Understanding why only finite sets have maximum/minimum

Question

I'm trying to understand intuitively why only finite sets are guaranteed to have a maximum and minimum. I believe, formally, the argument is that a finite set is closed (it contains all of its limit points, namely none), so it contains its supremum, which we therefore call a maximum, and similarly its infimum, which we therefore call its minimum.

If I have a set $a_1, \ldots, a_n$, I can surely take their maximum or minimum, but once I index by the natural numbers and try to take the max or min of $\{a_n \mid n \in \mathbb{N}\}$, I can't guarantee that I can do this. Surely $[0,1]$ has a minimal and maximal element, $0$ and $1$, respectively. But $(0,1]$ has no minimal element, $[0,1)$ no maximal element.

What exactly is intrinsic to infinite sets that makes this the case? Is it as simple that they need not be closed? In the $(0,1]$ case, I would prove that it has no minimum by assuming for a contradiction that it has one, say $x$, and then averaging $0$ with $x$. I can always find such an element, no matter how small I make $x$, because the set is infinite. But that sounds a bit "hand-wavy" to me.

Answer 1

It is not correct that only finite sets have a maximum and minimum. The condition for a set to have a maximum and minimum is compactness, not being finite. Compact sets can intuitively be thought of as those that are closed and bounded, so all finite sets are compact. Simply being closed is not sufficient, because $\mathbb{R}$ is closed, but has neither a maximum, nor a minimum.

Answer 2

As Sandejo pointed out, compactness also guarantees existence of a minimum or maximum. But it's not necessary. For instance, $[0,2]\backslash\{1\}$ is not compact but has a minimum and a maximum. There are no really useful equivalencies to the existence of both a maximum and minimum that I know of.

But about the second part of the question: what makes infinite sets different from finite ones? Well, finite ones can be covered by induction, while infinite ones can't. We can inductively prove that a weakly ordered set${}^\ast$ with $n$ elements, $n\in\mathbb N_+$, has a maximal element: For $n=1$ it's obviously true. The one given element is maximal. Then for $n+1$, we partition our set into two subsets, one containing a single element $M$, the other containing $n$ elements. The $n$ element set has a maximal element $N$. If $N\geq M$, then $N$ is maximal. Otherwise $M$ is maximal. This last step is not hard to do, and I leave it to you to formalize it, if you're interested in doing so.

Anyway, induction only allows us to prove things for finite $n$, so this approach fails for sets with an infinite number of elements. And as it turns out, it fails because the statement is simply not true for infinite weakly ordered sets.

${}^\ast$ a weakly ordered set is a set with an ordering relation $\leq$ such that $\leq$ is transitive and total (any two elements of the set are comparable, meaning that at least one of $a\leq b$ or $b\leq a$ is true).