Preference Relation and Utility Function - Problem with inductive proof I have a problem with an inductive proof of the following result.
Theorem:
If $X$ is a finite set, a binary relation $\succ$ is a preference relation iff there exist a function $u:X\rightarrow R$ such that
$$
x \succ y \hspace{1cm}\text{iff}\hspace{1cm} u(x) > u(y) \hspace{3cm} \text{(1)}
$$
where, by definition, $\succ$ is a preference relation when it is asymmetric and negative transitive. 
Focusing on the sufficient condition, it is written on the book I am reading:
"If $X$ has $n$ elements and $\succ$ is a preference relation on $X$, then there exists a function $u: X \rightarrow R$ such that $\text{(1)}$ holds. To do so, first I must prove that this result is true for $n=1$. In this case $X$ is a singleton, say $\{x\}$, and I define $u(x)=1/2$, then neither $x \succ y$ nor $u(x)>u(y)$ is possible for $x,y \in X$ (the former because $\succ$ is asymmetric), so $\text{(1)}$ holds trivially".
Now, even if I know it sounds mathematically naive, I don' understand a couple of things.


*

*Where does this $y$ comes from? Considering we choose $n=1$, there is no $y$, and even if it is mentioned in relation to $\text{(1)}$ (which should be the case), still why do we have to add that "...$x \succ y$ [...] is possible for $x,y \in X$ (the former because $\succ$ is asymmetric)"? In particular, why do we have to mention asymmetry?

*Can we say that for $n=1$, whetever we define $u(x)$, the results holds? Because, if this is the case, again I don't see why mentioning asymmetry in the previous spot.

On $1$: You seem to be making an unwarranted assumption that $x\ne y$. There's no such assumption in the theorem. In fact, since $n=1$, necessarily $x=y$, and asymmetry is used to show that in this case $x\succ y$ does not hold (since otherwise $y\succ x$ would also hold, contradicting asymmetry).
On $2$: Yes, the value of $u(x)$ doesn't matter.
