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.
https://en.m.wikipedia.org/wiki/Well-ordering_theoremWut?
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.
https://en.m.wikipedia.org/wiki/Well-ordering_theoremWut?
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 isnâ€™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.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 existence of an uncountable ordinal may be proved in ZF. We've been over this I thought. You're not making sense.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 isnâ€™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.
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.The existence of an uncountable ordinal may be proved in ZF. We've been over this I thought. You're not making sense.
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."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 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.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.
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 â€œinfinite ordinal sequenceâ€ 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?
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.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?:
On this we agree.pps -- I don't know what the first error is
The first error is that the quoted statement does not follow logically from anything that's gone before.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?