Another Opus Here

Jun 2014
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Thank you! Mom and baby are healthy and resting comfortably. I have the rest of the week off to feed, burp, and change diapers to take the edge off mom. :)

I have read through what you wrote, made sense of some of it, and developed a much better general understanding. I had also watched a few college lectures on Youtube to try and get a better understanding of ordinals, and in particular, the class of limit ordinals but in general the prof’s dropped off before getting helpful for my purposes, even when students started to poke at similar issues (e.g., Does that mean $\omega_1$ is countable? Well no, but I can’t explain why...).
 
Jun 2014
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To be perfectly clear, we’re saying that $\phi(\omega_1)$ is not independent of the axiom of (countable) choice. If we adopt some choice, there is no such sequence $S$. If we do not adopt choice, there may (possibly) be a sequence $S$. This confuses me. We’re talking about the ordinals, which should basically follow the same construction in any model as far as I can tell, and so it seems to me there should be a truth value for $\phi(\omega_1)$ that is independent of the axiom of choice.

Finite sequences sufficient to meet requirements #2 and #3 are a dime a dozen. E.g.:
$$S = \{x \in \omega_1 : x \leq \omega\}, \{x \in \omega_1 : x > \omega \}$$
We can easily create a sequence of three elements meeting requirements #2 and #3, four elements, five, and so on. Apparently we can also have an uncountable sequence (pardon the lay-speak), $\omega_1$ itself under the standard $<$ ordering being an example, that suffices, so on that note I’m happy to make my 1st restriction less restrictive. That is, let $S$ be any infinite sequence (countable or uncountable). The proof proceeds the same from there, only this time we know $S$ exists. Now the axiom of choice must be negated... or if not we have a proof that the assumption of the existence of $\omega_1$ leads directly to a contradiction and the continuum hypothesis is solved? How about that?

I'll keep thinking on and off here...
 
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Aug 2012
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To be perfectly clear, we’re saying that $\phi(\omega_1)$ is not independent of the axiom of (countable) choice. If we adopt some choice, there is no such sequence $S$. If we do not adopt choice, there may (possibly) be a sequence $S$. This confuses me. We’re talking about the ordinals, which should basically follow the same construction in any model as far as I can tell, and so it seems to me there should be a truth value for $\phi(\omega_1)$ that is independent of the axiom of choice.
Why? By the same logic one could argue that there should be a truth value for the axiom of choice that's independent of the axiom of choice.

You yourself brought up the right example. Without choice there's a model in which the reals are a countable union of countable sets. I also ran across in my reading the fact that (absent choice) $\omega_1$ is a countable union of countable sets. Once you have that decomposition, you essentially have the partition you want. And with choice, such a decomposition can't exist; and (proof still needed) therefore there's no such partition.

Anyway no hurry, write when you can. I believe the key concepts are cofinality and König's theorem. And König's theorem is equivalent to the axiom of choice.

But the fact that the existence of your partition is equivalent to the axiom of choice isn't any more mysterious than any other equivalent, like every vector space having a basis or every surjection having a right inverse.
 
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poof, figured out my confusion, sorry
 
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How about sharing? The ordinals are interesting. What did we learn?
Those familiar with the thread up until now may find this post redundant save for my attempt to rewrite restriction #1.

We find $\phi(\gamma)$ is true for an ordinal $\gamma$ if and only if we have a suitable $S$ where, again, $S$ is a sequence of sets of ordinals that must adhere to three restrictions (the first loosened per above):

1) $S$ = an infinite ordinal sequence of sets, where “infinite ordinal sequence” implies the sequence can be either a traditional sequence of order type $\omega$ or of any ordinal greater than it, so long as the order type has no greatest element under the standard $<$ ordering (let $s_i$ denote an element of $S$ where $i$ is an ordinal number representing the position of $s_i$ within $S$),

2) $\{ s_i : s_i \in S \}$ is a partition of $\gamma$, and,

3) where $s_i, s_j \in S$, we have $s_i \prec s_j \iff i < j$ (the relation $\prec$ is defined in the original post).

For purposes of proving $\phi(\omega_1)$ true, select $\omega_1$ as the order type for $S$:
$$S = \{1\}_1, \{2, 3\}_2, \{4,5,6\}_3, \dots, \{\omega, \omega +1, \omega +2, \dots \}_{\omega}, \{\omega \cdot 2, \omega \cdot 2 +1, \omega \cdot 2 + 2, \dots, \omega \cdot 3\}_{\omega + 1}, \dots, \text{ (inclusive of all ordinals in } \omega_1 \text{)}$$

Either statement #1 or statement #2 below is true:

1) There exists an element $s_i \in S$ such that $|s_i| = |\omega_1|$.

2) There does not exist an element $s_i \in S$ such that $|s_i| = |\omega_1|$

Assume statement #1 is true. Then the least upper bound of $s_i$ is both uncountable and less than $\omega_1$, resulting in a contradiction:
$$(\text{sup }s_i < \omega_1 \text{ and } |\text{sup }s_i| = |\omega_1|) \implies \omega_1 \text{ is not the smallest uncountable ordinal.}$$

Assume statement #2 is true. Then we have an uncountable set, $\omega_1$, being equal to a countable union of countable sets. Assume the axiom of choice and statement #2 implies that $\omega_1$ is countable, contradicting the definition of $\omega_1$.

With the axiom of choice, there would be no smallest uncountable ordinal (and because there is no smallest, none at all), so the (general) continuum hypothesis would be solved.

If the axiom of choice is negated, it may be consistent that $\omega_1$ is uncountable despite being a countable union of countable sets.

Sound good?
 
Aug 2012
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With the axiom of choice, there would be no smallest uncountable ordinal (and because there is no smallest, none at all), so the (general) continuum hypothesis would be solved.
You lost me here, but if you're happy I'm happy. I don't see the relation to GCH. I should probably leave this alone at this point.
 
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Jun 2014
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You lost me here, but if you're happy I'm happy. I don't see the relation to GCH. I should probably leave this alone at this point.
Axiom of choice implies all sets may be well ordered in ZF. Great, there are no uncountable ordinals either with the axiom of choice, so case closed.