wtf

pps -- By the end you claim T = S. But that's absurd. I already showed you the example of the 0 <* 1 <* everything else order in which S = (0,1), the uncountably infinite open unit interval; and T = {x : 0 <* x <* 1} is empty. If you would take the time to understand this example you would find it enlightening.

ps -- You define T = {x : a <* x <* b} and then in your next sentence you claim a and b are elements of T. That's ridiculous.

I found a mistake. I found the same mistake multiple times over the past several months that this thread has been going on. Your claim is that there is NO well order of the reals. Your proof consists of picking SOME order and showing it's not a well-order. But you have to prove that EVERY order is not a well-order. So you have to consider the case where T might be empty. And since you never consider that case -- indeed, since you give no appearance of even understanding the point -- you have not proven your claim. BTW pasting an image makes it impossible for me to reply line by line. But your approach is wrong. You have to prove that <* can NEVER be a well-order. You can't cherry-pick some <* where T is nonempty. Because there are perfectly obvious <* orders, and choices of a and b, for which T is empty. Your proof fails in your second sentence.

T CAN be nonempty but it MAY be empty. Your proof fails already in the second sentence.

> The proof is by contradiction. Let S =(a,b) where a,b∈ R, 0< a <b. Let T= {x: a<*x<*b} where <* is a total order relation other than, <. I define S this way so all z∈S are in T. But that is TOTALLY FALSE. You have S = (a,b) in the standard order. But in the <* order, T may be empty. I've already shown you the counterexample. In the orders 0 <* 1 <* everything else, we have S = (0,1), the USUAL open unit interval. But T is the set of reals strictly between 0 and 1 in the <* order, and T is empty. We've had this same conversation many times already in the past few weeks. In your example with a = 0 and b = 1, S is the usual unit interval, and T is empty. Do you understand that? > I define S this way so all z∈S are in T. But that is nonsense. S is the usual standard open unit interval, and T is empty exactly because of the way I defined <*. Which part of this is confusing you? I have given the same example to the same argument maybe six or eight times already over the past couple of months. If the <* order is: "0 <* 1 <* everything else" then T = {x : 0 <* x <* 1} is empty. Try to name a real that's in T under the order "0 <* 1 <* everything else". That's a perfectly legitimate total order and T is empty. There's nothing between 0 and 1 in this order. > S is an denumerable set so T is also. No, S is uncountable as is perfectly obvious, it's the open unit interval in the standard order. And T is empty because there's nothing between 0 and 1 in the <* order. But your eternal error is that you think T is S or even related to S. It's not. T is an open interval in the <* order. It's empty by the way I defined <*. The fact that S has lots of reals in it means nothing. <* is an entirely different order. Don't you see that there are NO REALS AT ALL between 0 and 1 in the <* order? > I am proving by induction we can generate a denumerable string of minimums of every subset of T arising from after the previous minimum was deleted. You could never do that because T MAY BE EMPTY.

My "0 < 1 < everything else" example was for the purpose of falsifying this particular point. You said "Both of you" agree S is nonempty. I don't know who both of you is, but I certainly don't agree S is nonempty. I gave an example in which S is empty. Now of course you can run your argument on the "everything else" part by picking a and b in the everything else portion of the ordering. BUT I AM NOT USING THAT EXAMPLE except to falsify your original claim that S is nonempty. Suppose the order is 0 < 1 < 2 < everything else. Then you might have picked 1 and 2 as your a and b and your S is still empty. Suppose you pick a and b in the everything else part of this new order. Then I'll give the example 0 < 1 < 2 < 3 < everything else and you might have picked 2 and 3 as a and b and S is empty. So what you did is take a specific counterexample I gave to your specific claim, and pretend that I gave the example as some general refutation of your argument. It's not. It's a specific refutation of a specific claim. But the 0 < 1 < everything else order is an example of a CLASS of counterexamples that will refute your argument no matter how you clip an initial segment. Your argument fails because no matter what a and b you choose, there might not be any other reals between them in the <* order. Your order might be: Everything less than pi <* pi <* pi + 1 < everything else. Then your a and b might be pi and pi+1, and your interval is empty. So I don't even need initial segments. The discrete portion of the order might happen in the middle. You can never be certain that your (a,b) interval in the <* order isn't empty. > I need to look up and see if the union of any collection of any countable sets is still countable and maybe use this result to help prove my claim. You need the axiom of countable choice to proof that. In the absence of countable choice it's not true in general. Although it's true for specific cases like the snaking diagonals proof of the countability of the rationals. https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable

How do you know your set can't be well ordered? That's something you are required to prove. My example falsifies your proof and any similar proof. I have taken YOUR EXAMPLE and shown that the set S might well be empty, making your proof fail. > I'm putting my argument together to post. Standing by to receive your transmission.

> What you are saying then is that you already assume S is WO and hence u can get the result it is WO. No that is not true. My order is 0 <* 1 <* everything else. Then the set of reals strictly between 0 and 1 is empty and has no minimum. It's not a well-order but it falsifies your argument. I'm not using a well-order, I'm not assuming anything is a well-order. But I'm not assuming S is well-ordered. You claimed S is nonempty. I showed an example where S is empty. > I'll have to think about your claim that the set we are arguing about is empty. What is the set of integers strictly between 0 and 1? > If u can give an example set where u say its empty then u can deny my claim. I gave the example. Twice in the same post. A third time just above. You have the order 0 <* 1 <* everything else. Then the set of reals strictly between 0 and 1 is empty. That's the example that falsifies your argument. > I'm presenting a different set. No matter what a and b you pick, I'll just give you the order a <* b < everything else, and then the set of reals strictly between a and b is empty. > What puzzles me is what is the purpose of all this. Ordinals are connected with proof strength in logic. They're used to develop the von Neumann hierarchy of sets. They're used to study well-orders. They're an interesting source of counterexamples in topology. For example the first uncountable ordinal, with the order topology, is compact but not sequentially compact. That's because sequences are "too short" to reach their limit. They're the basis of transfinite induction. Lots of other uses in math. The ordinals were discovered by Cantor in his investigations of the zeros of Fourier series. That's an interesting historical point. Set theory came directly out of the study of the physics of heat transfer. https://en.wikipedia.org/wiki/Von_Neumann_universe https://en.wikipedia.org/wiki/Transfinite_induction https://en.wikipedia.org/wiki/Sequentially_compact_space#Examples_and_properties > Anyway, I'll think about all this. Do you understand the 0 <* 1 <* everything else order, and how there are no reals strictly between 0 and 1 in this order?

I'll respond to the quoted text portion of your post, which I think was written by you as a question. Also my previous post has a copy/paste error, the second half where it starts again is the correct post. > Let S be a subset of R defined as S={x: a<*x<*b} . Both of you here agree S exists and is nonempty. I agree to no such thing. You haven't said what a and b are. * Suppose [math]a \geq^* b[/math]. Then [math]S[/math] is empty. * Suppose [math]<^*[/math] happens to be a well-order. Then perhaps [math]a <^* b[/math] and there is nothing between them. Perhaps [math]b[/math] is the successor of [math]a[/math]. * But maybe [math]<^*[/math] isn't a well-order. It doesn't have to be. What if the order is [math]a <^* b[/math] followed by all the other real numbers in their usual order? For example let's order the reals as 0, 1, everything else. Then your [math]S[/math] is empty. So I do not agree to what you say unless you tell me what [math][/math] and[math]b[/math]are, and enough about the order to know whether there is anything between them. > Here we have a is the greatest lower bound of S with respect to <^*. Ok. > You say all subsets of R must have a minimum, M. Under the assumption that [math]<^*[/math] is a well-order, all nonempty subsets have a minimum. Note well. Nonempty subsets. > Now M is in S and a<*M. Not necessarily. Let [math]<^*[/math] be the order on the reals: [math]0 <^* 1 <^* [/math] everything else. Let [math] S = \{ x \in \mathbb R : 0 <^* x <^* 1 \}[/math] Then [math]S[/math] is empty and does not have a minimum element. > Therefore the set {x: a<*x<*M} is empty. There is no such M. It doesn't exist. There's nothing between 0 and 1. It's like looking for an integer between 0 and 1. There isn't one. > There can be no elements between a and M. There is no M. > Carrying this further ... is pointless. M doesn't exist. Your set [math]S[/math] is empty whenever there's nothing strictly between [math]a[/math] and [math]b[/math] in the given order.

The passage you quoted and attributed to me is not mine. In fact I couldn't find it in the thread at all. Is it new text from you that you wanted me to respond to? Let me just talk about [math]\omega + \omega[/math]. When you ask, " How do you determine when one sequence ends and the next begins?" you're thinking about it the wrong way. In fact I don't even know what you mean. It's just an order relation where the even numbers, say, precede all the odds. An ordinal number represents a well-ordered set. Consider the following alternative order [math]<^*[/math] defined by: * If [math]n, m[/math] have the same parity (evenness or oddness) then [math]n <^* m[/math] just in case [math]n < m[/math] in the usual order; and * If they have opposite parity, then the even one is [math]<^*[/math] the odd one. You can verify that this is a well-order. It's two copies of the standard order [math]\omega[/math], one after the other. We denote it [math]\omega + \omega[/math]. So if I give you 5 and 46, then we know that [math]46 <^* 5[/math]. It's not any more complicated than that. You don't have to "find the point" where the first [math]\omega/math] ends or anything like that. It's just an alternative order where all the evens precede all the odds, and within each group everything's in standard order. And you have to verify it's a well-order. Having done that, you can name it. Another way to look at it is that just as $\ The passage you quoted and attributed to me is not mine. In fact I couldn't find it in the thread at all. Is it new text from you that you wanted me to respond to? Let me just talk about [math]\omega + \omega[/math]. When you ask, " How do you determine when one sequence ends and the next begins?" you're thinking about it the wrong way. In fact I don't even know what you mean. It's just an order relation where the even numbers, say, precede all the odds. An ordinal number represents a well-ordered set. Consider the following alternative order [math]<^*[/math] defined by: * If [math]n, m[/math] have the same parity (evenness or oddness) then [math]n <^* m[/math] just in case [math]n < m[/math] in the usual order; and * If they have opposite parity, then the even one is [math]<^*[/math] the odd one. You can verify that this is a well-order. It's two copies of the standard order [math]\omega[/math], one after the other. We denote it [math]\omega + \omega[/math]. So if I give you 5 and 46, then we know that [math]46 <^* 5[/math]. It's not any more complicated than that. You don't have to "find the point" where the first [math]\omega[/math] ends or anything like that. Another way to look at it is that we get to [math]\omega[/math] by taking the "limit" of the ordinals 0, 1, 2, 3, ... and we get to [math]\omega + \omega[/math] by taking the limit of [math]\omega + 1, \omega + 2, \omega + 3, \dots[/math]. If we want to we can formalize the definition of limit in this context. It's just the set-theoretic union of all the preceding ordinals, which turns out to be the exact same thing as the set of all the preceding ordinals. How can those two things be intuitively understood as the same? Because ordinals are transitive sets. That means that every element is a subset and vice versa. That's ordinal numbers in a nutshell.

This site supports Mathjax with the "math" and "\math" tags in square brackets surrounding your markup. In (1) your proof is fine but I'd reword it for clarity. You wrote, "I assumed [math]A = \{2,3,4,5]\}[/math] is the subset of the power set ..." I would reword that as, "Suppose [math]A = \{2, 3, 4, 5\}[/math]. Then [math]A \cap A = A \neq \emptyset[/math]. Or you could just say, For any nonempty set [math]A[/math], we have [math]A \cap A = A \neq \emptyset[/math]. That would be sufficient for me. Also note that you meant that [math]A[/math] is an element of the power set. It's a subset of the integers, but an element of the power set.

Can you explain what you mean? For example suppose I take [0,1] and reorder it in the reverse standard order. So that a <* b iff b < a. Then the reordered set looks like this: 1-----3/4----1/2----1/4----0 I don't see how you get a square from this. > Edit: I might say (S,<*) is S with a and b removed and = {x: a<*x<*b}. What exactly is S? What are a and b? It would help if you clearly define your notation. If S is some set, then S = S throughout your discussion. And if S is an ordered set with standard order <, then (S, <*) must be the EXACT SAME SET with a different order. You can't change the meaning of S in mid-argument. You could say that T is S with a and b removed, but you can't say that S loses elements by virtue of reordering. That's bad notation and confusing exposition. What is a reordering of a set? A reordering is just a bijection from the set to itself. The set is exactly the same before and after. A reordering is just a choice of some bijection from the set to itself. The elements remain the same.

Then {x : 0 <* x <* 1} is no longer of interest in that notation, is that correct? What exactly is (S,<*)?

When you take the lower bound it's with respect to <*. Of course (0,1) has no smallest element in the usual order. That's irrelevant once we're talking about <*. A set is well ordered by <* if every nonempty subset has a smallest element with respect to <*. You keep confusing <* with the usual order < and that has been the source of your mistake since your very first post. If S = (0,1) with the usual order, then take <* to be 0 <* 1/2 <* 1/3 <* 1/pi <* (everything else) and now 1/2 is the smallest element of {x : 0 <* x <* 1} with respect to <*. But you keep using the same symbol 'S' to mean the set (0,1) with the usual order and also {x : 0 <* x <* 1} and that is why your argument doesn't work. But consider this example. Say <* is defined as: 1 <* 0 <* 1/2 <* (everything else). Then {x : 0 <* x <* 1} is empty because it makes no sense. We have 1 <* 0.

My <* order is as follows: 0 <* 1/2 <* 1/3 <* pi <* -sqrt(2) <* 47 <* 1000 <* -5/2 <* 1, (everything else in its usual order) Of course this is not a well-order because past 1, it's a dense order. But it STARTS OUT as a well order. What is S = {x : 0 <* x <* 1}? It's obviously S = {1/2, 1/3, pi, -sqrt(2), 47, 1000, -5/2}. In this case S has a greatest lower bound, namely 1/2. It's true that 0 is a lower bound, but 1/2 is the greatest lower bound. Why do you think 0 is the greatest lower bound? It's clear that 1/2 is a lower bound for S and is greater than 0. So 1/2 is the greatest lower bound of the set S. It's a) A lower bound for S; and b) Greater than every other lower bound for S. > Is wtf's ordering above acceptable? That is, 1, 1/3, 1/5,...1/2, 1/4, ... etc I've always wondered how u get from numbers in the first sequence to the next sequence. Might this be in the study of ordinals? In a case like his sequences the order might be as f(n), f(n+1), ... To see how to get from the first part to the second part, consider the even-odd order on the naturals: 0 <* 2 <* 4 <* 6 <* ... <* 1 <* 3 < 5 <* ... That's a nice well-order that looks like two copies of the naturals one after the other. Every even number is <* every odd number. You don't have to "get from" one part to the other. You just define your <* any way you like, and show that it's a well-order. > Might this be in the study of ordinals? Yes. The even-odd order is the ordinal ω + ω. ω, lower-case Greek omega, is the ordinal corresponding to the usual order on the natural numbers. ω + ω is two copies of the usual order, one after the other. But why wouldn't my order be "acceptable?" You can define an order any way you like. It's entirely arbitrary.

Yes. If <* is a well-order, then S={x: 0<*x<*1} might be empty, or it might be nonempty. If it's nonempty, why can't 1/2 be its least element in the <* order? After all, why can't a well-order (of R or of (0,1)) start with 1/2? It's a well-order, it has to start with something. Might as well start with 1/2. Also for clarity, can you please tell me if <* well-orders all of R? Or just (0,1)? I can't determine that from your exposition. It would make things more clear if you'd just say which you mean. But in general, let's pretend we want to at least START a well-order of R. Maybe we can't finish, as you claim. But we can at least start: 0, 1/2, 1/3, pi, -sqrt(2), 47, 1000, -5/2, 1, ... Then your set S doesn't contain 0, but it does contain 1/2 as its first element. In fact in this case S has exactly 7 elements, with 1/2 being the smallest. In this case S = {1/2, 1/3, pi, -sqrt(2), 47, 1000, -5/2}. Do you agree?

Most def. But how do you know that the smallest element of that set in the <* order isn't 1/2, say? Does <* well-order the reals? Or just (0,1)? That's unclear to me in your exposition. If the former, what if the well-order is 0, 1, 1/2, pi, -47, etc. Then {x : 0 <* x <* 1} is the empty set. Either way, if {x : 0 <* x <* 1} is nonempty, can you explain why 1/2 can't be its least element? Suppose <* well-orders (0,1), where the notation (0,1) means the open unit interval in the usual order. Suppose the <* order starts with 1/2, 1/3, 1/pi, -2/3, etc. Why can't that happen? And continuing, what if it starts with 1/2, 1/3, 1/4, 1/5, ... That's countably many, that's your m, n, o from earlier. But after all those you have 1/pi, 2/pi, 3/pi, ... And after all those, 1/e, 2/e, 3/e, ... To visualize the beginning of that well order, imagine the natural numbers well ordered like this: 0, 3, 6, 9, 12, ..., 1,4, 7, 10, 13, ..., 2, 5, 8, 11, 14, ... That's still a countable set, because it's just a rearrangement of the natural numbers. But it's not the usual order type. It has the order type of three copies of the naturals, one after the other. It turns out that there are uncountable well-ordered sets. That can be proved even without the axiom of choice. You'd do well to study the ordinals. I suggested that a couple of months ago. I reiterate the suggestion. Your m, n, o argument shows that you don't understand the ordinals. If you spent a day reading up on the ordinals all this would become clear to you.

That's the best you can do? As moderator setting the example for the tone around here?

If I decline to respond to a mention, I'm being disrespectful. If I do respond, then I'm continuing to post page after page. No win with you, is there? I prefer not to say any more about emergence because I truly have nothing else to say. I'm satisfied knowing that David Chalmers considers emergence meaningful; and that frankly I don't. I'm perfectly ok with that. He's a smart guy and a clear writer, I'll take another run at his papers down the road. What exactly was the purpose of your post? To tell me that it upsets you if I post and upsets you if I don't? Ok. Well ... I hope you find a way to deal with that. Clearly nothing I could do would make the slightest difference.

Quite a disingenuous remark given the context. All the best.

I've already been accused of "bitching" about emergence and told that my very presence in this thread inhibits intelligent conversation of AI. (I've seen no evidence of the latter). I respectfully decline to play. I've said several times I don't want to discuss it. I've said my piece and I wouldn't want to inhibit all the insightful and intelligent commentary on AI that I was apparently preventing by my mere presence. Wow. Now you ARE using evolution as a proxy for emergence aka "stuff we don't understand." The bottom line is that when I do respond to my mentions, I get accused of "bitching" and of inhibiting all the world-class commentary on AI that would otherwise ensue if I would just STFU. Then when I DO announce I have nothing else to say on the topic, people complain about that too. Well make up your minds.

I hope nobody minds if I don't reply to my most recent mentions regarding emergence. I've said what I want to about the topic. FWIW I read one of David Chalmers's articles about emergence and I had the same objections. He's a superstar philosopher but all I could see was that he was defining emergence as "stuff we don't know." We fully understand how trees become tables, so that's not emergence. But we don't understand how brain goo becomes self-awareness, so we call that emergence. I'm not moved in the least. I found the Chalmers piece depressing in that regard. I'm just done talking about it. Also yes I did remember that someone pointed out that only the surface of the water is wet. I didn't forget that, I just ignored it because it doesn't bear on the point that wetness is not a property of hydrogen or oxygen. And also someone brought up a good point about evolution. Do I find evolution less murky than emergence? Yes. It's a mechanism. It makes sense. We even see it in petri dishes and in bacteria that evolve resistance to antibiotics. So evolution is farther along the continuum from metaphysical speculation to science than emergence is. I'm perfectly well aware that my misgivings are not the majority opinion. But (as I keep saying) emergence is not a major concern of mine. I have many other unpopular opinions as well.

I will retract the dogmatic claim and retreat to "in terms of nature, there are arguably no regularities at all. I can defend that. So ok the identicality of atoms is a regularity. In the historically contingent theory of atoms. Besides, atoms aren't identical at all. Each element's atoms are different from the other elements. And even among, say, hydrogen atoms, one could lose or acquire an electron, isn't that right? You could then say well all electrons are the same and I could counter with Wheeler's fanciful idea that the reason all electrons are identical is because there is only one of them in the universe and we just observe it whizzing back and forth in time. https://en.wikipedia.org/wiki/One-electron_universe But if that's all you've got for regularities in nature, I don't think that's much. When people say that they like to think of hexagons in honeycombs, the sun rising in the east every morning, and such. Not abstract thingies like electrons which aren't really particles at all but rather probability waves. Again falling back on the symmetry in the theory but not necessarily in nature. > I came up with the most interesting definition of regularity for this discussion: appearance or behaviour governed by an exact set of rules or regulations Sounds like a computational theory of reality. We have no idea if such a thing is true. Exact set of rules? Hilbert's dream, an exact set of rules for mathematics. Gödel destroyed that hope. Still people cling to the hope of an exact set of rules for nature. "Dreams of a Final Theory," an equation for the world that you can write on a t-shirt. Nice dream. Not yet a reality and arguably never. ... in fact, this was already at least equally interesting definition of regularity, or maybe even more interesting. So basically, are there laws in nature, or not. Right. Maybe there are. Maybe there aren't. Maybe I'm a Boltzmann brain, a momentary coherence in an otherwise chaotic and random universe. https://en.wikipedia.org/wiki/Boltzmann_brain

I did read your post as if you were equating nature itself with our historically contingent mathematical models of nature. You didn't argue otherwise or defend your point. I do reiterate that there are no regularities in nature that we can prove are regularities with absolute certainty. Of course our contingent physical models do have regularities. That is not at all the same thing. Sure, it's a matter of philosophy. The philosophy of the limits of science. You can't know for sure that there are regularities in nature. You agree or disagree?

Our latest historically contingent physical theory models symmetry. How can we know that it's "really" there in nature itself? Perhaps our theoretical symmetry is only an approximation to something deeper and more complicated. We can't possibly know. Physical science is limited by our ability to measure. You disagree? You believe human made physics is absolutely true? It never has been before throughout history. You mean we just got lucky and nailed it this century? How will such an idea hold up a century from now? What do you think?