wtf

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wtf last won the day on March 6

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  1. > I need to look at this wiki ordinal stuff but I suspect it won't change what I'm saying. I have in mind to write a bit of exposition that develops the ordinals in a very simple and natural way. My aim is to bring to life as best I can, a visualization of the space of the countable ordinals that might or might not be interesting to anyone who reads it, now or in the future. I'll give myself a couple of days to get this together, and meanwhile I'll fight the urge to post in this thread. Not saying I'll succeed. It's like flicking your tongue at a bad tooth. > I told you 3 times already "Ignore what I said before my last assertion". Yes and this is exactly the problem. Please read carefully what I'm about to write. Your pattern is to post a "definitive" version of the proof. Then someone finds an error, and you post, "You're right, that was an error." Now what NEVER happens is that you post a repaired version of your post. So what we have is a post containing a proof; followed by one or more posts saying, "Yes I see there's an error." So in YOUR mind there's some canonical version of the proof, which YOU think I see, but I assure you, I DON'T. I have no idea what the proof in your mind is; because you have not written it down. It's all in your head and frankly the ideas in your head about ordinals are fuzzy and malformed. You asked a long time ago for someone who "knows this material" to jump in. I am not a math prof as you asked once, just another forum denizen who loves math more than math loved me. But I know ordinals and I am qualified to tell you that you (a) Have not got a proof in general; and (b) Have not told me what your latest version of the proof is. That is my statement to you. > Quit trying to play like u don't understand my meaning. I am not doing that and I WOULD NOT do that. I'm earnestly trying to understand your idea. If I tell you I don't follow your argument and/or don't even know what your latest argument is; it's with the intention of giving you feedback about your mathematical exposition. If you would recognize that I'm putting my mathematical experience at your service, instead of being someone who is deliberately playing dumb to make you feel bad; all this would go a lot better. Of course nothing I could ever say would definitively prove that my intentions are of the sincere and hopefully helpful kind. If you prefer to think the worst of me, well ... actually you might not be alone around here! LOL. I will agree that I can speak sharply. If an argument is garbage I say so. That is not ever meant as a personal judgment. But I am qualified to judge proofs in this particular area. If I say your argument is garbage, you should try to do better. You should ask me WHY it's garbage and I'll try to explain it. Why do I insist that you write down a coherent proof? It's so that your ideas will get sharper. Your intuitions will get sharper. Writing down a proof makes you think right about the mathematical abstractions. You don't write proofs for others. You write them for yourself. Your ideas are muddled and the reason I know that is that you haven't written down a coherent proof. Ok I'm going to go work up a little writeup on the countable ordinals as they pertain to the question of well-ordering the reals without the axiom of choice. I should add that I"m trying not to show umbrage, but you should be aware that I was startled by your remark and did not respond to it kindly. You are wrong about my playing dumb. You are wrong about imagining you've expressed an argument. Those two things are related. And -- on the chance that I am wrong and you have in fact presented a perfectly coherent argument, which I'm too dumb to see -- then you should just repeat it for my dimwitted benefit; and in so doing, you would show that you appreciate my effort to understand you. I know, that's a lot to ask. I'm a dreamer.
  2. Can you please point me to the exact post that contains your proof? There are so many different versions. The proofs I've seen, I've debunked. I do see you often posting a proof then later saying you made a mistake, but not going back and reposting a complete proof. It's impossible for me to have any idea what you are referring to when you refer to "your last message," which most definitely did not contain a proof. You haven't got a proof unless you post one. You did say that you thought the [math]\omega + \omega[/math] example gives you problems. But then what about [math]\omega + \omega + \omega[/math] and so forth. The upward limit of that process is [math]\omega^\omega[/math]. You can keep going with that idea to get a countable tower of [math]\omega[/math] exponentiated with itself, which is a countable ordinal called [math]\varepsilon_0[/math]. And you can keep on going. These are all countable ordinals. The upward limit of all the countable ordinals is the first uncountable ordinal [math]\omega_1[/math], whose existence can be proved WITHOUT using the axiom of choice. Nothing you've written handles any of these cases. You have no proof. https://en.wikipedia.org/wiki/Epsilon_numbers_(mathematics) https://en.wikipedia.org/wiki/First_uncountable_ordinal
  3. Correct. You just proved that the usual order on the reals isn't a well order. However you have NOT shown that NO linear order on the reals can be a well order. Your proof fails. I looked back a couple of pages but could not find the supposed new version of the argument. I see no argument at all being made. Of course it is true that you can find a countably infinite ascending sequence of reals in a given linear order. For example let [math]<^*[/math] be defined as: all natural numbers [math]<^*[/math] everything else Now you have a countably infinite well-ordered subset. But we could extend it up through all the countable ordinals and you still wouldn't have a proof. You have not presented a proof that there is NO possible well order on the reals. But our current problem is that you have not presented a complete argument in quite some time; and I do not understand the purpose of the things you're saying. You take the trouble to point out that the usual order on the reals isn't a well-order? Why? We all agree that it isn't. Why bring it up now? Please give a complete argument. Give me something to work with. Nothing you say makes sense. In YOUR mind you think you're filling in details to something you've already explained. From where I sit there is no official version of your argument that I can refer to. I have no idea what you are talking about. Please: a) Point me to the post in this thread that expresses the coherent argument you're assuming I know; or b) Give such a coherent argument here. Write a complete argument so I can understand what you're thinking so I can respond. What is the point of your one-liners? I have no idea WHY you are telling me these things.
  4. I have no comments on matters of physics. > For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed. Ok. The usual order on the reals is a total order. We can form the set of all reals > 0 for example. What of it? Where is your argument? Premises, reasoning, conclusion. I see no argument.
  5. That's just not a coherent argument. I can't understand what you are trying to say.
  6. I didn't see an argument. Can you repeat it? All you said is that 0, say, is less than or equal to all natural numbers; and 14 is less than or equal to all of them except for some. But so what? There's no argument there. > BUT, we cannot come up with an order that well orders all sets. We can't "come up" with one, but we can easily prove that one exists in the presence of the axiom of choice. And even without choice, the reals might be well ordered even if some other set isn't. You haven't got a proof.
  7. > If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. Ok. Take the natural numbers. Then 0 is the smallest element. And S' might be all the numbers greater than or equal to 14. Ok. What does that prove? Of course there are always UPWARD chains, even infinite ones depending on the order. And can someone please point me to the definitive version of the proof? I seem to have missed the new proof. > For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too. Wait what? A moment ago m was the minimum of S, now you have some a that's smaller. What is the argument you are making?
  8. The idea of the proof seems to be that given an arbitrary linear order, you define [math]T[/math]. If it's not empty, you do your construction. If it is empty, you pick another interval.I have two points to make. * First, it has not been proven that this procedure will eventually find a nonempty interval. A lot depends on how the intervals are chosen. There are some details to be filled in if someone wants to claim they must eventually be able to choose a nonempty interval. I believe it's trickier than it looks. * Secondly, suppose you do find a nonempty interval. Then the argument as I understand it says that there's a smaller element, ad infinitum, giving a countably infinite sequence of ever smaller reals. This is impossible if the linear order happens to be a well order. Every downward chain of elements of a well-ordered set must be finite. That's because (as I noted earlier) there can be no infinite downward chains of set membership; and when it comes to ordinals, their order relationship is set membership. That's what it means that ordinals are inductive sets. Now discount you keep objecting that I'm "assuming" [math]<^*[math] is a well-order. No I'm not. I'm using it as follows. YOU claim that ANY linear order can NOT be a well order. I am pointing out that if ht DOES happen to be a well order, your proof doesn't work! So the only way for your proof to work is to assume that the linear order is not a well order. In other words you have to assume the thing you're trying to prove. The idea seems to be that given an arbitrary linear order, you define [math]T[/math]. If it's not empty, you do your construction. If it is empty, you pick another interval.I have two points to make. * First, it has not been proven that this procedure will eventually find a nonempty interval. A lot depends on how the intervals are chosen. There are some details to be filled in if someone wants to claim they must eventually be able to choose a nonempty interval. I believe it's trickier than it looks. * Secondly, suppose you do find a nonempty interval. Then the argument as I understand it says that there's a smaller element, ad infinitum, giving a countably infinite sequence of ever smaller reals. This is impossible if the linear order happens to be a well order. Every downward chain of elements of a well-ordered set must be finite. That's because (as I noted earlier) there can be no infinite downward chains of set membership; and when it comes to ordinals, their order relationship is set membership. That's what it means that ordinals are inductive sets. Now discount you keep objecting that I'm "assuming" [math]<^*[/math] is a well-order. No I'm not. I'm using it as follows. YOU claim that ANY linear order can NOT be a well order. I am pointing out that if [math]<^*[/math] DOES happen to be a well order, your proof doesn't work! So the only way for your proof to work is to assume that the linear order is not a well order. In other words you have to assume the thing you're trying to prove.
  9. Which set? You keep claiming T has a min when in fact I keep showing T might be empty. But now you say "this set" has no min. It would be helpful to me if you could repeat which set you are talking about at any given time.
  10. I said IF <* happens to be a well order, your proof would fail. How do you know it isn't?
  11. What about a well-order? Then there is no nonempty set that has no minimum element. How do you know <* doesn't happen to be a well order? You're claiming as a fact the very thing you are trying to prove.
  12. > I do think that the error you've identified is a relatively small error that can be worked around. It's interesting that you think that. I think the error is absolutely decisive and not repairable. Perhaps we are understanding the argument of @discountbrains differently. I take things to be thus: We play a game. I give discount (for short) a proposed linear order (= total order, same thing) [math]<^*[/math]; and discount proves that it can not possibly be a well-order. In this way discount would show that there is no well-order. Whatever [math]a[/math] and [math]b[/math] he picks, I choose the linear order: [math]a <^* b <^* \text{everything else in its usual order}[/math] In this order it's clear that [math]T = \{x \in \mathbb R : a <^* x <^* b\}[/math] is empty. Therefore there is no [math]z_0[/math] to start the "induction" (which of course isn't really an induction and doesn't prove what discount says it does). Now I think this is decisive. The compiler in my mind has detected a fatal error and I generally don't bother to read the rest except at a very cursory level. It doesn't matter that discount might have picked [math]c[/math] and [math]d[/math], I'd give the corresponding example. Remember discount's claim is that NO linear order can be a well order. I only have to produce a single example that breaks the proof. I think you might be thinking something like this. Correct me if I'm misunderstanding your point. You think you can fix this by saying that ok, [math]T[/math] is empty, so we don't care about it. We'll take some other interval. But I can give you the order [math]a <^* b <^* c <^* d <^* \text{everything} [/math] So ok you say never mind THAT interval. But I can do this all day. I could put countably many natural numbers, say all of them, in front of the "everything else" part. Or I could put all the even natural numbers, then the odds, and then everything else. So this would still be a countable initial segment, but it would be [math]\omega + \omega[/math], two copies of the naturals. Or three. I can just walk these examples right up through the countable ordinals. You might NEVER find a nonempty interval! And what if [math]<^*[/math] happens to actually be a well-order of the reals? It's some uncountable ordinal. You might find some nonempty [math]T[/math]'s but it wouldn't mean anything. It is a fact that descending chains of ordinals are always finite. Otherwise since ordinals are transitive sets, you'd have an infinite descending chain of element-hood, contradicting the axiom of foundation. You can't have infinite epsilon-chains. So at best you can only iterate the construction finitely many times and it doesn't prove anything. But the mental image here is that if the reals are well-ordered, it does in fact look like a long ordinal-indexed "sequence" of discrete elements. That's what ordinals look like. They are literally "one thing after another" like bowling balls. So it's discount's conceptualization that's still incomplete. He still conflates the dense order with the possible well-order. I better stop writing and let you tell me if you think I'm understanding this correctly. I could be misunderstanding everything, I would not be surprised. My point is that my first example actually implies all the rest of that! It's a refutation of not only the proof, but of all possible attempted rescues of the proof by trying to choose a different interval.
  13. > Let S=(0,1)⊂ R with 0<a<b<1. For any total order relation, <*, let T={x: a<*x<*b} and let S\{a,b}⊂T. Now, if we believe <* well orders T ∃z0∈T This is the same damn error again. Why won't you take the trouble to understand it? Let <* be the total order: "a <* b <* everything else in its usual order". Then T is empty. Your proof fails on the first line. There is no such z0. You keep making this same error and I keep giving the same counterexample over and over. This is the Groundhog Day thread. Not a shred of progress has been made since the very first post from August 2, 2018. This has been going on over a year now.
  14. What I do is compose in a separate text file, then paste into a forum window when I'm done. That way I reduce the chance of some hungry message board eating my post. I for one would like to see a new, clean, latest version so that we're all sure we're referencing the same proof. Perhaps as you write you might consider writing down numbered sentences or paragraphs instead of running everything together. Just for readability. If nothing else, just put some horizontal space every sentence or two.
  15. I've already explained that I"m NOT claiming I have a well-order. As you ALREADY AGREED, my showing that T might be empty is a refutation of your proof that there is no well-order. > I will get someone's attention somewhere. My 'proof(?)' is what it is. You haven't got a proof. And you even agreed that I found a flaw in your claimed proof.