wtf

I apologize. I get carried away sometimes. No personal malice intended.

I did no such thing. I characterized your error, not you personally. I added detail regarding unrestricted comprehension so readers could Google the relevant facts. What you SAID was garbage. That is an objective fact. Not vitriol. And "I think in pictures, not algebra," is unconvincing coming from someone whom I've seen lay out brilliant technical responses to questions of engineering math. You're not lacking in algebra by any means. you're simply lacking some basics in abstract math. Basics easily studied on Wikipedia.

If you're wrong on the facts, admit you're wrong on the facts. Tossing insults doesn't reflect well on your character. Perhaps OP "thinks in pictures and not in actual mathematical facts." Why do you get a pass on a silly statement like that? I've seen you rip through engineering math problems with skill and precision on two forums. I'm curious as to why you don't introspect and say to yourself, "I'm great at engineering math. Maybe I should review Russell's paradox." I know for a fact that when you're working a differential equation you don't indulge yourself in flights of fancy. I'm sure you must agree with this point. You know what you're doing and you solve such problems with authority. If you are a little weak on set theory, you could study up. Not toss insults and claim that flights of imagination are more important than getting the math right. You would never make that argument when you do engineering math. Agreed?

Nonsense. Garbage. Wrong. @Studiot why are you doubling down on this error? What is an "entity?" Do you see that you are using unrestricted comprehension, so that Russel's paradox has EVERYTHING to do with this? My God man this is GARBAGE. There is no set of all finite sets. There is no set of all infinite sets. There is no set of all finite or infinite sets. And for exactly the same reason. You can't form sets via unrestricted comprehension. There is no technical term in math called an "entity." You are just making this up.

Jeez Louise @studiot, there is no universal set. That's Russell's paradox.

The definitions given were: * A set is finite if it can be bijected to a natural number 0, 1, 2, 3, ... * A set is infinite if it's not finite. That is a perfectly clear definition that's workable in practice. It's true that it was given in two different posts. @uncool mentioned the def of finite and then I added that a set is infinite if it's not finite. You can't pretend this wasn't clearly stated. By the way I'll just clarify a technicality. @uncool said that a set is finite if it can be bijected to the set {0,1,2,...,n-1} for some natural number n. I said that a set is finite if it can be bijected to some number n. In my definition I'm using the fact that in set theory, we actually define the number n as the set {0,1,2,...,n-1}. So the number 1 is modeled as {0} and the number 5 is modeled as {0,1,2,3,4}. You'll note that in every case the cardinality of n is indeed n. That's a nice feature of this particular way of defining the natural numbers. It's not the only way, there are others. But this is the standard if no other encoding is mentioned. So my formulation and @uncool's are the same because in standard math, the number n IS the set {0,1,2,3,4,...,n-1}.

A set that's not finite.

> So, if I am getting it correctly u are separating out the rational numbers and placing them 1st. Then u have the irrationals. Where do u start with them? I don't need to. I don't need to well-order the reals to show that your argument is wrong. You say to pick a nonempty subset S of the reals. Say it has a first element a. Then you say that S\{a} has no first element. I showed the counterexample to that claim. So your argument fails. It also fails no matter how many elements you take away, since what's left is always nonempty and you can simply pick an element from it as the least. > So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC. I said no such thing. I simply pointed out that even if the reals can't be well ordered, your argument fails. You complained that I was using well-ordering of the reals to defeat your argument. So I turned it around and defeated your argument WITHOUT assuming AC or the well-ordering of the reals. I don't need to well order anything to prove that your argument fails. > So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC. No. This type of argument requires AC. It says that we end up with an endless sequence of nonempty sets; and that we can keep picking an element from each one. That's AC. This is the intuition behind why well-ordering requires AC. But I have not made that argument. All I need is to put two discrete elements at the beginning of a set to show that your argument is false. S\{a} has a least element. That falsifies your claim.

@dcb I did not understand anything you wrote here. I reiterate that I have refuted your claim that you can always remove enough initial elements to leave a set with no smallest element. On the contrary that is false and I showed you why very clearly. > I'm wondering here where do you start with the first real number after you run out of your initial list? Anywhere! You just pick a next number you haven't picked yet.

Well that's sort of my point. I don't think OP's argument relates to self-similarity. However, that does not mean I think I'm right and you're wrong. On the contrary, I think OP's presentation is so convoluted and ambiguous, that my interpretation is not the only possible interpretation. For all I know you are understanding OP better than I am. But self-similarity is important in the sense that the open unit interval (0,1) is homeomorphic to the entire real line. That's kind of a cool thing about the real numbers. You can take a bounded interval like (0,1) and stretch it infinitely like idealized taffy, until it coveres the entire real line to infinity in both directions. And in this transformation virtually nothing is lost. You lose boundedness but the cardinality and the topology are identical. Whether any of this bears on the OP's interests, I can't say. Did you happen to read my earlier post showing that OP's argument doesn't work? If that makes sense to you, that's what I think OP is getting at. But if not, maybe I'm misunderstanding the OP. It's certainly possible. But basically OP takes a subset S of the reals (under any order). If S has a smallest element, take that element away. The remaining set, OP says, doesn't have a least element. I've decisively refuted that idea. That's my take on this whole business at this point. I stay in because I love the ordinals. There's a theorem that every countable ordinal can be order-embedded in the rationals in their usual order. That's a really cool fact. The proof looks complicated but not beyond a little diligence.

I was referring to your most recent post. I'm thinking there's enough vagueness in @discountbrains's exposition that we might be reading different things into it. I didn't think that your comments were relevant to my interpretation of the argument, which is why I said I didn't understand your point(s).

Not really, I didn't follow your point. Might just be me. (0,1) as an open interval is universal except among those who use ]0.1[ If you have a reference using (0,1) as a closed interval I'd be interested in seeing it.

I have refuted this claim even without the assumption that I can well-order the reals. I assume "another subset" refers to the same order with its first element deleted. My examples show that no matter how finitely many times you delete the first element, what remains may still have a least element. And you might even remove countably many initial elements and what's left still has a least element.

No prob, everyone confuses dense with dense in the reals. They literally mean two different things. Bad terminology, like "open set." Just made up to confuse people. I haven't read your earlier post past the business with (0,1). I need to review that. But meanwhile, just consider the following ordering on (0,1): 1/2, 1/3, 1/pi, 1/e, 1/sqrt(2), and everything else stays exactly where it was. And in the reals at large, everything stays were it is. So it's the usual order on the reals; but within (0,1), the first few elements of the order are discrete, and the rest are in their usual dense order. And clearly you can do this trick with as many discrete elements at the beginning as you like. Finitely many discrete values. But also infinitely many. For example suppose we order (0,1) such that we start with an enumeration of the rationals: q1, q2, q3, q4, ... Then after all those, everything else stays the same. So we have an order that's not a well-order, and isn't dense, and starts with a countably long infinite sequence of discrete elements. So now with these examples in mind, we can see that even if the reals aren't well-ordered; your argument STILL does not work. You want to remove the first element of some proposed well-order of S and claim that what's left doesn't have a first element. But my examples show that this claim is simply false. Your proof fails EVEN IF the reals can not be well-ordered. I've shown that the assumption that the reals may be well-ordered is no longer needed. I thank you for challenging me on this point. I don't actually need that claim. Even if the reals could not be well-ordered, your argument would fail because we can always order (0,1) in such a way that it has as many initial discrete elements as we like. Your idea of continually removing first elements can always be defeated by some counterexample analogous to the ones I showed. The open brackets are universally agreed on to mean the open interval, excluding its endpoints. I've seen ]0,1[ used for this (in France, perhaps elsewhere in Europe) to avoid ambiguity with the ordered pair (0,1). But when its meaning as an interval is certain, (0,1) is the open unit intervals. It does not include its endpoints.

If you think (0,1) is dense in the reals you need to dig out that Real Analysis text of yours and review the definition of dense in R. It's true that (0,1) is dense; but it's not dense in the reals. Do you know the defs? An ordered set is dense if there's a third distinct element between any other two. A subset of the reals is dense in the reals if every real is the limit of a sequence of elements from the subset. Easy to confuse the two. (0,1) in the usual order is dense, but it's certainly not dense in the reals. 2, for example, is not the limit of any sequence of elements of (0,1). More later, but you seem to have a fundamental misunderstanding here.

If you would clearly state a thesis and a claimed proof, I can point out your errors. I've been doing so regularly for weeks. I'm not "determined to find fault," I'm responding to your initial request for critique by someone who understands the material. Do you revoke your request? If so I'm happy to comply. ps -- Here is your direct quote from your initial post in this thread: Do you still want that? If not, I'll happily stop responding.

> If σ is the min for S then there is no min for the set I mentioned several times above for <*. You keep "mentioning" it but your claim is false. > u maintain I say the opposite. Isn't that exactly what you've been arguing? That the reals can't be well ordered? If that's not your claim, what is? Isn't "Where to submit my proof that the set of real numbers can't be well ordered?" a clue?

> No, no, I didn't say that at all. I think u'r the one confusing things. Since u take everything can be WO'd as dogma u think any statement not consistent with that must be negating it. I'm only testing the theory. I take the well-ordering of the reals as a logical consequence of the ZFC axioms; or alternately, a logical consequence of ZF + CH. That's the opposite of dogma. That's logical proof from axioms using standard principles of reasoning. You're not testing any theories. You're presenting convoluted arguments that, when carefully unpacked, don't hold water.

Your argument seems to be: Assume we can't well-order the reals. Then there is a nonempty subset of the reals that can't be well-ordered. That's circular. If you claim some subset of the reals can't be well-ordered, I can refute that argument by well-ordering the reals. So the burden is on you to show that can't be done. You have not done so. > FWIW, I would like u to look at a post I put on Physics Forum on this scienceforum board. I submitted a flaw in Einstein's special relativity. I got no responses contradicting me; all I got were personal attacks. Haven't looked at the board for months though. I used very simple straightforward arithmetic. I don't say the whole theory is wrong. Fatal flaw? I can't address different claims you made on different forums, especially when you didn't bother to link the thread. I will say that PhysicsForums is a much more strictly moderated forum than this one, and they have zero tolerance for alternative points of view not backed up by a rigorous argument. I can't comment further since I have no idea what post you might be talking about or what you said. Nor am I qualified to discuss alternative physics. Or standard physics for that matter. I do know that bowling balls fall down. My understanding of the latest theory is that we happen to live in a universe where bowling balls fall down. In some other universe, they fall up. This is what passes for scientific thinking in our postmodern age.

> You are being a great help to me. Thank you so much for the kind words. > This will keep you from saying I declared R is well ordered by some <* Shame on me for ever saying that. But that's the whole point. You claim it can't be done. But you haven't proved it can't be done. I'm perfectly well allowed to say the reals can be well-ordered, since it's a theorem of ZFC, unchallenged since it was put forth by Zermelo in the early 1900's. > There exist some subsets of the reals that don't have a minimum for any order relation. Well this is falsified by any well-order on the reals, since any nonempty subset of the reals must also be well-ordered by the same order. And why can't I say that? You haven't proved no such well order can exist. You can't start your proof by assuming there's no well-order, that's assuming the thing you are trying to prove. The burden is on you to prove no well order on the reals exists, since it is a theorem that there DOES exist such a well order. > Are u stating that if a set is WO the order produces a set like <1, 2, 3, 4,...>? If so this makes the set into a countable set and if the set is ((0,1), <) there are far more numbers that are not accounted for so we don't know whether they're less than the min or what. Good point. No, I'm just illustrating my use of angle brackets. Well-orders in general can be quite wild. For example there's the even-odd order on the naturals: <0, 2, 4, 6, ..., 1, 3, 5, 7, ..>. And there's the mod-3 order: <0, 3, 6, 9, ..., 1, 4, 7, 10, ..., 2, 5, 8, 11, ...>. You can see how you could keep going and eventually get countably many copies of the usual order on the naturals. It can be defined explicitly. Of course those are all countable well-orders. There are uncountable well-orders, and that can be proved even without the axiom of choice. > I used to go into little spurts from time to time over the years to try to come up with a WO for the reals and concluded that for any order relation definition there would always be a set satisfying exactly the opposite definition. See this stackechange thread on the question of whether there's a know well-order on the reals. https://math.stackexchange.com/questions/6501/is-there-a-known-well-ordering-of-the-reals The technical fact is that there is some model of the reals in which there is a definable well-order of the reals. But in the usual model, there isn't. It's advanced set theory. In ‎Gödel's constructible universe there is a definable well-order of the reals. https://en.wikipedia.org/wiki/Constructible_universe But nobody thinks the constructible universe is the right model for set theory. Philosophers and set theorists have been at this for a century. Plenty of literature out there but it's highly advanced math. For example see this. https://mathoverflow.net/questions/71337/definable-wellordering-of-the-reals Now Mathoverflow is a question/answer site for professional mathematicians. It's not expected that any of us mere mortals can even read the page I linked. But at least you can see that it's a hard question that mathematicians wonder about. Now, back to earth. YOU CLAIM there is no well-order on the reals. If that is true, then surely there is some subset of the reals that can't be well-ordered. I perfectly well agree. But you haven't proved your claim; and it contradicts known math, at least if you accept the axiom of choice. And even without choice, there is an uncountable ordinal. And if you accept the continuum hypothesis, then that uncountable ordinal must well-order the reals! So this is all a very subtle business that depends on the axioms you choose and even on the particular model of your axioms.

You want to redirect database requests? What are you trying to do exactly? It sounds like using some sort of proxy setup would be simpler.

Perhaps for the benefit of us casual spectators, I wonder if you can just outline as simply as you can, the connections among your geometry problem; prime numbers; and one-way functions. By the way I am a little fuzzy on one-way functions. My understanding is that a one-way function (in computer science) is an invertible mapping between two sets such that one direction is much easier to compute than the other. The classic example (again this is according to my limited knowledge of the subject) is that 3 x 5 = 15 is an easy problem. We learned the algorithm in grade school (before Common Core utterly destroyed the teaching of arithmetic fundamentals in the US). And it is a very efficient algorithm, as computer scientists define the term. Probably polynomial I'd guess. (You have a row for each digit, and each row is linear in the digits of the other number. So it's n^2 or something. Just off the top of my head.) On the other hand, 15 = 3 x 5 is a very difficult problem! The only known way to factor an integer is by trial division. That algorithm grows exponentially with the size of the input. In the CS biz, polynomial = good, exponential = bad. If what I said were true, this would be an example of a one-way function. Multiplication is easy and factoring is hard, and they are each other's inverse via the Fundamental theorem of arithmetic. That's the one that says Mrs. Screechy will rap you over the knuckles with her wooden ruler if you don't learn your times tables. However, here is the beautiful kicker as I understand it. We don't know for certain that factoring is hard! We haven't found a polynomial factoring algorithm, but we haven't proven one doesn't exist! [There's a quantum algorithm that factors in poly time. It's called Shor's algorithmThis is one of the most amazing things I've learned in the past few years]. And more generally, we do not know if any one-way functions exist! That's everything I've come across about this subject in my readings. Anything I've got wrong please correct me. So now, given all this, what does your diagram do?

My friend @discountbrains, every nonempty subset of a well-ordered set is well-ordered by that same order. When you understand this you will be enlightened.

You have to be careful to distinguish < from <*. Your argument certainly doesn't lead to your result, since S \{anything} always has a smallest element. It must since S is well ordered by <*. Of course S\{anything} may not have a least element under < but that doesn't prove anything. > but 'a' being the minimum of S makes it the greatest lower bound of {x: x ∈ S and a <* x, a ≠ x} which means there is no l. b. for {x: x ∈ S and a <* x, a ≠ x} greater than a, thus {x: x ∈ S and a <* x, a ≠ x} has no minimum with respect to <*. This is what my argument hangs on. Your argument is wrong. Please write it out line by line clearly. I already explain why you're wrong. S is well ordered by <*. > > but 'a' being the minimum of S makes it the greatest lower bound of {x: x ∈ S and a <* x, a ≠ x} which means there is no l. b. for {x: x ∈ S and a <* x, a ≠ x} greater than a, Simply false. Try that argument with {1,2,3,4,5,}. 1 is the least element. 2 is the least element if you delete 1. And so forth. You are confusing yourself by bringing in greatest lower bounds. We're talking about well-ordered sets, not topological completeness.

Ok. I see the subtle error you're making. Let's walk through this. Let's say that <* well orders the reals. And let's write out the first few elements of the well-order <*. Let's say they are: 1/4, 1/5, 1/pi, 1/2. It's convenient that they're's all in (0,1), but they don't need to be, this is just for illustration. I hope you can see that after the well-order <* is applied to (0,1), the elements of that set are still well-ordered. They must be, since any subset of the reals is also well ordered by <*. However, under <*, (0,1) is no longer an interval. Its points are scattered to the winds, distributed all over the real number line. So the notation (0,1) is a little misleading. Under <, the notation (0,1) indicates a particular subset S of the real numbers. Under <*, S is still the same subset. But it's no longer an interval. It's simply a well-ordered set whose elements jump around the real number line randomly, when you re-order the real line by <*. Ok, you are correct that there is some element \(x_1 \in (0,1)\) that is minimum with respect to <*. In fact we know what it is, it's 1/4, that's the well-order we're assuming for illustration. So now let's consider \(S \setminus \{x_1\}\). By the way note that's how you notate set difference, the larger set is on the left of the backslash, and the set you're subtracting is on the right. Minor notational point. So \(S \setminus \{x_1\}\) itself has a minimum under <*, which we'll call \(x_2\) and which has the value 1/5 in our example well-order. You see we can keep doing this, because (0,1) is well-ordered by <*. But its elements are scattered all over the place, and your INTUITION is leading you ASTRAY. You say: > BUT, we can plainly tell {a}\S contains no <* minimum number! But it does!! It's 1/5. Or in general, \(x_2\). And after you remove \(x_1\) and \(x_2\), then \(x_3\) is the smallest element under <* of what's left, in this case 1/pi. You have made a CLAIM and not forced yourself to supply a PROOF, so you have made an ERROR. \(S \setminus \{a\}\) in your notation DOES have a smallest element. It's just the next element of the well-ordering of (0,1) under <*. Do you understand this? You made a claim that's false. I suspect it's because you haven't quite intuited that (0,1) does get perfectly well-ordered by <* and its elements are scattered all over the place, throwing off your intuition. To sum up: If <* well orders the reals, it also well orders (0,1). But under <* you can no longer think of (0,1) as an interval. It's just a random set of points in the reordered real numbers. This is confusing your intuition.