Well-ordering of reals

[Ookke]

If < is any total order in R, isn't it always possible to construct an infinite sequence
x1 > x2 > x3 > ... > y for some limit point y in R.

It seems to me that {x1, x2, x3, ...} is then a subset of R that does not have
least element in this ordering. No total order in R can be well-order?

[Preno]

The axiom of choice says that such a sequence exists. You may or may not accept that axiom.

[Hurkyl]

If < is any total order in R, isn't it always possible to construct an infinite sequence
x1 > x2 > x3 > ... > y for some limit point y in R.
Why would it?

Do keep in mind that a total order on R does not have to bear much resemblance to the usual order.

[Ookke]

Why would it?

Do keep in mind that a total order on R does not have to bear much resemblance to the usual order.

It's just an intuition of that uncountable set is necessarily kind of dense for any order, not just the common order.

I had in mind two assumptions which seem quite natural:
1. Uncountable set of reals must have uncountable interval (a,b) for any order <.
2. It's always possible to split uncountable interval into two parts, both uncountable.

Then what follows, we can split (a,b) into two parts, select x1 from the upper part. Then split lower part again into two parts, select x2 from the new upper part, and so on. This way should be formed a sequence {x1, x2, ...} where x1 > x2 > ... for the order <, and the sequence does not have a least element.

I'm just developing the idea, but if this is already flawed, I would appreciate any point.

[Hurkyl]

2. It's always possible to split uncountable interval into two parts, both uncountable.
This is demonstratably false, by a rather slick argument. It depends on two facts:

(1) The class of all ordinal numbers is well-ordered
(2) There exists an uncountable ordinal

Therefore, there exists a smallest uncountable ordinal number (http://en.wikipedia.org/wiki/First_uncountable_ordinal); let's call it \omega_1.

Now, consider the (uncountable) half-open [0, \omega_1) of ordinal numbers. If you choose any ordinal \alpha in this interval, we have:
* [0, \alpha] is countable,
* [\alpha, \omega_1) is uncountable.
In particular, every Dedekind cut of [0, \omega_1) has a countable lower part.

[Ookke]

Ok, thanks. Had fun trying, now I need to learn some ordinals.

[Hurkyl]

Since you're researching it anyways -- it might interest you to look at another counterexample you can create using \omega_1: the long ray (http://en.wikipedia.org/wiki/Long_line_(topology)).

[Ookke]

Therefore, there exists a smallest uncountable ordinal number (http://en.wikipedia.org/wiki/First_uncountable_ordinal); let's call it \omega_1.


I found an interesting article about smallest uncountable set:
http://www.math.fsu.edu/~bellenot/class/su08/found/other/omega-one.pdf

From there, I understood that it's possible to have uncountable set X for which every seg(y) = \{x\in X | x < y\} is countable. This is exactly against an idea I had for proving my second assumption (yes it hurts, but I don't care).

But why is it impossible to have a sequence \{x_n\} in X that takes up the whole X, i.e. for any x, x < x_n for large enough n?

Existence of such a sequence seems almost obvious to me, and this would be enough to prove X as union of seg(x_n) countable. Or is it just that because of initial assumptions, we must deny this kind of sequences. Do you happen to know any "real-world" example where any sequence cannot take the whole set?

[Hurkyl]

But why is it impossible to have a sequence \{x_n\} in X that takes up the whole X, i.e. for any x, x < x_n for large enough n?
...
Existence of such a sequence seems almost obvious to me, and this would be enough to prove X as union of seg(x_n) countable.
And since we know \omega_1 isn't countable, that tells you why it's impossible.

If this seems obvious to you, that just means your intuition isn't equipped to deal with orders of uncountable cofinality (http://en.wikipedia.org/wiki/Cofinality). ("cofinality" measures this quality of orders)


Do you happen to know any "real-world" example where any sequence cannot take the whole set?
"Real world" is very much in the eye of the beholder. As I mentioned in my previous post, \omega_1 can be used to construct counterexamples, such as the long ray -- which tells geometers if they want to study more "normal"-looking spaces, they need to require that manifolds aren't merely Hausdorff and locally Euclidean... one must also insist on second countability (http://en.wikipedia.org/wiki/Second_countable).

I've personally explicitly used uncountable ordinals in proofs because my preferred method of applying the axiom of choice is to use the well-ordering theorem along with transfinite iteration1.

I've also explicitly used them to help understand nonstandard analysis. (In particular, to understand just how ugly the hyperreals look when studied externally)

1: This refers to a form of transfinite induction that I find particularly natural, but I don't think it's a standard term.