Relations, part 3

 [I continue writing in the form of a dialogue between Master (M) and Apprentice (A).]

M: As I promised last time we met, I'll explain to you today what is an ordering.

A: I've been waiting so eagerly for this! Please, go on!

M: Let's look at some set $X$. You now know what a binary relation on $X$ is. Let $<$ be a binary relation on $X$. By the way, since $<\subseteq X\times X$, we can have that $(x,y)\in <$ for some elements $x$ and $y$ of $X$. However, we usually write "$x<y$" instead of "$(x,y)\in <$".

A: Oh, yes, that's a standard mathematical notation. Now I understand what it formally means.

M: Good. Okay, so $<$ is a binary relation on $X$, but in order for it to be an ordering, it needs to have some additional properties. In fact, I'll first tell you what is a strict ordering. Binary relation $<$ will be a strict ordering just in case it is irreflexive and transitive. Being irreflexive means that for no $x$ in $X$ does it hold that $x<x$. In other words, no object is strictly smaller than itself. On the other hand, being transitive means that for all $x$, $y$, and $z$ in $X$, if it is the case that $x<y$ and $y<z$, then $x<z$. To rephrase it, if an object $x$ is strictly smaller than an object $y$ and if the object $y$ is strictly smaller than an object $z$, then the object $x$ is strictly smaller than the object $z$. Those are the conditions that strict orderings satisfy.

A: I see, I see. I'm wondering now, what are non-strict orderings than?

M: Staring from a strict ordering $<$, we can get another binary relation, $\le$, which is a non-strict ordering, or simply, "ordering". The meaning of "$x\le y$" is that "$x$ is smaller or equal to $y$". The natural formal definition of this relation is asserting that $x\le y$ holds if and only if $x<y$ or $x=y$. Properties of this relation are reflexivity, anti-symmetry, and transitivity. You already know what is transitivity. Reflexivity is might expect, that is, for all $x\in X$, it holds that $x\le x$. And finally, anti-symmetry is asserting that if $x\le y$ and if $y\le x$, then it must be the case that $x=y$. Does this make sense? If $x$ is smaller or equal to $y$ and if $y$ is smaller or equal to $x$, then it cannot be anything else but that $x$ and $y$ are equal.

A: Sure thing, sure thing. So you started from a strict ordering $<$ and obtained an ordering $\le$.

M: Yes, but you don't need to go that way. You can just say that some $⊴$ is an ordering on $X$ in the case that $⊴$ is a binary relation on $X$ which is reflexive, anti-symmetric, and transitive. You can then go on to define a binary relation $⊲$ by asserting that $x⊲y$ if and only if $x⊴y$ and $x\ne y$. It will turn out that $⊲$ is now a strict ordering! You see, orderings and strict orderings are very closely related!

A: I see! So then, if $m$ and $n$ are natural numbers, we'll have that $m\le n$ just in case that $m$ is an element of $n$ or that $m$ and $n$ are equal?

M: Well observed, young one, well observed!