discountbrains 29 Posted August 3 I thought the Journal of the American Mathematical Society might evaluate my proof so I sent it to them. Their secretary said it didn't meet the criteria for an article in their journal so they can't publish it which is quite understandable. It's only 4 lines. It's on Quora, but no one has commented on it. Maybe I should send it to a math professor or two. This is really bazaar; I can't find anything wrong with it. Much work has been done for nearly a century saying what I did is impossible and math is not open to consensus-its either right or wrong. I only considered the case of the possibility of there existing a function to well order the reals. No other way is useful to consider. I may paste it here, but I want answers from people who really know this stuff. 0 Share this post Link to post Share on other sites

wtf 72 Posted August 3 (edited) People who "really know this stuff" know that you're wrong, since they know the proof of the well-ordering theorem. That leaves those of us who know this stuff but also enjoy sparring with cranks alternative thinkers. So post your proof. Or don't. It's wrong regardless. But speaking for myself, I'd like to see your proof. Especially if it's only four lines. By the way here's the proof sketch of the well-ordering theorem. Consider the collection of well-ordered subsets of the reals. Verify that this collection satisfies the hypotheses of Zorn's lemma. Zorn then asserts the existence of a maximal well-ordered subset of the reals;, and this must in fact be the entire set of reals. Done. Edited August 3 by wtf 0 Share this post Link to post Share on other sites

John Cuthber 3416 Posted August 4 Interestingly, WIKI explains that "The standard ordering ≤ of any real interval is not a well ordering" in a single clause "since, for example, the open interval (0, 1) ⊆ [0,1] does not contain a least element."And I wonder if that's what's misled the OP 0 Share this post Link to post Share on other sites

wtf 72 Posted August 4 (edited) 1 hour ago, John Cuthber said: Interestingly, WIKI explains that "The standard ordering ≤ of any real interval is not a well ordering" in a single clause "since, for example, the open interval (0, 1) ⊆ [0,1] does not contain a least element."And I wonder if that's what's misled the OP Yes that's a good point. In fact on checking the Wiki article my proof sketch is a little inaccurate. When I said to consider all the well-ordered subsets, the question is, which well-order? The answer is that we actually Zornify on the collection of all possible well-orders on all subsets of the reals. It's a minor but important modification of what I initially said. The OP missed a chance to call me out on a nontrivial inaccuracy but didn't. Wonder if he's gone. Too bad, I love a nice chat about Zorn's lemma, well-ordering, the axiom of choice, and other aspects of nonconstructive math. But yes, as you note, the well-order of the reals is decidedly NOT the usual order. Edited August 4 by wtf 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 5 Exactly, the Zorn's Lemma and all these other theorems are dependent on the Axiom of Choice which of course is an axiom. I at this time just can't figure out what's wrong with my proof. The A of C has always troubled me some and I've heard it's trouble some others. Should I go ahead and paste my proof now or teaze a little longer? Certainnly there must be something wrong and I'll look ridiculous.....Next time I'll post it; I need to look over it omce more. Its been on Quora for several months and got no comments. 0 Share this post Link to post Share on other sites

wtf 72 Posted August 5 (edited) 59 minutes ago, discountbrains said: Exactly, the Zorn's Lemma and all these other theorems are dependent on the Axiom of Choice which of course is an axiom. I at this time just can't figure out what's wrong with my proof. The A of C has always troubled me some and I've heard it's trouble some others. Should I go ahead and paste my proof now or teaze a little longer? Certainnly there must be something wrong and I'll look ridiculous.....Next time I'll post it; I need to look over it omce more. Its been on Quora for several months and got no comments. Well, who would benefit here? You want someone who knows this material to critique your proof. I don't need to see your proof to know it's wrong. I sketched the standard proof and your comment is that AC "troubles" you. Whatever. If you deny AC you lose the well-ordering theorem and you're free to do that. Of course then you'll have a vector space without a basis, a set that's infinite yet Dedekind-finite, and you'll lose all of probability theory. A small price to pay for your "trouble." ZF minus Choice is perfectly consistent (if ZF is). So you have every right to adopt it as your personal axiom system. You're the one wanting assistance from someone who knows this material. Do you have something substantive to say or do you just want to jerk people around? I did search Quora and couldn't find your question, which you didn't bother to link. You won't get any more responses from me until you say something interesting. Edited August 5 by wtf 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 5 I could have just said look for "What's wrong with my thinking?" by George Heyer on Quora. 0 Share this post Link to post Share on other sites

wtf 72 Posted August 6 (edited) I found your Quora post here. https://www.quora.com/Where-is-my-flaw-in-thinking Here is the copy/paste for the benefit of readers keeping score at home. ======= let <* and ≤* be total order relations on ℝ. Let S be any complete (in relation to <*) subset of ℝ , let z ε S such that z ≤* x ∀x ε S. Then ∃T = S - {z}. And, T is complete also. So, ∀x ε T, z <* x. Hence any such z is not a least element of T. This implies there is no way to Well Order ℝ This shows for any possible well ordering of ℝ there still exists subsets of ℝ such that they have no least element in relation to the ordering. ======= I found two substantive and fatal errors. One, how do you define completeness for an arbitrary total order? Completeness is defined in terms of Cauchy sequences, and Cauchy sequences depend on a metric. If you are only given a total order, you can induce a topology on that order, called the order topology. https://en.wikipedia.org/wiki/Order_topology But the order topology may or may not be metrizable. If it's not, then you can't define a metric and then you can't define completeness. Secondly, just consider S = [0,1] in the reals with the usual order. Then z = 0 and all you're doing is showing that 0 is not in (0,1]. And in this case your T is not complete. I wrote a detailed, marked-up response. Evidently this site no longer supports MathJax. so I'll just upload some screenshots of my marked up doc. If you have any comments or questions let me know, but I'm not going to do any more screen shotting for this. Perhaps post this on a MathJax-friendly forum. If this site does support MathJax but I just couldn't figure it out, my apologies to the moderators. Edited August 6 by wtf 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 6 I'm going to look more closely at your argument. I knew I should have defined more precisely what I mean by 'complete". I'm not saying 'completeness'.I got my definition from an introductory math analysis book where they say a set is 'complete' (need to check the exact definition) if all the numbers of the reals are dense in it. In other words, it has all the numbers from the reals, x, such that a<x<b or it has no gaps in it. I know I need to be more formal than this. Of cpurse these are intervals with respect to <* rather than <. This is all I need to show because this is supposed to be a collection of all subsets. I believe I found a much easier way to write math than any of this stuff. I go to 'type math symbols' and just insert their symbols in my text in their space below then copy and paste the whole thing. 0 Share this post Link to post Share on other sites

wtf 72 Posted August 6 (edited) There 3 hours ago, discountbrains said: I'm going to look more closely at your argument. I knew I should have defined more precisely what I mean by 'complete". I'm not saying 'completeness'.I got my definition from an introductory math analysis book where they say a set is 'complete' (need to check the exact definition) if all the numbers of the reals are dense in it. In other words, it has all the numbers from the reals, x, such that a<x<b or it has no gaps in it. I know I need to be more formal than this. Of cpurse these are intervals with respect to <* rather than <. This is all I need to show because this is supposed to be a collection of all subsets. I believe I found a much easier way to write math than any of this stuff. I go to 'type math symbols' and just insert their symbols in my text in their space below then copy and paste the whole thing. There are two related but different definitions of "dense," and completeness has NOTHING TO DO WITH EITHER. * A set is dense if there's a third (distinct) point between any two points. So the rationals are a dense set and so are the reals. * A set X is dense in a set Y if every element of Y has elements of X arbitrarily close to it. For example the rationals are dense in the reals. * A Cauchy sequence is a sequence that should "morally" converge. For example the sequence 1/2, 1/3, 1/4, 1/5, ... should converge, and in the real numbers it does, to 0. But in the open unit interval (0,1), it does NOT converge, because there's nothing for it to converge to. Its limit, namely 0, isn't in the set. * A set is complete if every Cauchy sequence converges. The real numbers are complete, the unit interval [0,1] is complete, and the half-open unit interval (0,1] isn't. Note that completeness is a metric property and not a topological one. The open unit interval (0,1) and the real numbers are homeomorphic, that means they are identical topologically. But the reals are complete and (0,1) isn't. These are all perfectly standard definitions and any book on real analysis will define them the same way. Your posted proof is a train wreck full of misunderstandings and logic errors. Also, I'll assume you meant to say, "Thank you wtf for putting some effort into this to help me when nobody else in the world did" but simply forgot because you're so busy with other important things. ps -- Your approach is hopeless, by the way. Well-ordering has NOTHING to do with denseness or completeness. A well-ordered set looks discrete. It's got a first element, a second, a third, etc. A well-order can't be dense, because for example there are no integers between 2 and 3. In a well-ordered set, given any element there's a successor, with no other element between them. So your approach simply can't work. Likewise, completeness can't work either, since it's based on a metric, and most well-ordered sets have no metric defined on them. For example any transfinite ordinal number is well-ordered, but there's no metric defined on it. So the concepts you're using to attack the problem are the wrong tools. I do encourage you to learn more real analysis, it's a wonderfully interesting subject that will forever clarify your understanding of the real numbers. But flailing away with half-understood and misunderstood concepts, in order to disprove a standard theorem that's already been proven and accepted as true for well over a century, is not the most productive way to go about learning. Edited August 6 by wtf 2 Share this post Link to post Share on other sites

discountbrains 29 Posted August 6 Well thanks, wtf. The real idea behind my effort is that I've always noticed that when you try to come up with a well ordering for the reals there is always a set who's definition is the very thing that violates the ordering. My use of intervals is not a good example. Let me see if I can come up with a construction that works. Let me try this: Suppose A has a l.e. a with respect to <*. Now comsider B=A\{a}. Then a is not the l.e. of B. Your WO hyp says B has a l.e. b and a<b since b is inA. A and B are dense so there exists an x such that a<x<b . x can't be in B..... Anyway, this all leads to a contradiction. I would say the WO says all subsets of the reals defined in relation to <* have a l.e. in telation to <*. So, I think I can speak of them as interval in relation to <*. I hope I don't sound like I'm going around in circles. Yes, I need more thought on this; no doubt you'll tell me I need a lot more thought....OK, now I think I know what you are saying: If any set is WO it is turned into the form a_{1}.a_{2}, a_{3}, ..., but I'm saying for any WO there will always exist a set that can't be turned into this structure. I realized I haven't really proven anything. While I believe any ordering can select a group of numbers it can't find a l.e. for I have to show just how this works. 0 Share this post Link to post Share on other sites

wtf 72 Posted August 6 (edited) 2 hours ago, discountbrains said: Well thanks, wtf. The real idea behind my effort is that I've always noticed that when you try to come up with a well ordering for the reals there is always a set who's definition is the very thing that violates the ordering. My use of intervals is not a good example. Let me see if I can come up with a construction that works. Let me try this: Suppose A has a l.e. a with respect to <*. Now comsider B=A\{a}. Then a is not the l.e. of B. Your WO hyp says B has a l.e. b and a<b since b is inA. A and B are dense so there exists an x such that a<x<b . x can't be in B..... Anyway, this all leads to a contradiction. I would say the WO says all subsets of the reals defined in relation to <* have a l.e. in telation to <*. So, I think I can speak of them as interval in relation to <*. I hope I don't sound like I'm going around in circles. Yes, I need more thought on this; no doubt you'll tell me I need a lot more thought....OK, now I think I know what you are saying: If any set is WO it is turned into the form a_{1}.a_{2}, a_{3}, ..., but I'm saying for any WO there will always exist a set that can't be turned into this structure. I realized I haven't really proven anything. While I believe any ordering can select a group of numbers it can't find a l.e. for I have to show just how this works. [Replying point by point by hand, since the forum software doesn't seem to allow interspersed replies. Is there a better way to do this?] > Well thanks, wtf. You're very welcome. > The real idea behind my effort is that I've always noticed that when you try to come up with a well ordering for the reals there is always a set who's definition is the very thing that violates the ordering. Yes perfectly well understood. You will never come up with a well-order of the reals and neither with anyone else. It's a consequence of the Axiom of Choice (AC), so it is inherently nonconstructive. There are models of set theory in which there is a definable well-order of the reals. There are also models in which there is no definable well-order of the reals. It's all a matter of highly technical set theory. For us mere mortals, it's enough to know that AC is logically equivalent to the well-ordering theorem, and we will never ever visualize or imagine a well-ordering of the reals. Now, perhaps we should reject AC. Because AC does imply some very weird things, like the Banach-Tarski paradox, the existence of nonmeasurable sets, and the Infinite Hat problem (if you haven't seen this it's a total mind-blower). https://en.wikipedia.org/wiki/Hat_puzzle#Countably_Infinite-Hat_Variant_without_Hearing So maybe we should reject AC. But there's a problem. The negation of AC also has a lot of bizarre consequences, like a vector space without a basis; an infinite set that's Dedekind-finite; and the nonexistence of the Cartesian product of a collection of nonempty sets. Since we're damned if we do and damned if we don't with respect to AC, mathematicians choose to adopt AC because it gives more ways to prove interesting theorems. This shows that mathematics is not about certainty. It's about historically contingent aesthetic choices as to what constitutes valid mathematics.Whenever we use AC we are right at the heart of the philosophy of math. In the end there's no right or wrong. AC is mainstream today. In the future, who knows how people will regard nonconstructive math. They might reject it entirely. > My use of intervals is not a good example. Let me see if I can come up with a construction that works. Let me try this: Suppose A has a l.e. a with respect to <*. Now comsider B=A\{a}. Then a is not the l.e. of B. Your WO hyp says B has a l.e. b and a<b since b is inA. A and B are dense so there exists an x such that a<x<b . x can't be in B..... Anyway, this all leads to a contradiction. Ok there's an awful lot of confusion to untangle. [I don't mean for that to sound personal in any way. I am only critiquing your argument you originally requested. I share your intuitive sense of all this. A well-order of the real numbers is absurd on its face. We're in perfect agreement on that. We're just at different stages of acceptance of nonconstructive math. It's a psychological process]. Let me say this to start. Dense orders and well-orders are mutually exclusive. If a total order is dense, it's not a well-order. And if it's a well-order, it's not dense. At some level I think that's where you're coming from. You're showing that a dense order can't be a well-order. Your proofs are inaccurate but your intuition is perfectly correct. You're then concluding that no total order can be a well-order. That doesn't follow. All you're proving is that a dense order can't be a well-order. That is true! Also the converse. No well-order can be a dense order. You might have a go at writing a formal proof of that. > I hope I don't sound like I'm going around in circles. I think your instinct is correct, that no dense order can be a well-order. Your proofs are muddled and you should think about how you could prove this. Your other issue is that you are concluding that NO total order can be a well-order. But all you've proved is that no DENSE order can be a well-order. That's true, but it doesn't preclude some non-dense order from being a well-order, and that's exactly what happens. > Yes, I need more thought on this; no doubt you'll tell me I need a lot more thought Thought and study. The trick is to learn to get very rigorous and picky with the symbols. By the way that's where LaTeX or MathJax comes in handy. I know how to copy and paste symbols, in fact http://math.typeit.org/ is one of my favorite sites. However, and just take this as a suggestion, if you are interested in studying math ongoing, you should take the trouble to learn LaTeX. It produces beautiful output and it's much easier to read than the copy/paste stuff. It's the default for all mathematical writing these days in those online venues that support it. > ....OK, now I think I know what you are saying: If any set is WO it is turned into the form a_{1}.a_{2}, a_{3}, ..., but I'm saying for any WO there will always exist a set that can't be turned into this structure. Not actually following what you're saying here. But if you drop your belief that every linear order has a dense subset, you'll see that this is exactly the counterexample to your mental picture. A well-order is a linear order that has NO dense subset. You've actually given the proof. If x is the smallest element, then if you delete x there's no smallest element if the set was dense in the first place. So a well-order can have no dense subsets. You proved that! You know your mathematical sense is pretty good. You need to learn to do the formal symbolic part. That's what Real Analysis does. They run you through so many proofs that the symbology becomes second nature. If you're so inclined, you should methodically work through your book and do lots of exercises. > I realized I haven't really proven anything. You haven't proven your overall thesis that a well-ordering of the reals can't exist. But you have proved that a dense linear order can't be a well-order. That's a good thing to know. But what if there's some linear order that doesn't have a dense subset? That order is a candidate for our mythical well-order. You haven't covered that possibility. > While I believe any ordering can select a group of numbers it can't find a l.e. for I have to show just how this works. Not clear what this means. Well I wrote a lot of words, I hope some of it was helpful. I really encourage you to learn more real analysis and more set theory too. Feel free to ask questions. Edited August 6 by wtf 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 7 Yes, it's inconceivable how a dense set can be turned into a 'discrete' set of numbers-what I was trying to depict there. I haven't found a way to stop this absurd autocorrect on this Chromebook either.On thinking of this more I concluded we could consider any subset whatever no matter the structure with <*. Does that need to be proved or taken as an axiom?. Such a set I constructed doesn't have a l.e. But, guess what, saying any set can exists implies a set satisfying the AC exists. So, again this says the whole notion is undecideable. Spoiler 0 Share this post Link to post Share on other sites

wtf 72 Posted August 7 3 hours ago, discountbrains said: > Yes, it's inconceivable how a dense set can be turned into a 'discrete' set of numbers-what I was trying to depict there Well it's not a bad idea for the first try. But now we see there's no use even thinking about dense orders. They are not candidates for being a well-order or proving there isn't one. > I haven't found a way to stop this absurd autocorrect on this Chromebook either. Do you like your Chromebook? I need a new computing device and they look interesting. Then again do I really want to be even more tied in to Google? > On thinking of this more I concluded we could consider any subset whatever no matter the structure with <*. I'm afraid I don't know what you mean. Consider any subset of what, for what? Are you talking about the powerset of the reals, the set of all possible subsets of the reals? That's a big set. What does it have to do with you idea? You lost me here. Please me much more clear and specific. > Does that need to be proved or taken as an axiom?. Whatever it is you're talking about, you have not made it clear to me. "consider any subset whatever no matter the structure with <*" strikes me as meaningless. Can you please clarify what you are talking about? > Such a set I constructed doesn't have a l.e. I have no idea what set you've constructed or what you are talking about. The last time I looked, we had agreed that a dense linear order is not a candidate for being a well-order. Now whatever you are talking about makes no sense to me. I don't know what argument you're making or what you're trying to show. > But, guess what, saying any set can exists implies a set satisfying the AC exists. Any set can exist? What does that mean? And what is "a set satisfying the AC?" That also is inaccurate and murky. AC is a property about all collections of sets. A given set can't satisfy or not satisfy AC. Are you perhaps a little fuzzy on what AC actually is? > So, again this says the whole notion is undecideable. We already agree that AC is independent of the rest of math; but is universally accepted as one of the axioms of math on pragmatic grounds. I'm afraid this most recent post of yours does not make any sense to me at all. Please do clarify your idea, starting from the beginning and going step by step. 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 7 I gave an answer using a library computer and didn keep track of the time limit. It was completely lost. Iĺl have to do it again. Iḿ thinking Iḿ sticking to my original statements. As I said before if a order relation, <*, WOs the set of reals there has to be a l.e. according to <* in every possible subset of the reals. If every possible subset cannot be constructed there is no guarantee a subset made up of elements each from each subset of the reals exists either. I like this Chromebook yes and no. There are a number of limitations, but its easy to carry anywhere. There is no delete' key.You have to use backspace. I think I finally found out how to disable this auto correct. 0 Share this post Link to post Share on other sites

wtf 72 Posted August 7 (edited) 48 minutes ago, discountbrains said: I gave an answer using a library computer and didn keep track of the time limit. It was completely lost. Iĺl have to do it again. Iḿ thinking Iḿ sticking to my original statements. As I said before if a order relation, <*, WOs the set of reals there has to be a l.e. according to <* in every possible subset of the reals. If every possible subset cannot be constructed there is no guarantee a subset made up of elements each from each subset of the reals exists either. I like this Chromebook yes and no. There are a number of limitations, but its easy to carry anywhere. There is no delete' key.You have to use backspace. I think I finally found out how to disable this auto correct. > I gave an answer using a library computer and didn keep track of the time limit. It was completely lost. Iĺl have to do it again. Sorry to hear that. On the other hand it's been my longtime experience that whenever I write a lengthy screed and manage to lose it, my second effort always comes out much better. > Iḿ thinking Iḿ sticking to my original statements. But your original statement is wrong. You showed that no dense linear order can be a well order. But you didn't account for the case of some non-dense linear order. > As I said before if a order relation, <*, WOs the set of reals there has to be a l.e. according to <* in every possible subset of the reals. Every nonempty subset. Picky picky! > If every possible subset cannot be constructed there is no guarantee a subset made up of elements each from each subset of the reals exists either. The powerset axiom guarantees the existence of every subset of the reals. These subsets do not need to be "constructed," whatever you might mean by that. I do not have to explicitly exhibit each of the subset of the reals to know they exist. https://en.wikipedia.org/wiki/Axiom_of_power_set But you seem to be arguing against AC. That's pointless. You're perfectly free to adopt its negation, that gives a consistent set theory. There's no argument to be had. If you reject AC then well-ordering is false and we're done. You lose a lot of modern math and a good chunk of physics but you're not wrong. Only out of step with contemporary mathematical practice, which I perfectly well agree is historically contingent. > I like this Chromebook yes and no. There are a number of limitations, but its easy to carry anywhere. There is no delete' key.You have to use backspace. I think I finally found out how to disable this auto correct. I'll check it out. Google already owns my life anyway. Resistance is futile. Edited August 7 by wtf 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 7 I´m going to delete my post. What I was really trying to say is any well ordering is defined in a certain way (even though we may not know how its defined). Given any WO one can define a set (or rather a set exists) which has exactly the wrong properties for it to have a least element according to this well ordering. What I did does not show this. Iĺl have to say Iḿ a bit rusty at this and have not thought about this in depth in a lot of years. 0 Share this post Link to post Share on other sites

wtf 72 Posted August 7 12 minutes ago, discountbrains said: I´m going to delete my post. What I was really trying to say is any well ordering is defined in a certain way (even though we may not know how its defined). Given any WO one can define a set (or rather a set exists) which has exactly the wrong properties for it to have a least element according to this well ordering. What I did does not show this. Iĺl have to say Iḿ a bit rusty at this and have not thought about this in depth in a lot of years. If you leave the thread as it is, someone curious about this subject might come by in a few years and learn something. I think your instinct was right. You did show that a dense order can't be well-ordered. The point wouldn't be to give up on your studies. The point would be to learn more. As I said, your instincts are good. I'm trying to encourage you to learn a little more, not to give up. Remember that when Wiles announced his celebrated proof of FLT, someone found a significant error in his work. It took Wiles over a year to patch the problem and get a completed proof. You did a good job. You had an insight, you proved something interesting, and you realized that you had a logic error. Wiles did the exact same thing! If you are rusty at math, well you are less rusty now than you were before. Keep at it and soon the rust will turn to shiny stainless steel. 1 Share this post Link to post Share on other sites

wtf 72 Posted August 8 (edited) I ran across something that I didn't realize before, but it makes perfect sense. The well-ordering theorem is equivalent to AC. The well-ordering theorem says that ANY set can be well-ordered. That implies of course that the reals can be well-ordered. However, suppose we deny AC. Then there is SOME set that can't be well-ordered ... but it might not be the reals. That is: There is a model of set theory in which AC is false, and there is SOME set that can't be well-ordered, yet the real numbers are well-ordered! So I was wrong before when I said that you can deny AC and thereby prevent a well-order of the reals. On the contrary, the reals can be well-ordered (in some model of set theory) even in the absence of AC. Edited August 8 by wtf 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 8 (edited) WAIT A MINUTE! HOLD THE PHONE! I was right all along. I just need to put more detail in my proof. This proof is by contridiction. For suppose we say we have a WO, <*, for ℝ then for any S⊂ℝ there is an a∈S such that a≤*x for all x∈S. Now <* totally orders ℝ so for any a,b∈ℝ a<*b or b<*a. If T=S\{a} all x∈T,x≠a are the same as those in S and are ordered the same. We have T={x:a<*x and x∈S}. But, where is the l.e. of T in relation to <*? There is none. Therefore <* is not a WO of ℝ. About the subsets needing to be dense I threw that in at the beginning so there would be no argument about holes in the subsets. WO says ALL subsets have to have a l.e. Just pointing out one subset that doesn’t have a l.e. is sufficient. Yeah, I know I don like google either. First of all I don like the name. I was unhappy they bought youtube-about the best site on the web. Without the auto correct this leaves off the ´. Cannot even type an example. Edited August 8 by discountbrains 0 Share this post Link to post Share on other sites

taeto 15 Posted August 8 (edited) 25 minutes ago, discountbrains said: But, where is the l.e. of T in relation to <*? There is none. This would look more meaningful if you also explain why you think there is no least element of T, assuming T is not empty. Edited August 8 by taeto 1 Share this post Link to post Share on other sites

discountbrains 29 Posted August 8 T in this case has the same order relation <* as S, We sometimes write (S,<*) and (T,<*). There is no l.e. in T with this order relation is there? We cannot make up a new order relation for each set we encounter. About AC imples WO: WO implies AC which implies Zornś Lemma implies Maximality Principal (the one with ever chain has a greatest element) and I don know what else. WO implies AC. They all imply each other or are equivalent. We proved all this stuff in a topology cjass I took or the professor did. Years later I wrote out proofs of my own to see it I could do it, Its prtty abstract stuff. Oh, and the Continuum Hypothesis is also in these equivallent statements. 0 Share this post Link to post Share on other sites

taeto 15 Posted August 8 (edited) 28 minutes ago, discountbrains said: T in this case has the same order relation <* as S, We sometimes write (S,<*) and (T,<*). There is no l.e. in T with this order relation is there? We cannot make up a new order relation for each set we encounter. That is precisely the point: \( (\mathbb{R},<^*) \) by assumption is a particular kind of order relation with the property that each of \( (S, <^*) \) and \( (T,<^*) \), or any other \( (X,<^*) \) for that matter, has a least element wrt \( <^* \) so long as they are nonempty. It seems you are thinking that we have one order for \(S\) and then after removing the least element of \(S\) we have to start making an order for \(T\), which is \(S\) with one element removed. That is not the case. You really do keep the same order. Edited August 8 by taeto 0 Share this post Link to post Share on other sites

wtf 72 Posted August 8 (edited) 1 hour ago, discountbrains said: T in this case has the same order relation <* as S, We sometimes write (S,<*) and (T,<*). There is no l.e. in T with this order relation is there? We cannot make up a new order relation for each set we encounter. About AC imples WO: WO implies AC which implies Zornś Lemma implies Maximality Principal (the one with ever chain has a greatest element) and I don know what else. WO implies AC. They all imply each other or are equivalent. We proved all this stuff in a topology cjass I took or the professor did. Years later I wrote out proofs of my own to see it I could do it, Its prtty abstract stuff. Oh, and the Continuum Hypothesis is also in these equivallent statements. @taeto has already addressed and falsified your latest idea. I'll just add that you can get some clarity by running your argument on the natural numbers {0, 1, 2, 3, ...}. The naturals are well-ordered and 0 is the first element. So let T = {1, 2, 3, 4, ...}; that is, the naturals with 0 removed. Now what is the least element of T? It's 1. That's how well-ordered sets work. Wouldn't your argument "show" that the naturals aren't well-ordered? As @taeto already asked, why do you think T doesn't have a least element? > Oh, and the Continuum Hypothesis is also in these equivallent statements. No no no. CH is independent of all of them. In fact if you assume CH you get a nice well-ordering of the reals that you can almost visualize. The first uncountable ordinal can be proved to exist even without AC. If CH is true, the first uncountable ordinal well-orders the reals. And with a bit of study, you can visualize the first uncountable ordinal. It's just the set of all possible well-orders of the naturals. Equivalently, the first uncountable ordinal is the set of all countable ordinals. Edited August 8 by wtf 0 Share this post Link to post Share on other sites

discountbrains 29 Posted August 8 OK, I see what I did. That a set can be WO depends on the AC. The two are equivelent statements. What I showed is WO does not exist naturally on its own. Its like saying if we can live on air we do not have to grow food and we know the first part is impossible. What I did was just an exercise. Of course something like this could never overturn 100 years of history. Of course it cannot be that simple. If a set is WO then there you have your set of numbers for the AC. Yes, its clear the natural numbers can be WO, but my thesis was that not every set can be WO. If you accept the AC then of course you say they can. 0 Share this post Link to post Share on other sites