Does every locally finite acyclic directed set embed into a linear order locally isomorphic to the integers? (Edit: extend, not merely embed.)

Question

Let $S=(S,\prec)$ be a set together with an acyclic binary relation, generally nontransitive. $S$ is locally finite if, for every element $x\in S$, the sets $\{w|w\prec x\}$ ("direct past of $x$") and $\{y|x\prec y\}$ ("direct future of $x$") are finite.

A linearly ordered set $(L,\le)$ is locally isomorphic to $\mathbb{Z}$ if every element of $L$ has a maximal predecessor and a minimal successor. The simplest example "larger than $\mathbb{Z}$" is $\mathbb{Z}\cup\mathbb{Z}$, in which each element of the first copy of $\mathbb{Z}$ precedes each element of the second.

An embedding of $(S,\prec)$ into $(L,\le)$, if it exists, is a set map $f:S\rightarrow L$, such that $f(x)<f(y)$ in $L$ whenever $x\prec y$ in $S$; i.e., it preserves $\prec$ strictly.

The question is then "does every locally finite acyclic directed set $S$ embed into a linearly ordered set $L$ locally isomorphic to $\mathbb{Z}$?"

Example: Let $(S,\prec)$ be the set (I will call it "Jacob's Ladder") with elements $\{x_i\}_{i\in\mathbb{N}}\cup \{y_{-j}\}_{j\in\mathbb{N}}$ and relations $x_i\prec x_{i+1}$, $y_{-j}\prec y_{-j+1}$, and $x_i\prec y_{-i}$ for all $i,j$. $S$ is locally finite; its transitive closure is $\mathbb{N}\cup-\mathbb{N}$, and it embeds into $\mathbb{Z}\cup\mathbb{Z}$ in the obvious way.

I have a messy ad hoc "proof" in the affirmative, but I have little confidence in it. The question seems simple enough that I'd expect it to be already known one way or the other. Can anybody help? I'd be satisfied with the countable case. EDIT: this particular "proof" doesn't work, but I still don't know the answer to the question.

EDIT: As Joel points out, the question is trivial as I first stated it. What I really want is an extension of a locally finite acyclic binary relation to a linear order isomorphic to the integers except possibly for a unique maximal and/or minimal element. The example of extending Jacob's Ladder to $\mathbb{N}\cup-\mathbb{N}$ illustrates this. Is it possible in general?

Answer 1

$\newcommand{\P}{\mathbb{P}} \newcommand{\L}{\mathbb{L}} \newcommand{\Z}{\mathbb{Z}}$

The answer is yes, and you don't need the locally finite hypothesis. Also, you may assume that the embedding is injective.

Theorem. Every acyclic relation $\langle S,\prec\rangle$ admits an order-preserving injective map from $S$ to a linear order locally isomorphic to $\Z$.

Proof. Suppose $\langle S,\prec\rangle$ is an acylic relation. Let $\leq$ be the reachability relation on this digraph, that is, the transitive closure of $\prec$. This is a partial order on $S$, because $\prec$ is acyclic. Since every partial order extends to a linear order, we may inject $\prec$ order-preservingly into a linear order $L$. Further, every linear order $L$ sits inside a linear order locally isomorphic to $\Z$, simply by taking $L$ copies of $\Z$, that is, by embedding $L$ into $\Z\times L$ with the usual order (reverse lexical order). This new order looks locally just like $\Z$, since what we have done is replace each point in the linear order with a copy of $\Z$. QED