Extension theorem on acyclic relations By Sziplrajn's Theorem, we know that every partial order $\succsim$ (i.e. reflexive, transitive and antisymmetric relation) on a nonempty set $X$ can be extended to a linear order (i.e. a complete partial order) on $X$, where an extension of $~\succsim$ is a preorder $~\trianglerighteq$ such that for all $x,y \in X$, $x\succsim y$ implies $x\trianglerighteq y$.
I am looking for a variant of this theorem in which the extension should simply be a preorder (reflexive and transitive), and the initial relation would only be required to be acyclic (an acyclic relation is one for which there exists no list $(x_1, x_2, \dots, x_n)$ with $x_i \in X$ for all $i\in \{1,\dots,n\}$ and $x_1\succsim x_2 \succsim \dots \succsim x_n \succsim x_1$)
So can any acyclic relation be extended to a preorder?
 A: The question was not precise enough. In the way it is formulated, there is a trivial example. Let $ x \sim y$ stand for [$x \succeq y$ and $y \succeq x$]. Then for any $X$, the relation
$$ x \sim y, \qquad  \text{ for all } x,y \in X,$$
is a trivial preorder of $X$ that extends any acyclic relation on $X$.
Now, if we reinterpret the question as :

For any reflexive, acyclic and antisymmetric relation $\succ$ on $X$, does there exits a linear order of $X$ which extends $\succeq$?

we can still answer positively. First let $\blacktriangleright$ be  the transitive closure of $\succ$. The relation $\blacktriangleright$ exists because $\succ$ is acyclic : by definition, for any $ x_1\succ x_2 \succ \dots \succ x_n $ with $x_i \in X$ for $i = 1, \dots, n$, we don't have $x_n \succ x_1$. If $x_1 \succ x_n$, then $x_1 \blacktriangleright x_n$. In case $\neg(x_1 \succ x_n)$, because we don't have $x_n \succ x_1$, we are free to chose $x_n \succ x_1$.
Then because $\blacktriangleright$ is a partial order, Sziplrajn's Theorem applies and we can extend $\blacktriangleright$ to a linear order $\triangleright$, which is clearly an extension of $\succ$.
