Jump to content

discountbrains

Senior Members
  • Content Count

    180
  • Joined

  • Last visited

Everything posted by discountbrains

  1. OK, of course its necessary for me to be precise. When I speak of (a+b)/2 I mean for a to be the min of an <* ordered set and b to be the very next number. So, in our above example 3 is not in our set with with min a=2. (a,b) is meant to be {x: a < x < b} I hope I pretty much covered everything. I'm looking forward to seeing examples and corrections. Its fun to play with them. If I'm wrong, I'm wrong.
  2. 1, 3, 2, 4, 5, 6, 7, 8, ... Good example, (2 +4)/2 <* 2. In this set 2 is the min, therefore (2 + 4)/2 =3 and is not in the set. (1 + 3)/2 = 2 so its in the set with min=1. All I really need to do is consider the set (a,b) and let x=(a+b)/2 with the usual order <. Therefore x is definitely in (a,b) and hence x is in ((a,b), <*). So, I don't have to say If x is in a set with some minimum with respect to <* which is also in (a,b)-ie, min(a,b) and x are both in (a,b). You gave an example of where my claim doesn't work, but I presented a specific kind of subset.
  3. I was going to say "thank you" to your post about 3 back because u said the arithmetic worked the same no matter what order one uses. Then I saw the rest of your post. I'm working on another proof of my claim. I think there may be several possible proofs. I'll continue also trying to present my proof more formally.
  4. 1, 3, 2, 4, 5, 6, 7, 8, ... Now 2 <* 4 but (2 + 4)/2 is not between 2 and 4. Any set of positive integers can be ordered any way you choose to get the results you're looking for. 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.
  5. I was doubting myself thinking (a + b)/2 might not be in the same set that has a as its minimum. But, we must consider all subsets of R and certainly such sets exist and the WO <* must work for all sets. "But (a+b)/2 may not be between a and b if + retains its usual meaning and the ordering is nonstandard." I'll have to think about that. If 'its always greater than b' creates a contradiction and its not between a and b where is it? It has to be one or the other since <* is a linear order relation on R. Otherwise could it be its always less than a and never in S.
  6. And what is your full argument in the rest of your post? I agree with your latest post which I can't find. For example, 8+5 might = 2 or 15 in an ordering other than <. But, then I started thinking in this case how could 2(a + b)/2 not = a + b and 2b not = b + b etc. So, for the rest of my argument assuming any x in S with a <* x, we have a <*b <* x which implies a<* b and b <* a which is a contradiction. Hence, there could be x's between a and b with a WO which contradicts what a WO should look like. I may not be completely through though. I need to prove there exists an x between a and b. I started thinking I could prove my original assertion another way. I'll work on this some more. What I just said above is anytime x=(a + b)/2 its between a and b; otherwise, of course, if x is in S x could be greater than b.
  7. I believe I want to state my claim in a different way: Let S be the subset above containing a and b with a<* b. Consider (a + b)/2 which is in R. If a <* (a + b)/2 and (a + b)/2 is in S then b <* (a + b)/2 which leads to b <* a and a <* b as above. But, as you said this arithmetic may not work with <*. I'm thinking now its really independent of any ordering. Anyway, I can prove my original claim using Cantor's method. We note he only really generated one new number, but that's not the point. The assumption is this was a list of ALL numbers. In my case the new generated number can't be compared to any number in the list because its not in the list.
  8. As for the 2nd question I pondered this for a while myself. I think you can perform these +, -, etc operations independently of the order relation just like you do in balancing both sides of an equation to maintain equality. I can't give a specific theorem that's dependent on WO, but I suspect some <* is assumed and then further on something is manipulated with the ordinary + and - as if it doesn't matter. I believe I meant x is in S. I'll have to be more clear about that for myself. Some ideas I've played with end up with an element that's not even in the set in question.
  9. I will now prove there is a dense nonempty subset, (S, <*), of R that can't be well ordered by <* no matter what <* is. Proof: If S is well ordered every subset of it has a minimum element. Suppose for S itself its 'a' and we can form {a}\S which also has to have a minimum element let it be 'b'. We could go on deleting number from each new subset and of course this would result in the discrete string of numbers mentioned several times above. thus a <* b and there cannot be any x with a <* x <*b. We can write a +a <* a + b and get 2a <* a + b or a <* (a + b)/2. It must be that b <* (a+b)/2 because (a + b)/2 can't be between a and b. Thus, 2b <* a+b which implies b <* a and this is a contradiction (the rules for addition and whatnot should be valid for <* because this is the supposed well ordering that theorems based on well ordering also need). Sets containing such exhibited numbers certainly do exist and are subsets of R and their elements can't be arranged as a WO would. So, R can't be well ordered. I believe I can also write out a proof following very similarly the Cantor method for his uncountable sets.
  10. I made a mistake. I found my error yesterday. I simply asserted or assumed a set with the form [a, b) with respect to the well ordering <* exists. I believe I need to prove it exists. Otherwise, all subsets with the WO would look like {a1, a2, a3,...} and if an<*an+1 there is no x such that an<*x<*an+1 for any subset of R. I will try to prove there are.
  11. I keep saying over and over again all I need to do to show there is no WO is to exhibit a set for any possible WO you choose is not well ordered. I need only choose the easist example. I decided that if there exists a well ordering for the reals its really a philosophical question like God exists. To construct a WO is impossible. This is why I said 'a method'. OK, I'll go over my thinking again since I'm getting so much push back.
  12. NO, you are WRONG. I specifically chose sets that are dense. Of course, the natural numbers can be WO. You are making the same mistake as wtf did. You don't understand the problem. The WO theorem states there is at least one ordering that well orders the reals. That is there is some <* that can identify a l.e. for ANY subset of R. Of course, this is not the same number for all subsets. I don't think I saw any explanation from wtf showing me my errors. He at first kept referring to the AC. If u leave the AC out you have nothing. I question the logic of my own assertion: That is, if u assume u have a method of choosing a l.e. for any subset of R and exhibit a set with its l.e. another subset of this set can naturally be generated for which a l.e. can't be found using your well ordering <* method. This doesn't say the same specific l.e. works for both. My logic is like Wack-a-Mole. I'll call this my "Wack-a-Mole" proof. This is an answer to someone new. I'm not seeing their post right now.
  13. I'm sorry, I misunderstood what wtf was saying. I was accusing him as saying the same thing I was being accused of. No, we all know the reals or (0,1) can't be simply ordered like a1, a2, a3, ... Like someone said, its way more complicated than that. We know there are an infinite number of disjoint and intersecting subsets of these sets. I kinda want to think of them as multidimensional which I don't have a definition of. I would like someone to focus more directly on my original claim at the start and show its wrong.
  14. I know this post is directed at me. Let me be clear, the a1, a2, a3 ... sequence was asserted by wtf in one of his posts! Don't accuse me of saying it. I wanted above to state this seems like a fantasy, but forgot to say it. " You can compare them by just saying that they are different, by the construction". I was just saying the obvious about the usual order.
  15. In the version of Cantor's proof in my book if a digit ann=1 then the corresponding digit of the constructed number is 2 and if its not equal to 1 then the the digit=1. So, a typical constructed number might look like 0.1211121... Suppose there is a number ai =0.3472000000.... then clearly 0.1211121...< ai and its in (0,1). But who knows how to compare the two numbers with the new ordering?
  16. Oh my goodness. No, no, no. I'm not trying to prove anything about countability. I mean use the same method Cantor used and construct a numer that's not a member of ypur list. If the number is not in the list how can we find its place in the list? How can we compare it to any of the numbers? Typo above 'A + {a1, a2, a3,...}' should be A = {a1, a2, a3,...}. I have vision problems. And, yes, 1,3,5,... is also clearly a well ordering and so are many other sequences due to the definition of WO..
  17. OK, here we go. I'm just answering wtf at this point. Let A = (0,1) with the ususal ordering. Yes, it seems obvious like he says if a set A is WO by some other ordering then it can be able to be written as A + {a1, a2, a3,...}. This means it can be arranged in one of Cantor's array like things. Now we can do like he did and contruct a diagonal can be chosen to this arrangement and ask if this element is in A. At first I thought 'of course'. Now I see this needs proof. Such a diagonal would show a contradiction.
  18. Ok, the question just popped in my head. I didn't give it any thought. There are, I think, many ways to construct a bijection between N and Q. "Nice to know that those with greater depth of knowledge of these matters than mine have my back." Yes, I know its universally true its always safest to be on the side of conventional wisdom. I ought to try it sometime, but if everyone did this there would be no advances in anything.
  19. I quite well understand this. Of course they are mutually exclusive. Also, its hard to imagine how any set that is dense with respect to < or about any ordering could be ordered in such a way that it can be enumerated by 1,2,3,... There are just too many numbers that still pop up in between. I was going to add in my post above that intuitively it seems you can pick one number out of any set in a collection of sets. But, actually doing this is something else. For one thing, which number are you going to choose? Maybe I should ask another question: Can one show that any set dense with respect to one ordering is therefore dense for all possible orderings? And, yes studiot, I also think I've found the best way to understand a new concept is to see a few examples of it. Its maddening sometimes to try to figure out what a theorem is saying by just looking at it alone.
  20. Can't find an 'edit' button. Ran out of time yesterday. wtf kept correcting me saying look at an example like the natural numbers and observe these clearly can be WO. I was very surprised at his answer since it sounds like he is a professor or something. This is clearly incorrect: the thing is like what I said above which is 'not finding one example of a set WO'ed by some.<*'. The example he gives is not dense anyway. Very interesting I get 29 negative points while he and everyone else gets many positive points. I get tons of negative points on every board on the net I post. People just don't like anyone who doesn't go with the crowd-its really, really ridiculous. Its people like me who are the REAL pioneers in everything. But, if I'm wrong please write a valid proof.
  21. I'm now revisiting this thread. Like I said several times before to prove my assertion all you have to do is exhibitjust one set that can't be WO by <* assuming you actually found a WO , <*, for a set that contains it.
  22. I need to generalize my <* ordering. It needs to be an ordering on the whole set R because it has to be the possibility of any ordering. And, it certainly could be that a<*r<*c<* b<*....In fact, any number of elements not in T could be mixed up in there, but that doesnt mean there is a z in T with z≤*x for all x in T. I might write a typical ordering of mine as ....<* a<*r<*c<* b<*p<*... where r and p are not in R. wtf, do you mean that Cantor rational number thing? 1 2 3 4 ... 1/2 2/2 3/2 .... 1/3 2/3 3/3.... . . . This, of course, is where you start in the upper left corner and count back and forth diagonally showing thereś a one to one onto relationship to the natural numbers and doesnt apply to what im saying. Clearly this is a WO set. I dont know why u asked this. I was going to show what I think is a way to reorder any set of real numbers in a dense or çontinuous way, but will do that later. Im going to take a break from this now.This kinda gets to me after a while. If anyone has any comments or questions I will still read them.
  23. t is not in T therefore the reordering is not being applied to t. But, thats a good question. I believe I came up with 3 different ways to prove my claim. I will have to think about this. Now, just where is t in relation to the others? Yes, you are right; I will have to define <* so it works for the whole set R.
  24. Run your argument on the naturals ordered by the usual <. Is there such a subset S? Did I not say <* is not the same ording as <? Again, I say there are subsets of R with this property. Thats all I need to show; clearly the natural numbers are not going to work. Are u saying if one can find a <* such that even the natural numbers satisfy my claim that negates it?
  25. No numeric values need considering for the elements of T. Just the relative positions of the elements to each other.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.