Another Opus Here

Jun 2014
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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 $\omega_1$.
This is the full statement you say is wrong. How so? The class of ordinals within a model of ZFC would imply that $\omega_1$ cannot exist because either statement #1 is true or statement #2 is true and both statements assert that $\omega_1$ cannot exist. Statement #1 asserts that $\omega_1$ cannot exist and is independent of the axiom of choice. Statement #2 asserts that $\omega_1$ might exist given the negation of the axiom of choice, but that $\omega_1$ cannot exist given the axiom of choice.
 
Aug 2012
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This is the full statement you say is wrong. How so? The class of ordinals within a model of ZFC would imply that $\omega_1$ cannot exist because either statement #1 is true or statement #2 is true and both statements assert that $\omega_1$ cannot exist. Statement #1 asserts that $\omega_1$ cannot exist and is independent of the axiom of choice. Statement #2 asserts that $\omega_1$ might exist given the negation of the axiom of choice, but that $\omega_1$ cannot exist given the axiom of choice.
Let's take this up later. It's not productive for you to keep repeating something that's manifestly wrong, and for me to become progressively more annoyed.

Earlier you said you'd found your mistake. I should have left it at that. I drew you out to see if you'd learned anything and that was an error on my part. If you were satisfied earlier we could leave it at that.
 
Jun 2014
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Earlier you said you'd found your mistake.
...typo on a particular post that I edited and re-posted, but not overall, assuming there are any.

Ok. Baby girl turned a week old today and then I came down with a case of the sniffles so I got banned from being near baby and instead have been doing all the laundry, dishes, etc.

I really do understand what was wrong with my original post. I thought, hey, we can't have a regular sequence that will be unbounded in $\omega_1$, but maybe we don't need one. At least, we can assume there isn't one and derive a contradiction. That works too, and that is exactly what I did. I assumed all the standard stuff and derived a contradiction... Set theory has been waiting for that one axiom that would tidy things up. Right now it's just a game of assume this and prove that but assume this and instead prove that. We haven't found the answer. I think we assume the axiom of choice, then assert both statements #1 and #2 (one of which must be true) imply that $\omega_1$ cannot exist, and we put to bed the long standing issue of the continuum. I'm all ears if you find something wrong and as always, I'm happy to be the crank until then.
 
Aug 2012
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...typo on a particular post that I edited and re-posted, but not overall, assuming there are any.

Ok. Baby girl turned a week old today and then I came down with a case of the sniffles so I got banned from being near baby and instead have been doing all the laundry, dishes, etc.

I really do understand what was wrong with my original post. I thought, hey, we can't have a regular sequence that will be unbounded in $\omega_1$, but maybe we don't need one. At least, we can assume there isn't one and derive a contradiction. That works too, and that is exactly what I did. I assumed all the standard stuff and derived a contradiction... Set theory has been waiting for that one axiom that would tidy things up. Right now it's just a game of assume this and prove that but assume this and instead prove that. We haven't found the answer. I think we assume the axiom of choice, then assert both statements #1 and #2 (one of which must be true) imply that $\omega_1$ cannot exist, and we put to bed the long standing issue of the continuum. I'm all ears if you find something wrong and as always, I'm happy to be the crank until then.
Do you accept the simple proof that using Choice there's an obvious uncountable ordinal, namely the one order-isomorphic to the well-ordering of the reals? If so, then you have an error and perhaps you can make some effort to find it. Yes? Meanwhile I have some other things to do and will leave this thread alone for a while. I haven't said anything new in several posts.
 
Jun 2014
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Do you accept the simple proof that using Choice there's an obvious uncountable ordinal, namely the one order-isomorphic to the well-ordering of the reals?
I would never think to question it. I'd call someone who did nuts. Then I proved otherwise. Sorry, is what it is. ... You can't say my result is wrong and ignore the proof. You have to say what about the proof is wrong. Then, as always, I'll gladly admit I'm wrong just like I always do. :p
 
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I have no such obligation. Why don't you take a couple of days and see if you can find your own logic error?
You certainly don’t have an obligation, but then why post at all? This isn’t a novel I’ve written here. You could have gone through every inch of the proof 20 times by now. :ninja:
 
Aug 2012
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You certainly don’t have an obligation, but then why post at all? This isn’t a novel I’ve written here. You could have gone through every inch of the proof 20 times by now. :ninja:
I feel that I have already expressed my opinion. Your argument appears sound as far as it goes, till near the end you make a leap and out of nowhere claim that you've resolved GCH and that the axiom of choice precludes the existence of an uncountable ordinal. Neither of these statements follow from what's gone before. When I ask you to make an actual argument, you send me a growly emoticon and a snide remark.

I post as long as it's interesting and fun. Those conditions are not presently met.

I feel some sense of responsibility as being a bit of a diversion from your real-life adventures of late. So I didn't want to be silent. But rereading this last post didn't make me want to take another run at your exposition. So I'm just posting something. Hope you're having a good week.
 
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Neither of these statements follow from what's gone before.
I walked you through it and you just said let's talk about it later. I'll make it easy for you.

1) Do you agree that either statement #1 or statement #2 must be true?

2) If statement #2 is true, which we are free to assume for the sake of argument, then $\omega_1$ would be equal to a countable union of countable sets. Do you agree?

3) A countable union of countable sets is countable within a model of ZFC. Do you agree?

4) Where $\omega_1$ is by definition the least uncountable ordinal, it cannot be countable, so we know that statement #2 cannot be true when the axiom of choice is assumed. Do you agree?

5) If statement #2 is not true, then statement #1 must be true. Statement #1 implies that $\omega_1$ is not the least uncountable ordinal though, so it also cannot be true. Do you agree?

6) If either statement #1 or statement #2 must be true, but neither one is, then we have an issue on our hands here. Do you agree?
 
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I feel some sense of responsibility as being a bit of a diversion from your real-life adventures of late. So I didn't want to be silent. But rereading this last post didn't make me want to take another run at your exposition. So I'm just posting something. Hope you're having a good week.
This is my second kid. The older one is almost eight, so I'm not a rookie in the whole parent business, but at the same time it's been long enough since I took care of an infant that I'm getting the new parent experience all over again to an extent. I am loving every minute of it so far. I am of course bias, but she is just soooooo freaking adorable!!

I know this thread is a bit of a brain teaser, but you've always helped me before Maschke and taught me an awful lot over the years. So did CRGreathouse and many others. You guys are great in that you help others take on this tough subject by demonstrating that it can be fun. I apologize if I've offended you. I trust that by now you know I am a silly goose.

I'm not sure I ever caught your name over the years. Mine's Brendon. I hope you are having a good week too.