Another Opus Here

Jun 2014
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I know the theorem. If you make an argument I'll read it. Your last two claims, absent an argument, look like trolling. There's no mathematical relationship between the claim you made and the Wiki page you linked.
The axiom of choice implies all sets may be well ordered. This means each can be bijected with an ordinal. The earlier proof shows it isn’t possible to have a smallest uncountable ordinal. Where there is no smallest, there are no uncountable ordinals due the hierarchy of the ordinals. If each set can be bijected with an ordinal and there are no uncountable ordinals, then each set is countable.
 
Aug 2012
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The axiom of choice implies all sets may be well ordered. This means each can be bijected with an ordinal. The earlier proof shows it isn’t possible to have a smallest uncountable ordinal. Where there is no smallest, there are no uncountable ordinals due the hierarchy of the ordinals. If each set can be bijected with an ordinal and there are no uncountable ordinals, then each set is countable.
The existence of an uncountable ordinal may be proved in ZF. We've been over this I thought. You're not making sense.

"'The earlier proof shows it isn’t possible to have a smallest uncountable ordinal."

Nonsense. Should I take you seriously and push back, or just assume you're not getting any sleep this week? Hard to know how to take your recent posts, which are quite divorced from mathematical sense.

"If each set can be bijected with an ordinal and there are no uncountable ordinals, then each set is countable."

That's just false. $\omega_1, \omega_2, \dots, \omega_\omega, \dots$ are all uncountable ordinals with or without choice. You're just totally misunderstanding the well-ordering theorem and I'm not sure if you want to engage or just state a sequence of misunderstandings.
 
Jun 2014
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The existence of an uncountable ordinal may be proved in ZF. We've been over this I thought. You're not making sense.
Perhaps, but not in ZFC. If you are going to object to my proof you need to find the first thing that doesn't make sense. Confinality was applicable with a sequence of order type $\omega$ but I fixed that. Go back.
 
Aug 2012
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Perhaps, but not in ZFC. If you are going to object to my proof you need to find the first thing that doesn't make sense. Confinality was applicable with a sequence of order type $\omega$ but I fixed that. Go back.
I'm not going back. You have confused yourself. You are making nonsensical claims. I'm happy to engage on things that at least make a little sense, but not your latest claims, which are "not even wrong."

Do you understand (at least in outline form) the proof in ZF that there is an uncountable ordinal? With choice the proof's even easier since the reals can be well-ordered. By claiming there are no uncountable ordinals you're not letting me take what you say seriously enough to respond.
 
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Jun 2014
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I'm not going back. You have confused yourself. You are making nonsensical claims. I'm happy to engage on things that at least make a little sense, but not your latest claims, which are "not even wrong."

Do you understand (at least in outline form) the proof in ZF that there is an uncountable ordinal? With choice the proof's even easier since the reals can be well-ordered. By claiming there are no uncountable ordinals you're not letting me take what you say seriously enough to respond.
I am making big claims, yes, but I also back them up. What, you don't have 5 minutes to spare? I don't care if I'm wrong, it's interesting and who know, even a blind squirrel finds a nut sometimes.

What's the first thing wrong below?:

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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I am making big claims, yes, but I also back them up. What, you don't have 5 minutes to spare? I don't care if I'm wrong, it's interesting and who know, even a blind squirrel finds a nut sometimes.

What's the first thing wrong below?:
I'm sorry, I haven't got five minutes for this. I slogged through it the other day and once was enough. You have a nice insight about cofinality and the fact that KÅ‘nig's theorem is equivalent to choice.

But if you assume choice, the reals can be well-ordered and are order-isomorphic to some ordinal, which must be uncountable. So your proof is wrong and your thinking is wrong to be repeating such an obviously false claim.

ps -- I eyeballed your quoted argument and I see no sense in it. You agree $\omega_1$ is an uncountable ordinal then you deny it then you name-check GCH without explanation. I don't get where you're coming from.

pps -- I don't know what the first error is, but this statement's an error.

"With the axiom of choice, there would be no smallest uncountable ordinal"

With the axiom of choice the reals are order-isomorphic to some uncountable ordinal hence there's a smallest one. So you have an error there. Do you at least agree that this is a proof in ZFC that an uncountable ordinal exists?


ppps -- Your statement "With the axiom of choice, there would be no smallest uncountable ordinal" DOES NOT FOLLOW from what's gone before. That's an error.
 
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Jun 2014
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pps -- I don't know what the first error is
On this we agree.

I also agree with you uncountably many times less one out of uncountably many times on this next statement:

"With the axiom of choice, there would be no smallest uncountable ordinal"

I know the basics. I haven't lost it. I will respectfully admit failure when we find that first error, ok?
 
Aug 2012
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On this we agree.

I also agree with you uncountably many times less one out of uncountably many times on this next statement:

"With the axiom of choice, there would be no smallest uncountable ordinal"

I know the basics. I haven't lost it. I will respectfully admit failure when we find that first error, ok?
The first error is that the quoted statement does not follow logically from anything that's gone before.