I find this weird, anyone any thoughts?. Let's start by defining awell-order.

A set S is said to bewell-orderedif each of its subsets has a least element. A moments thought tells you this implies that each element of S has an immediate predecessor. No problems here, I trust.

For example, the "counting" (aka natural) numbers are well ordered; 0 is the least element in N and 2 < 3. Allow me a little abstraction. I will say that, given the binary operation ≤, the follow relations must apply:

x ≤ x;

if x ≤ y and y ≤ x, then x = y;

if x ≤y and y ≤ z, then x ≤ z.

No difficulties here either, I imagine.

Let's now consider the set R of real numbers, with, say (0, 1) a subset in its standard topology. Is R well-ordered? What is the least element in this subset? Obviously it's not 0, what is it? Think of a real number ever so slightly more that 0, then half it, half it again and so on. You'll never find a least element, right? Moreover, what's the immediate predecessor to 1? (No, don't open that can of worms!)

So, thewell-ordering theoremstates that it isalwayspossible to place a well-order onanyset, which will, of course, include R. What gives? Here's my take, anybody any better ideas?

Well, we know this: as 0 ∉ (0, 1), 0 cannot be the least element of this subset. But we also may assume 0 < 1, so there must be some element x of R with 0 < x < 1 that is the least element in (0, 1). Can we say what it is? No. Does it exist? We must assume so.

Likewise, for y < z in (0, 1), we must assume there is some r ≠ y, r ≠ z such that r immediately precedes z. Can we say what it is? No. Does it exist? It must!