppose $P$ is a poset, $L$ is its lattice of lower sets, and $f:P\to L$ is an order-preserving surjection.  Let me first sketch the argument intuitively; if you just want a briefer formal proof you can skip down to the bottom.  The idea is to inductively rule out all possible "sizes" of $P$.  First, can $P$ be empty?  No, it can't, since $L$ always has at least one element, namely the empty set.  So since $f$ is surjective, there is some $p_0\in P$ such that $f(p_0)=\emptyset$.
OK, now can $p_0$ be the only element of $P$?  No, because once we know $p_0\in P$, there is a second lower set in $L$, namely $\{q\in P:q\leq p_0\}$.  So let's pick $p_1\in P$ such that $f(p_1)=\{q\in P:q\leq p_0\}$.
OK, now can $p_0$ and $p_1$ be the only elements of $P$?  Well, consider the lower set $D_2=\{q\in P:q\leq p_0\text{ or }q\leq p_1\}$.  We can't have $f(p_0)=D_2$, since $D_2$ is not empty.  And we can't have $f(p_1)=D_2$, since $f(p_1)=\{q\in P:q\leq p_0\}$ and $p_1\in D_2$ so that would imply $p_1\leq p_0$.  But then since $f$ is order-preserving, we'd have $f(p_1)\subseteq f(p_0)$, which is impossible since $f(p_0)$ is empty and $f(p_1)$ is not.
Maybe we're done now: maybe $P$ has no elements besides $p_0$, $p_1$, and $p_2$.  Nope: consider the lower set $D_3=\{q\in P:q\leq p_0\text{ or }q\leq p_1\text{ or }q\leq p_2\}$.  You can prove that $D_3$ cannot be equal to $f(p_0)$, $f(p_1)$, or $f(p_2)$, so there must be some new element $p_3\in P$ such that $f(p_3)=D_3$.
And so on.  By repeating this procedure inductively, you can keep getting more and more distinct elements of $P$.  You can continue this transfinitely, to define distinct elements $p_\alpha$ for all ordinals $\alpha$.  This is impossible, since $P$ is a set.

OK, now here's the argument written more formally.  We will construct a sequence of elements $p_\alpha\in P$ for each ordinal $\alpha$ by transfinite recursion.  Having defined $p_\beta$ for all $\beta<\alpha$, let $D_\alpha=\{q\in P:q\leq p_\beta\text{ for some }\beta<\alpha\}$ (the lower set generated by the $p_\beta$).  Since $f$ is surjective, there exists some $p\in P$ such that $f(p)=D_\alpha$.  Define $p_\alpha$ to be such a $p$.
I claim that $p_\alpha\not\in D_\alpha$ for all $\alpha$.  We can prove this by induction: suppose $p_\beta\not\in D_\beta$ for all $\beta<\alpha$ but $p_\alpha\in D_\alpha$.  Then $p_\alpha\leq p_\beta$ for some $\beta<\alpha$.  Since $f$ is order-preserving, this means $f(p_\alpha)\subseteq f(p_\beta)$.  By construction, $f(p_\alpha)=D_\alpha$ for all $\alpha$, so $D_\alpha\subseteq D_\beta$.  But $p_\beta\in D_\alpha$, so this gives $p_\beta\in D_\beta$, which contradicts the induction hypothesis.
Thus $p_\alpha\not\in D_\alpha$ for all $\alpha$.  But $p_\beta\in D_\alpha$ for all $\beta<\alpha$ and so $\beta<\alpha$ implies $p_\beta\neq p_\alpha$.  Thus $\alpha\mapsto p_\alpha$ is injective.  This is a contradiction, since $P$ is a set and this map is defined for all ordinals.
(Note that this argument uses the axiom of choice to pick a $p_\alpha$ such that $f(p_\alpha)=D_\alpha$ in each step.  However, you can avoid using the axiom of choice by making all the choices at once.  That is, instead of picking just one new element $p_\alpha$, let $P_\alpha$ be the set of all $p$ such that $f(p)=D_\alpha$.  You then correspondingly need to define $D_\alpha=\{q:q\leq p\text{ for some }\beta<\alpha\text{ and some }p\in P_\beta\}$.)
