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: 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?

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?

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.

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: 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?