I have a problem with the logic behind the fact that, given a nonempty set X, the diagonal relation $D_X := \{ (x,x) : x \in X \}$ is a partial order on $X$.
More specifically, my problem is with the fact that it looks to me that the result could be easily estabilished in the following way:
Tentative "Proof":
Reflexivity: The relation $D_X$ is reflexive by definition.
Transitivity: The relation is vacuously transitive, because there are no ordered pairs of the form $(x, \cdot)$, where the dot represents any element different from $x$.
Antisymmetry: The relation is vacuously antisymmetric for the same reasons that apply to transitivity.
In other words, as I pointed out in other previous questions, I have problems with the "range of application" of vacuously true statements, i.e. I don't really know when I should actually stop to apply this line of reasoning.
[For example, for a while I thought I could actually prove this is a linear order, out of this applying the vacuous truth anywhere. However, I realized that from $$\forall x, y \in X ((x,y) \in D_X \vee (y,x) \in D_X )$$ we get $(x,y) \notin D_X \Longrightarrow (y,x) \in D_X$, where the vacuous truth of the antecedent does not apply.]
Notice that I have an alternative proof (one I have no doubts about), definitely less concise, that avoids any reference to vacously true properties.
Thus, getting to the questions, does what I wrote (including the linear order excursus) make sense?
Is my tentative proof sound?
Any feedback is more than welcome. Thanks a lot!
The term "vacuously" means that there are no counterexamples. That is when we have an implication "If ... then !!!", but the "..." part never happens. But this is not the case, it's just happened that there are no interesting cases to verify.
This is because the definition of a transitive relation is not "For every $x,y,z$ distinct we have ..." but rather "For every $x,y,z$ ...". The case $x=y=z$ is trivial, and so non-interesting. But it means that the statement does not hold vacuously as you said.
If $(x,y)$ and $(y,z)$ are both in $D_X$ then by definition $x=y$ and $y=z$, therefore by transitivity of the equality relation $x=z$ and so $(x,z)\in D_X$ as well.
So there is something to verify here. It's just that this something is very trivial to check.