discountbrains

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About discountbrains

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  1. 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? Maybe u could say 21/2, 31/2, ... Then what? I tried this stuff years ago and don't think it works. 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. So, we don't need an axiom like this at all. If u could do this u ought to get the Clay Prize. Don't think it is offered for this because its considered unprovable. I wonder if some theorems that depend on the AC could work for just the sets you need and you need only talk about a finite number of sets. I need to try to find theorems that depend on the AC on the internet. I know I've seen one or two in my studies, but don't remember them.
  2. "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." Yes, your assumption is right, it would always be whatever the chosen order relation is. And, after all your nth number in the list we see we can generate an n+1th number (and of course we would have a 1st one) so this would be by math induction at least a countable number of sets. I'm wondering here where do you start with the first real number after you run out of your initial list? To add to what I said earlier if u claim <* happens to be in fact a well ordering then it therefore has the property of creating a discrete set of numbers which negates my claim this is analogous of saying the Bible says this is the way it is therefore other things can't be true.
  3. I made a mistake. You're right. When one considers the power set they're talking about distinct subsets. Here we are talking about the same subset. The only difference is the order of the elements is different. Yes, I know, I was a bit loose with my terminology about dense sets. I didn't consult my book. I really meant any two elements of the set has anther element between them. I'm left with needing to prove your ordering of my original set is impossible....If I can.
  4. OK, I see what's happening here. There is a subtle disagreement in our logic and my approach is unorthodox. At least I've never seen anything like it. My thought is: Given any nonempty, noncountable, subset of R that is dense in R ( (0,1) is an example) for any function or algorithem or whatever that produces an order relation, <*, that identifies a min for this subset another subset with this minimum deleted has no minimum for this <*. You say pf course it has a min because if the <* is a well ordering it will order all subsets like {a, b, c,...} and each min can easily be found. I believe the power set for R contains any kind of even unimaginable subsets of R and such a subset exist which implies there is at least one subset with no min no matter what the order. I defined this subset in a previous post. Maybe I need to go ahead and try to prove your discrete order can't exist. I'm still trying to modify one of my failed attempts to prove this. Or should it suffice to say clearly a set of the form I described exists?
  5. You are determined to find fault with my assertion. You say I'm starting off with one claim and then u accuse me of starting with the opposite claim. For answer to taeto, I believe I left going back to my original conclusion on physics forum. I'll have to go back and see what I said.
  6. wtf: I had a book called "Axiomatic Set Theory" I copied because it was out of print. I know where it is. I believe it contains a lot of the technical stuff u'r talking about such as ZFC set theory models etc. I used to study it. I take it it was the gold standard on set theory and meta mathematics. Why didn't u tell me this 50 posts ago? Any nonusual order <* might order the interval S = (0,1) like for example, <0.5, 0.0000002, 0.46, a, b,...> where a ∈ (0.2, 0.4), b ∈ (0.75, 0.54)-reverse order. If σ is the min for S then there is no min for the set I mentioned several times above for <*. This is true no matter how the new ordered set looks. That's just the way it is even though its not consistent with pretty well established set theory for years. I have to add u at one time claimed I declared the reals to be well ordered and another time u maintain I say the opposite. I just can't win.
  7. 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'll look up the link. Yes, I see what goes on on these boards. They DO NOT like challenges to conventional wisdom. They like questions from people less well versed in the subject so they can appear smart in answering. That stack exchange I don't even visit anymore. Oddly, supposedly intelligent people act literally like 10 year olds- so frustrating..
  8. "YOU CLAIM there is no well-order on the reals". You say I begin by stating something as fact and then use the statement to prove itself; thus, making a circular argument. I don't know how I;m doing that. I just went through a simple illustration to show that there exists at least one set that doesn't have a min with any order relation. This is like its established that x is defined to have certain properties if its divisible by 3 and I show x is not divisible by 3. 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?
  9. "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 are being a great help to me. You are helping me refine my thoughts and I'm gradually sneaking up on a proper statement of my theorem. I'll change my statement to: There exist some subsets of the reals that don't have a minimum for any order relation. This will keep you from saying I declared R is well ordered by some <* which keeps you from claiming what follows in your argument. Then I just proceed as I have stated many times above. I'll just say if S has a <*min the rest follows. 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. Don't know if I ever said what motivated me to make my claim. 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.
  10. I have long been aware the numbers could be scattered all over the place. That's why I wrote S\(a) = {x: x ∈ S and a <* x, a ≠ x} and not as an interval. EDIT to my above post: I should check with my analysis book, 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. OK, I SEE WHAT U ARE SAYING. You're saying if we assume <* to be a WO of reals than it can order (0,1) as you said. I can accept this and this leads to not my result. But, I am saying it also leads to my result which leads to a contradiction.
  11. I should have stuck to my original argument at the very start. You people got me side tracked; its all your fault. I started with: consider the subset of the reals S = ((0,1), <), < the usual order. If R is well ordered there must be a number a in S such that a ≤* x where x is any number in S and <* is the well ordering. Now if we delete a from S we have {a}\S = {x: x ∈ S and a <* x, a ≠ x}. BUT, we can plainly tell {a}\S contains no <* minimum number! We claimed <* well orders R, but clearly {a}\S ⊂ R and it has no least element which contradicts our assumption <* WOs R. NOTE: if <* is an order relation on R it applies to all elements in R; therefore, if <* is the order chosen for S it also applies to {a}\S. You can't pick a different order relation to give you what you want for every set you're considering. We defined {a}\S using <* and its claimed here that WO guarantees {a}\S has a minimum and the AC guarantees it. But, we can't find it! I don't have to show S exists; its already the same numbers of the interval (0,1) with we started with; its just reordered by our supposed <* order relation. I'll get to your latest questions, but the above is the most straightforward proof. PLEASE NOTE: All I need to do is show there is one set that no matter which order you choose no least element (minimum) can be found. wtf seems to think I need to prove the opposite (apparently) that all subsets have no minimum; hence, he keeps bringing up natural and rational numbers.
  12. Yeah, I know. Yeah, I know. I just threw that in. It seems to me though that whatever some sort of addition + means x + (-x) = 0 and 2x should be x + x.
  13. I discovered my Cantor type proof has some problems. I'll repeat my (a + b)/2 proof: Let S =(0,1), R being well ordered means there's an a which is a min(S, <*). Consider {a}\S, There is a minimum, b, for {a}\S and there can't be any x in S with a <* x <* b. Now, (a + b)/2 ∈ (S, <) so a <*(a + b)/2 and it can't be less than b; therefore, it must be greater than b. We must have b <* (a + b)/2. Now for some arithmetic: Suppose b = 0.4 and 2b = -2 and suppose b = 6 + 7 +5 where + = +*. In my case this doesn't matter because b + b <* a + b is the same as b + (b - b) <* a. b - b = 0 and b + 0 = b. Therefore, b <* a which is a contradiction. Finally this says assuming all subsets of the reals can be ordered so they have a least element leads to a contradiction. I'll think about your rational number thing. I like my latest reordering proof. I need to clean it up and make sure its perfectly clear. The problem with the Cantor type proof is I can show there is a countable set of minimums for subsets of S = (0,1) and another number exist in S that's not in this set of minimums, but there may need not be any reason why it should be in the set of minimums.
  14. I'm going to prove the reals cannot be well ordered another way: Let S be any subset of R such that for any a,b ∈ S there is an x a < x < b. Let ai be the min of S since its claimed every subset has a minimum. Construct {ai }\S. There is also a min for {ai}\S. Call it ai+1. There are x, y, z, ... such that If these two ak s are minimums of S and {ai }\S, x, y, z,... can't be between them; there are no x such that ai <* x <* ai+1. Therefore, any such x, y, z, ... with ai < x < y <z,...< ai+1 must be greater than ai+1 . But, what this does is reverse every element of any subset S or R. That is, now for any x, y, z,... the order is now z <* y <* x or I'll write .... <* z <* y <* x. This says any time we try to WO R we can't avoid merely reversing the order and the numbers are dense in R as before, yet well ordering requires them to be separated. And, we get another contradiction here. Its left to the reader to show any required reordering of the numbers from the usual order < results in a totally reverse order. Try anytime you have a < x < b for any a and b moving x such that a <* b <* x and see what you get. Moderators 3253 19504 posts Location: 珈琲店 R I suspect you are mixing up two different idioms : "singing from the same hymn sheet /book" and "to be on the same page Oh, no, no, No! Think about it, everyone in church has their hymn books out and are singing and you got someone singing louder than everyone else and in conflict with everyone because they're on the wrong page. I think this is hilarious.
  15. I believe we are "not singing on the same page" (the original expression which people don't seem to get and is actually humorous to visualize). Maybe I should generalize my subset S from being (a,b) to any subset. Here is your example S = <1, 2, 3, 4, ....> in the usual order where 1 < 2 < 3 < 4 <... in this order I write (1 + 2)/2 = 3/2, (2 + 3)/2 = 5/2, ...None of these numbers are in S. You write S = < 1, 3, 2, 4, 5,...> which of course is well ordered. All my (a + b)/2 were calculated using the usual order and all come out as fractions and therefore not in S so we can't talk about their order in S. Early on I stipulated (a + b)/2 must be a member of the set S. I'm taking S to be the very same set (a,b) as it was with the usual order. Every x in S is the very same x as it was originally. Order doesn't change that. I'm just playing with your example here; I'm not trying to make any further point about what I previously stated. I keep believing I can come up with yet another way to prove my original claim. I welcome seeing more counterexamples.