Solving Ordinal Arithmetic: X Countably Compact but Not Compact

[Gear300]

Problem Statement:
Show that the set X of all ordinals less than the first uncountable ordinal is countably compact but not compact.

Let μ be the first uncountable ordinal.

The latter question is easy to show, but I stumbled upon a curiosity while attempting the former. In showing the former, I simply tried to show that every infinite subset of X should have a limit point (or in particular, an ω-accumulation point) in X. And so, in doing this, I needed to ensure that any infinite subset with μ as a limit point has another limit point in X. I reasoned that the first ω ordinals of this subset should only span a countable range of ordinals, since each of their co-initials are countable and a countable union of countable sets is countable. Any neighborhood of μ, however, is uncountable, so the limit point of the first ω ordinals of this subset cannot be μ. But when I considered the following set -

The sequence {ωⁿ} = ω, ω², ... , ωⁿ, ...

- it was hard to discern a limit point other than ω^ω. Aside from what the exact nature of the first uncountable ordinal is chosen to be, there should still be a limit point somewhere before ω^ω. So simply put, what ordinals exist between ω^ω and the sequence I presented?

 

[stevendaryl]

Problem Statement:
Show that the set X of all ordinals less than the first uncountable ordinal is countably compact but not compact.

Let μ be the first uncountable ordinal.

The latter question is easy to show, but I stumbled upon a curiosity while attempting the former. In showing the former, I simply tried to show that every infinite subset of X should have a limit point (or in particular, an ω-accumulation point) in X. And so, in doing this, I needed to ensure that any infinite subset with μ as a limit point has another limit point in X. I reasoned that the first ω ordinals of this subset should only span a countable range of ordinals, since each of their co-initials are countable and a countable union of countable sets is countable. Any neighborhood of μ, however, is uncountable, so the limit point of the first ω ordinals of this subset cannot be μ. But when I considered the following set -

The sequence {ωⁿ} = ω, ω², ... , ωⁿ, ...

- it was hard to discern a limit point other than ω^ω. Aside from what the exact nature of the first uncountable ordinal is chosen to be, there should still be a limit point somewhere before ω^ω. So simply put, what ordinals exist between ω^ω and the sequence I presented?

That limit is certainly $\omega^\omega$. Why do you think it might not be?

To say that the set of countable ordinals is countably compact is to say (I think) that for any countable collection, the limit is also a countable ordinal. To say that it is not compact is to say that this is not true for an uncountable collection (the limit is not a countable ordinal).

 

[Gear300]

That limit is certainly $\omega^\omega$. Why do you think it might not be?

To say that the set of countable ordinals is countably compact is to say (I think) that for any countable collection, the limit is also a countable ordinal. To say that it is not compact is to say that this is not true for an uncountable collection (the limit is not a countable ordinal).

I figured that since I am supposed to show that the set X of all ordinals prior to the first uncountable ordinal is countably compact, the set { ωⁿ } should have a limit point in X, since it is infinite and each ωⁿ is a countable ordinal. But ω^ω is an uncountable ordinal, so it isn't so much that I am denying that ω^ω is a limit point, but rather that there should be some other limit point in X for the hypothesis to hold. The neighborhoods being used are of the form (α,β) ⊆ X.

 

[stevendaryl]

I figured that since I am supposed to show that the set X of all ordinals prior to the first uncountable ordinal is countably compact, the set { ωⁿ } should have a limit point in X, since it is infinite and each ωⁿ is a countable ordinal. But ω^ω is an uncountable ordinal

No, it's not. It's countable.

 

[stevendaryl]

No, it's not. It's countable.

Exponentiation means something different for ordinals than for cardinals. For cardinal exponentiation, $\alpha^\beta$ means the cardinality of the set of all functions from $\beta$ into $\alpha$. For ordinal exponentiation, $\alpha^\beta$ is defined here:
http://mathworld.wolfram.com/OrdinalExponentiation.html

By definition, $\omega^\omega =$ the smallest ordinal greater than $\omega^n$ for every $n$

 

[Gear300]

Exponentiation means something different for ordinals than for cardinals. For cardinal exponentiation, $\alpha^\beta$ means the cardinality of the set of all functions from $\beta$ into $\alpha$. For ordinal exponentiation, $\alpha^\beta$ is defined here:
http://mathworld.wolfram.com/OrdinalExponentiation.html

By definition, $\omega^\omega =$ the smallest ordinal greater than $\omega^n$ for every $n$

I must have been trying to think outside the box. I had completely forgotten about the anti-lexicographic nature of the ordering. Thanks. Your answer has been enlightening. I may as well add this as a supplement:

https://math.stackexchange.com/questions/278992/how-to-think-about-ordinal-exponentiation