Why are upper bounds only defined on subsets of preordered sets?

Question

So I recently was reading through spivaks calculus and just learned the least upper bound property. I noticed that in the definition the sets in consideration are always subsets of $\mathbb{R}$ This made me curious as to why only subsets of $\mathbb{R}$ were considered and why they have a notion of upper bounds.

Having looked up the definition of what an upper bound of a set is on Wikipedia I came across this rather confusing definition.

an upper bound or majorant of a subset S of some preordered set (K, ≤) is an element of K which is greater than or equal to every element of S.

I do not know what a preordered set is but im willing to guess that $\mathbb{R}$ is an example of one .

So basically my question is what is a preordered set and what about them makes them able to have upper and lower bounds?

Thanks in advance.

Answer 1

A pre-ordered set is a set $X$ with a relation, often denoted by $\le$, that obeys two simple axioms: $x \le x$ for all $x \in X$ and transitivity: If $x \le y$ and $y \le z$ both hold, then so does $x \le z$. If you think of $x \le y$ as "$y$ is at least as big as $x$", these properties make intuitive sense. A common example is set inclusion, so $X$ is all subsets of $\Bbb N$ say and $x \le y$ is defined to mean $x$ is a subset of $y$. Then $x=\{0,1\}$ and $y=\{2,3\}$ are not "comparable", neither $x \le y$ nor $y \le x$ holds, but the set $\{x,y\}$ has a least upper bound, namely $z=\{0,1,2,3\}$: both sets are smaller: $x \le z$ and $y \le z$ and there can be no smaller common upper bound. One can easily show that this $X$ with this (non-linear, unlike $\Bbb R$) pre-order also has the least upper bound property.

It's an important defining feature of $\Bbb R$ that it has this lub property, it's what distinguishes it from $\Bbb Q$; it "fills up" the rational number line with relevant real numbers like $\sqrt{2}, \pi$ and $e$.. That's why it's used a lot in analysis. But the phenomenon of preordered sets is everywhere in maths and having the lub property can be very relevant there as well.

Answer 2

$\mathbb R$ is a well-ordered set. For every $x,y \in \mathbb R$ either $x > y$ or $y<x$ or $x=y$ And, for every non-empty subset of $\mathbb R$ that subset inherits the ordering. So, those sub-sets are well-orderd.

However, just because the subset is well-ordered, doesn't mean it has the least upper bound property. The set of rational numbers such that $x\in\{\mathbb Q|x<\sqrt 2\}$ does not have a least upper bound in $\mathbb Q.$ i.e. $\sqrt 2$ is the least upper bound, but $\sqrt 2$ is not in $\mathbb Q$ yet there are infinitely many elements in $\mathbb Q$ that are arbitrarily close to $\sqrt 2$

Partially ordered sets and pre-ordered sets are not as restricted as well ordered and totally ordered sets.

There will be $x,y \in S$ such that $x>y$ however it may not be true for every $x,y$

For example, consider a tree. We might say $x>y$ if $y$ if there is a path down the branch from $x$ to $y.$ But, if $x,y$ lie on different branches we cannot say anything about how $x$ relates to $y.$

Hope this helps.