# The Well-Ordering Principle

• Feb 17th 2006, 02:26 PM
topsquark
The Well-Ordering Principle
So they asked me, "Dan, what do you do for fun on a Friday night?" I answered "Well..."

I was thinking about Well-Ordered sets today and a thought struck me. I know that (assuming the Axiom of Choice) there exists as well-ordering for any set. Some sets are easy to well-order and others, like the reals, seem rather impossible.

My question is this: Has anyone ever managed to define a well-ordering on any set that satisfies the continuum hypothesis in some (other) topology? I suppose the question is roughly the same if I ask whether a set with the cardinality of the reals has ever been well-ordered (though I am aware that the two questions aren't exactly the same).

Just curious.

-Dan
• Feb 17th 2006, 10:12 PM
rgep
I'm not quite sure what you mean by a set satisfying the Continuum Hypothesis: that's an axiom about the non-existence of cardinals between those of the natural numbers and the reals. Anyway, you can certainly construct well-ordered sets of the same cardinality as the reals: for example the set of all binary sequences (maps from N to {0,1}) with the lexicographic order (compare two sequences from the zero-th term up and order them by their values where they first differ).
• Feb 18th 2006, 04:59 AM
topsquark
Quote:

Originally Posted by rgep
I'm not quite sure what you mean by a set satisfying the Continuum Hypothesis: that's an axiom about the non-existence of cardinals between those of the natural numbers and the reals. Anyway, you can certainly construct well-ordered sets of the same cardinality as the reals: for example the set of all binary sequences (maps from N to {0,1}) with the lexicographic order (compare two sequences from the zero-th term up and order them by their values where they first differ).

Sorry, I didn't mean continuum hypothesis, I meant "linear continuum." I hadn't heard of binary sequences. So if there exist well-ordered sets with the cardinality of the reals, why can't we set up a bijection between the two sets and define a well-ordering on the reals? The last I had heard no one had yet been able to define a such a well-ordering. Or am I mistaken?

-Dan
• Feb 18th 2006, 02:48 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Sorry, I didn't mean continuum hypothesis, I meant "linear continuum." I hadn't heard of binary sequences. So if there exist well-ordered sets with the cardinality of the reals, why can't we set up a bijection between the two sets and define a well-ordering on the reals? The last I had heard no one had yet been able to define a such a well-ordering. Or am I mistaken?

-Dan

Maybe you can say that since any set which has the same cardinality as the countinuum is isomorphic to the countinuum and since the countinuum cannot be well-ordered thus any set whose cardinality of the countinuum cannot be well-ordered.

It seems that $\aleph_0$ is the only infinite set which can be well-ordered.
• Feb 18th 2006, 03:47 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Maybe you can say that since any set which has the same cardinality as the countinuum is isomorphic to the countinuum and since the countinuum cannot be well-ordered thus any set whose cardinality of the countinuum cannot be well-ordered.

It seems that $\aleph_0$ is the only infinite set which can be well-ordered.

Presumably any set may be well-ordered. It is equivalent to the Axiom of Choice. I don't know the proof, but logically if you wanted to well-order something like the reals you would be making an infinite number of arbitrary choices, hence invoking the Axiom of Choice. But as you mentioned, it is hard to imagine well-ordering an uncountable set.

However, the reals are a linear continuum only because of the ordering we impose on them, so with a new ordering...

I don't know. Anyway, I was just wondering if anyone had heard of something that had been done since my textbook went to press. Thanks!

-Dan