My question comes from personal investigation.
Suppose you have a poset $(X, \le_X)$. I would like to associate to all elements $x \in X$ a value $v(x) \in V$, where $(V, \le_V)$ is a totally ordered set, in such a way that for all $x,y \in X$ $$x <_X y \ \ \Rightarrow \ \ v(x) <_V v(y)$$ holds. I am wondering if this is always possible ($V$ is not fixed).
This is equivalent on asking the following:
Let $X$ be a poset. Can we find a totally ordered set $V$ and a strictly increasing map $v:X \to V$?
I realized that this question may be restricted to posets of the form $(2^X, \subseteq)$, since every poset $(X, \le_X)$ can be embedded in $(2^X , \subseteq)$ via $x \mapsto \{ y \in X:y \le_X x\}$.
Moreover, if $X$ is finite, the answer is yes, since we can define $$v: 2^X \to \Bbb{N}_{\ge 0}$$ as the cardinality function.
So, it remains the case when $X$ is an infinite set. And here I got stuck. So the question is:
Let $X$ be an infinite set. Does it exist a totally ordered set $V$ and a strictly increasing map $$v: 2^X \to V$$ ?
