From what I've been taught in school, the well-ordering principle states that every non-empty set must have a least element. To me, the least element of some set $X$ is an element $a$ such that, for all $a \in X$, $a < x$.
This principle makes to me sense for most sets, but it not for a set with one element, as a consequence of my own definition of a least element. Say $X$ is a set with $1$ element. By the well-ordering principle, $X$ must have a least element, so there must be an $a \in S$ such that, $\forall x \in X$, $a < x$. But that can't be, since the only element in $X$ is $a$, and to say that $a$ is less than all elements in $X$ would be saying $a < a$, which is impossible! So, by this notion, $X$ does not have a least element.
This leads me to think that a set should have at least two elements in order for the well-ordering principle to apply–that way, two elements could be compared with respect to their sizes. However, whenever I share this with my peers and professors, they tend to regard my deduction as "wrong" somehow, without sufficiently explaining why.
The closest I have gotten to understanding why the well-ordering principle applies to sets with only one element is looking at the axiom of choice, which if I'm not mistaken is equivalent to the well-ordering principle. According to Wikipedia, the axiom of choice reads:
"For any set $X$, there exists a binary relation $R$ which well-orders $X$. This means $R$ is a linear order on $R$ such that every nonempty subset of $R$ has a member which is minimal under $R$."
Looking into the definition of a linear order, its first axiom is $a \leq a$. At first, I thought this might satisfy my question, but looking closer at the definition, there is still a concept of a "minimum," which to me does not make sense without at least two elements, for the reasons given prior. Moreover, the definition of a well-order seems to rely on the notion of a least element, which again brings me back to my prior confusion.
So, is there anything wrong with my reasoning? Is my definition of the well-ordering principle or of a least element of a set flawed, or different than that of what most people have? Why is it that the well-ordering principle applies for all non-empty sets, and not just for sets with at least two elements?
Best Answer
You have not distinguished an order $\le$ and a strict order $<$. Set theorists usually do not distinguish these two since one induces the other, but you should distinguish in this context. You can see that your definition of the least element does not make sense if we understand $<$ to mean a strict order: If $a$ satisfies $a<x$ for all $x$, then we have $a<a$, a contradiction.
When we say "Every set has a well-order," it means for every set $A$ there is a well-order $\le$ over $A$. Every well-order (or a linear order, in general) induces an associated strict order $$a<b \iff a\le b\land a\neq b.$$ Conversely, for every strict order $<$ we can associate an order by letting $$a\le b \iff a<b\lor a=b.$$
If you define the minimum as an element $a$ satisfying $a\le x$ for all $x$, then there is no contradiction. Clearly, every singleton $\{a\}$ has a well-order given by $${\le} = \{(a,a)\}.$$ And we can go one step down! Even the empty set has a well-order, and the empty relation is the well-order.