The following is taken from chapter one of Efe Ok's book on Order Theory: Let R an acyclic binary relation on a nonempty countable set X. Prove or disprove: there is a map $f:X->\mathbb{R}$ with $f(x)>f(y)$ for each $x,y$ such that $xPy$ (i.e. $xRy$ and $yRx$ is false.)
My proof:Define $x\geq y$ iff $x=y$ or $(x,y)\in P^k$ for some $k$. Then we have $x \geq x$ and $z\geq y\geq x$ implies $z\geq x$. Asymmetry follows from the fact that $(x,y)\in P^k, (y,x)\in P^l$ implies $(x,x)\in P^{k+l}$, which violates acyclicity. So indeed, $\geq$ reflexive, transitive, asymmetric.
Let $X$ a countable set, $(c_n)_{n\in \mathbb{N}}$ the Cantor diagonalization of $\mathbb Q$. By countability $X=${$x_1,x_2,...$} for some sequence $(x_n)_{n\in \mathbb{N}}$. Let $X_n:=${$x_1,...x_n$}. We proceed by induction, covering the base case of $X_2$. Let $f(x_1):=0.$ If $x_2\geq x_1$ we let$f(x_2)$ the first $c_n$ greater than $0$ and if $x_1\geq x_2$ or $x_1,x_2$ not $\geq$ - comparable we let $f(x_2)$ the first $c_n$ less than zero. This is well defined due to asymmetry, transitivity of $\geq$. So then we have a map $f$ on $X_2$ (and trivially on $X_1$) such that for $(x_i,x_j)\in P$ we have that $f(x_i)>f(x_j)$, as desired.
Suppose then that such a map $f$ exists on $X_k$. Consider then $X_{k+1}=X_k \cup$ {$x_{k+1}$}. We need only find a desirable value for $x_{k+1}$. Let
$L_{-}:=max${$f(x_j)|x_{k+1}\geq x_j$} and $L_{+}:=min${$f(x_j)|x_j\geq x_{k+1} $}.
Suppose, for contradiction, that $L_{-}\geq L_{+}$. Then there exists $x^{+}, x^{-}$ with $f(x^{-})>f(x^{+})$ and $x^{+}\geq x \geq x^{-}$ Then by transitivity $x^{+}\geq x^{-}$. But then $f(x^{+})>f(x^{-})$. So indeed $L_{-}<L_{+}$. Then we let $f(x_{k+1})$ the first $c_n$ in $(L_{-},L_{+})$, the existence of which is guaranteed by density of $\mathbb Q$ in $\mathbb R$. Then for $x_i\neq x_j$ with $x_i P x_j$ we have that $x_i\geq x_j$ and so $f(x_i)\geq f(x_j)$. So indeed for any $n\in \mathbb{N}$ there is a map $f$ from $X_n$ into $\mathbb R$ with $f(x_i)>f(x_j)$ if $x_iPx_j.$
We use this construction then for $X$. Suppose, for contradiction, that this construction fails for $X$. Then there are $x_i, x_j$ with $x_iPx_j$ but $f(x_j)\geq f(x_i).$ Yet we know that the construction cannot fail on $X_{max(i,j)},$ a contradiction. So this construction works on the countable set.
So for $R$ an acyclic binary relation on a nonempty countable set $X$ there is a map $f:X\rightarrow \mathbb R$ with $f(x)>f(y)$ if $xPy$.
Questions: Does this proof work? It seems too long, complicated for what seems like a straightforward construction and problem. Am I overlooking something here? Thanks much.