wtf

I read this and cracked up laughing. I hope you meant this as a lighthearted joke. But I for one will be very happy if you can stick to one topic so that we may attain clarity. I'll get to the rest of this later. Just wanted to let you know you made me laugh, and hopefully that was your intention. Skimming ahead I found this line. On the contrary, I'm well aware that if the reals are well-ordered by <* AND you can find a single nonempty subset of the reals that is not well-ordered (ie that has no least element under <*) then you would have proved your claim. But this can not happen. Every nonempty subset of a well-ordered set is well-ordered by the same order. So your quest is futile. If the reals are well ordered by <* then so is every nonempty subset. I have pointed this out to you, I don't know, six or eight times. I can't keep this up much longer. I ask you directly: Do you understand that every subset of a well-ordered set must be well-ordered under the same order? Yes or no? And (again for the sixth or eighth time), I ask you to contemplate well-orders of the rationals because they are key to understanding the uncountable case. The rataionals are a set that is usually densely ordered, and you want to understand the nature of any well-order it admits. That is why I am telling you to understand the case of the rationals. If you understand the case of the rationals, you will see why you are wrong about the reals. In your first post you asked for someone who understands this material. You found me. And you refuse to listen or even engage with the specific points I'm making. It's frustrating. Do you understand that every nonempty subset of a well-ordered set is well-ordered by the same order?

I would like specific responses to the two points I made on your previous post. * If <* well-orders the reals; and S is any subset of the reals; then: 1) <* also well-orders S; hence S can NOT be a dense order; and 2) If < is the usual order on the reals, then the rationals are a dense set (under <) with no least element, no matter what order you put on R. Please clearly state your understanding or questions, agreement or disagreement, with these statements.

ps -- If you would contemplate a well-order on the rationals, you will see why your reverse idea is wrong. There's a first rational and a second and a third and so forth. But the well-order doesn't need to reverse the usual order or whatever you're trying to say. > Its left to the reader to show any required reordering of the numbers from the usual order < results in a totally reverse order. Well-order the rationals by, say, the Cantor pairing function. You'll see that you're wrong. https://en.wikipedia.org/wiki/Pairing_function And think about it. If we reverse the order on the rationals, they're still dense. The reverse order on the rationals isn't a well order. In the usual order, 1/2 < 3/2. In the reverse order, 3/2 <* 1/2. SO WHAT???? Exercise: Show that the reverse of a dense order is also dense.

Is there some reason why you jam everything together into one run-on paragraph that makes your exposition difficult to read? Can't you please write one or two ideas per line and put in some whitespace? Like paragraphs, but for math. Fortunately, there are fatal errors in the first couple of lines, so it's not necessary to read the rest of it. > Let S be any subset of R such that for any a,b ∈ S there is an x a < x < b. You have been meticulous -- for which I commend you -- in using < for the usual order, and <* for an alternate order. So I assume < is the usual order on the reals. In which case let S = the rationals ... > Let ai be the min of S since its claimed every subset has a minimum. ... and the rationals have no minimum. So your argument is dead in the water right here. It's over. I need read no further. * So, can I see if I can figure out if you mean something else? Perhaps by < you mean <* so that S is a subset of (R, <*). But then S can not be a dense order. Every subset of a well-ordered set is well-ordered. So if the reals are well-ordered by <*, no subset of the reals could be densely ordered by <*. So under this interpretation, where you meant to write <* instead of <, your proof is also dead in the water. * Ok perhaps you mean that the reals are well-ordered by <*, and S is a subset that is densely ordered by the usual order. Let S be the rationals again. And let's say they're well-ordered by <* as \(q_1, q_2, q_3, \dots\). But what can this have to do with anything? I didn't read the rest of it because there is no relationship between the usual order < and the well-order <*. I'm only mentioning this interpretation to ask if this is what you mean. That the reals are well-ordered by <* and S could be the rationals, dense in the usual order. Is this where you're going? And why? No argument you can make along these lines could work. I skimmed the next sentence or two and as usual you are taking a well-ordered set, claiming you have a subset that's densely-ordered by the same well-order, which of course doesn't exist; then you try to argue a contradiction. You keep making the same mistake over and over of confusing < and <*. If the reals are well-ordered by <*, so is every subset of the reals, under the same order. And any well-ordered set (or subset) can not be densely ordered by <*. So there is no S and your proof is nonsense after the first line or two. ps -- I skimmed the rest, your idea of reversing a well-order is nonsense. You're just confused about < versus <*. Again, taking S to be the rationals, they're densely ordered and have no minimum element under <. And they can't be densely ordered under a well-order <*. Two different things.

ps -- Ok. <1,2,3,4>. And there's no integer between 2 and 3. What of it? Surely that's not surprising in a well-ordered set where addition is defined. Can you write down a complete, line-by-line argument?

Oh I see. By "not in the set," you mean not defined. 5/2 is undefined in the integers. Ok, now that I understand your terminology, your argument is somewhat less incoherent but is still unclear. So now that I understand what you mean, can you clearly restate your argument? The fact that 5/2 is undefined in the integers surely doesn't show that the reals can't be well ordered. > I keep believing I can come up with yet another way to prove my original claim. Another way? You haven't shown one way.

Again in the spirit of playful pickiness, I will take a shot at a rational defense of a YES position on that issue. Rationals are pretty much the same as reals, in the scheme of things. It it surely the case in a literal sense that the rationals and reals are different sets. One is a proper subset of the other. It's also true that the rationals differ from the reals in many important respects. The rationals are countable, that's important. They serve different roles in math. The rationals let you solve all the linear equations like ax + b = 0. That was a big deal back in the day but we need a lot more from our numbers these days! Modern science requires a mathematical model of continuity. And that's what the reals do. Algebraically and topologically complete. The very idea of flow. I hope you will grant that I am not ignorant in the many profoundly important ways in which rationals are different than reals. Yes I argue that in the scheme of things, the rationals and the reals are more alike than different. They are very different mathematically. Sure. But if you're not into math, one mathematical abstraction is like another. In school some screechy old math teacher said the fractions were like 5/7 and the "improper fractions" are like 33 & 1/3, and that the real numbers were some "infinite decimals" and don't ask what that means because we won't tell you. I say this: The average human being walking around the planet; or even around a modern college campus; does not understand or remember or care about the difference between a rational and a real number. They don't even know what the words mean. So yes, among those of us who care, the rationals are countable and they're all algebraic and computable, and almost all reals aren't any of those things. The rationals are essentially the result of a finite process; and the reals are the result of an infinitary process. So yes, we regard the rationals and reals as being very different. But a cat is more like a giraffe than a giraffe is like the number 3. You see my point I hope. Cats and giraffes are different, if you are a biologist. Or in the process acquiring a pet. But they are much more alike, as animals, as either of them is to the number 3, or the Eiffel tower, or the concept of civilization. So I say that the rationals and the reals are MUC more alike than different. Only specialists think there's a difference worth mentioning. That is true of everything. A biologist will insist a cat is not a giraffe. Everyone else can plainly see they're both animals, and not bridges and television sets. If you are a research team from Mars sent to study earth, you will conclude that cats and giraffes are more like each other; and the rational and real numbers are more like each other; than any combination of cats and real numbers or giraffes and rationals could ever be. Well thanks for asking. And for reading.

LOL The "it is not relevant" defense. As if inaccuracy may be forgiven, as long as it is accompanied by irrelevance! That the story you're sticking with? [The tone at my end is playful joshing, sometimes that can be misconstrued but I'm just having fun here]. But yes I'll stipulate that the Peano axioms imply a well-ordered set. That's because of Axiom 8, which says that 0 is not the successor of any number. So Well-ordering is baked in to the definition of the natural numbers. https://en.wikipedia.org/wiki/Peano_axioms#Formulation

The counterexample is the integers. That is not a well ordering but each element has a unique successor. IMO this thread has veered off course in the sense that discussing the definition of a well-order, or the mysterious properties of the rationals inside the reals, doesn't really help us understand the Cantor set. But clearly everyone else is happy so I'll stay out of it.

> For x=ω you have y=ω+1, so no escape clause needed. Sorry about that, I read that wrong. But your definition is not the standard one, so I'm wondering if it's equivalent or imprecise. Will get back to you on that.

> A well-ordering is pretty much precisely the example of an ordering < such that if x is any given element, there is one particular element y such that x < y, Oh my. What is the immediate predecessor of the ordinal \(\omega\)? Or did that fall under the escape clause of "pretty much precisely," which means in this case "not at all?"

This para in the Wiki article on the Cantor set may shed some light. So there are lots of points in the Cantor set that are not endpoints of some middle third. What a tricky set. Also, "today I learned." Cantor didn't discover this set! It was discovered by Henry John Stephen Smith. https://en.wikipedia.org/wiki/Cantor_set

You haven't said anything meaningful here. You're very far from your original ideas. I simply do not have any idea what "our set" is. Is it like "our gang" from the 1930's films with those annoying rugrat kids? I want you to tell me WHAT SET you are talking about. Show me the set, tell me the order, and then make your point. You are constantly confusing the usual arithmetic with a nonstandard order. By refusing to notate your set with set brackets and then tell us what the order is, you are confusing yourself and giving me nothing to respond to. Your posts recently are incoherent. You were making much more sense (although you were still wrong) pages ago. You're not giving me anything to work with. > OK, of course its necessary for me to be precise. Yes. So do that. Here's the pattern I'm looking for. * Identify some set by listing its elements in set brackets. {x, y, z, w, q} * Tell me what the order is. <z, w, q, y, z>. * Then make your point about (a+b)/2, making sure you understand that this is the USUAL addition on the integers and does not necessarily apply to a nonstandard order. I think that when you force yourself to be explicit and precise, you'll see that your argument has problems. > I hope I pretty much covered everything. You didn't cover anything. I don't know what "our set" is and you constantly change around orders and make vague claims that are neither true nor false since it's unclear what they refer to.

> 1, 3, 2, 4, 5, 6, 7, 8, ... Good example, (2 +4)/2 <* 2. In this set is the min Could you please explain what "this set" refers to? I don't know what you mean. And 2 is not the min of the ordered set <3, 2, 4>, 3 is. I'm going to notate ordered sets differently than sets. The set {1,2,3} has no order on it and none should ever be inferred by how you list them. The curly braces say, "Unordered set." That's the convention. You have a couple of uses for parens and that's a little confusing. So I'm going to denote an ORDERED SET as <2, 1, 3> for example, as a shorthand for the set {1,2,3} with the order 2 <* 1 <* 3. > therefore (2 + 4)/2 =3 and is not in the set. What? 3 is not in WHAT SET? > (1 + 3)/2 = 2 so its in the set with min=1. WHAT SET? > Therefore x is definitely in (a,b) and hence x is in ((a,b), <*). But NO. This is as false as false can be. I assume by (a,b) you mean the INTERVAL of elements strictly greater than a and strictly less than b. That's what the notation means. If you want to include a and b, write [a,b]. Now in the usual order <1,2,3>, 2 ∈ (1,3). But in the order <1, 3, 2>, 2 ∉ (1,3). That is the counterexample to what you wrote.

There is no circle in nature. There is only one circle, the idealized, abstract one in our minds. Which is represented by the mathematical circle, the set of complex numbers of modulus 1. Or if you prefer, the set of points in the Euclidean plane a distance of exactly 1 from the origin. Ah, I reread what you said. That while I said the mathematical circle represents the abstract idea of a circle; you are calling these two separate things. I believe you are right. It's an assumption that the math circle represents the idea of a circle. But that's not an actual fact, only an assumption. Just as it's an assumption that the real number line may be identified with the line of geometry. Is that what you are saying? If so I agree.

> Any set of positive integers can be ordered any way you choose to get the results you're looking for. Exactly. Which is why I'm puzzled by your arguing in terms of the compatibility between standard addition and a nonstandard order. They are not compatible and we can always find such counterexamples. So why are you trying to reason this way? It's not productive and I think you can see that. > This in no way represents all sets that have to be considered. All one needs to do is exhibit just one and subsets of R with the right properties certainly do exist. Can you exhibit such a subset? You haven't so far.

I don't understand what you're trying to say. Consider the following well-order of the positive integers: 1, 3, 2, 4, 5, 6, 7, 8, ... Now 2 <* 4 but (2 + 4)/2 is not between 2 and 4. What contradiction is there? All we've shown is that the usual addition is not compatible with the nonstandard order. And why should it be? After all, addition is based on the standard order.

No, that's false. 8 + 5 is always 13 in any order, as long as the symbol '+' has not been redefined. You could have an alternate definition of +, which you might call \(+^*\), but that would not be the same as the usual +. I hope you can see that just as you are always careful to distinguish \(<\) from \(<^*\), you have to be just as careful to notationally distinguish alternative definitions of addition from the standard one. > how could 2(a + b)/2 not = a + b Of course 2(a + b)/2 = a + b always. But \((a + b) / 2\) may not be between a and b if + retains its usual meaning and the ordering is nonstandard.

Let \(\mathbb N \) be ordered as 0, 1, 2, 5, 4, 3, 6, 7, 8, 9, ... Call that order \(<^* \). Now \(2 <^* 3 \) is true, but \(2 + 2 <^* 2 + 3\) is false where \(+\) is the usual addition. In order for addition to be compatible with \(<^*\) you need to redefine it as \(+^*\) with new rules to account for the reordering. > I don't think I saw any explanation from wtf showing me my errors. Perhaps you should read my posts. I've explained your errors multiple times.

I don't get this at all. I gave a very reasoned response to what you wrote. You posted a textbook page on the method of finite differences that didn't mention your lowercase delta. And you contradicted yourself by first saying the upper and lower case deltas are Yank versus Brit usage; then you said it means something different, but you still haven't clearly defined it. I stand by what I wrote and regard it as a perfectly civil communication. I have no idea what you are objecting to. Other than debunking your vague, contradictory, and incorrect ideas. The weather in San Diego has been lovely recently. It's almost always lovely here. ps -- Wow. I'm at a loss. You gave two different explanations for upper/lower case delta that contradicted each other. You posted a page that you claimed supported your point, but (1) it didn't contain lowercase curly delta; and (2) it was about the method of finite differences, and not standard calculus. In addition you've shown ongoing confusion about the nature of limits. I've pointed out your errors in a respectful manner. pps -- Maybe you're upset that I noted that you're strong on engineering math and weak on abstract math but don't realize the latter. I've been reading your stuff for several years on two message boards and that's been my observation.

> This seems somewhat inflexible considering I have already posted an extract from a well recognised textbook using exactly that notation. Incorrect information that happens to be published doesn't make it any less incorrect. But I note that in the photocopied page you posted, lowercase curly delta does not appear. So what the bloody hell, as they say on your side of the pond? > I Note that (from a straw poll) American practice most often uses Upper case delta, whilst British and European practice uses lower case delta for the same entity. Oh. Well that is a perfectly sensible remark. But that doesn't explain ... > I advocate the reservation of the upper case delta for finite differences since these preceded the calculus of real variables in history and are still in important use today. > Hence the use of lower case delta for something entirely different. Can you see that you just contradict yourself? Is curly delta merely a Britishism for what we Yanks notate with uppercase triangle delta? Or is it a separate symbol with a separate meaning? In the former case that's all I need to know, it's perfectly fine. In the latter case, what exactly is the meaning? I claim it has no meaning and the way you are using it is as a physicist or engineer would who can build a bridge or make a quark omelette, but who didn't happen to take a rigorous class in real analysis. It's not wrong, per se, just ignorant, in the literal meaning of the word. It's not a pejorative, it just means you never needed the formal mathematical understanding, and you seem to be lacking it. You haven't engaged with this point. > How would you write the formulae for forward, backward and divided differences, Gregory's formula etc? Um, what? I've never heard of forward, backward, and divided differences. Or Gregory's formula. I get that you had kind of an old-fashioned classical education in calculus, which is cool. It's interesting to see what they used to teach. It's not what they teach these days. Nor is your understanding of limits correct. I don't know if it's because you were taught wrong, or because perhaps you just learned wrong. But limits aren't infinitesimals, nor are functions that go to zero infinitesimals. > In my opinion this sort of thing is quite enough use (and very good and compact use it is) for upper case deltas tl;dr: Now you're back to the Yank/Brit divide again? Which is it? Is lowercase curly delta is Britspeak for uppercase triangle delta, a FIXED nonnegative (in the case of y) or strictly positive (in the case of x) difference in position or y-value? Or is it a brand new symbol that indicates that you think limits are infinitesimals? And Studiot, the page you uploaded doesn't contain any curly deltas. It took up all that screen space yet failed to make any discernible point. ps -- I looked at your picture again. You said it illustrates a use for the uppercase triangle deltas. But the illustration you posted is about the method of FINITE DIFFERENCES. It's not continuous calculus at all. It has nothing to do with what we're talking about. pps -- Are you perhaps making some sort of finitist argument? That the method of finite differences is sufficient for all of calculus? That's one way to interpret why you would post an entirely off-topic page. We are talking about standard calculus based on the standard model of the real numbers. If you're talking about something else, please let me know.

I've already done that several times over. If you have a new idea, you haven't expressed it yet. Also, nobody is "accusing" anyone of anything. I've studied some math and you're interested in the well-ordering theorem. I'm happy to share what I know with you, and also use my mathematical training to point out logical errors in your reasoning. It's been a long time in this thread since you've posted any mathematical content. Perhaps you could just restate your argument clearly and we can go over it.

> So someone was wrong? Ideas go in and out of fashion. There's no right or wrong about it. > Upper case delta, (followed by a referent) is the difference between two fixed values (proabably in a table) of that referent. > So Δxmeans(xn−xn−1) > all three are fixed or constant. Yes, [math]\Delta]x[math] and [math]\Delta]y[math] are fixed, finite numbers. If you visualize this as Newton did, as the position of a point moving through space (in this case one-dimensional space), [math]\Delta]y[math] is a number representing a difference in position and [math]\Delta]x[math] is a strictly positive real number representing a difference in time. Surely I don't need to tell you this! Why are we having this conversation? > Lower case delta followed by a referent means an arbitrarily small increment in that referent. There is no such thing as lowercase delta. You seem to be confusing arbitrarily small (as in a variable) and infinitely small (as in an infinitesimal). There is no lower-case delta in standard math. Where did that idea come from? Is that something they use in engineering? > To be arbitrarily small it must be a variable. > As a variable it is an increment in the referent, which is also a variable. > So δx is an arbitrarily small increment in the variable x. If that is what you truly believe then you need to read a book on real analysis. You are simply misunderstanding what a function of a real variable is. You do of course understand that concept physically; but not mathematically. There is simply no such thing as an "arbitrarily small increment." That is indeed how freshman and physicists think about it; but it's not how the math is actually understood by mathematicians. It the way you express these ideas that makes me wonder if you are just misunderstanding what limits are. > To create the limiting process you refer to it is only necessary to append the following improper numbers to define a derivative I have no idea what it means. And to the extent that I do understand, it's wrong ."Only necessary to append the following improper numbers?" What is an improper number? I don't mean to be disputatious with you since we have more views in common than not. But I have noticed that you do bring an engineering orientation; and that's often not the right way to understand limits.

In that case I surely misunderstood your notation. Can you please clarify what you meant to say? It certainly seemed to me that you were saying that if you well-order the reals, you get an ordered set that looks like [math]\{a_1, a_2, a_3, \dots \}[/math]. That can't be true, because that is the definition of a particular countable well-order; and we know that the reals may only be given an uncountable well-order. That's because the reals are uncountable. A set may be reordered in many ways, but all orders have the same cardinality. Now if you did not mean to imply that meaning, just tell me what you meant.

Why won't you just explain the notational question I asked you several times? I don't get it. An infinitesimal doesn't "tend to zero." An infinitesimal is a particular number in the hyperreal number system. Infinitesimals don't get smaller and smaller. I'm getting the feeling you have an engineering understanding of what limits are. I only say that because you have a far greater grasp on engineering math than I do, but you seem to think that limits are infinitesimals. Am I getting this wrong? Limits are not infinitesimals. Limits made infinitesimals unnecessary. Limits caused infinitesimals to be banished from math [till their recent resurgence]. If by curly delta you mean capital delta, then delta-x and delta-y are not infinitesimals. Not in Newton's thought and not in modern thought. They're each strictly positive quantities. ps -- Re your comment that you know Latin. I can't top that. Should I just quit now?