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discountbrains

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Everything posted by discountbrains

  1. Are u _____ or what? READ MY POST!! I DID read yours. I was trying to convey I was not including your set. You keep repeatedly bringing up an example I'm not even talking about. I wrote I'm talking about subsets (or sets if u prefer) we don't already know a WO for which I was hoping was precise enough. Maybe this is the source of confusion for both of u.
  2. I'm sorry I'm wasting your time by my dumbness. I have always been thinking of a subset of for example S=[0,1] (with the usual order, <) which would be an interval or segment from [0,1]. Surely, if one considers {1/2, 1/3, 1/4,...} and they delete 1/2 there is a next number etc so it too a well ordered subset of S. This type of set I'm not thinking of Should I say "start with a set for which there is not already a known well order for?" If we use (0,1) then clearly < doesn't well order it. What order relation will? Here's the way I visualize it: What if you have a set you have no WO for and you pick some number, z, out of it and claim z is a <*min for it then you delete z. You then are left with all numbers from S greater than z with respect to <*. What is your <* min for this set. I am well aware of what wtf says. This is what motivated my whole notion, there is no WO for all sets. Here's my order which should better be put in some sort of an array: Lets use numbers in [0,1] we have 0, 1, 2, 3,..., 1/2, 1/3, ..., 2/3, 3/4,...,whatever,.... irrational numbers,..., then you get to 0.0023...,...0.000042... You just get to numbers you can't find the least of or they become incomparable. You finally end up with sets of numbers that you either can't find the smallest or you don't know which is bigger. There are several unusual ways of ordering numbers, but they always end up like this. As the sets of these groups of numbers go along it gets so you can't find a min. Think about it this way: You have your set, S, above with all its original numbers in it. You have your <*. You declare z the min of S. You delete z and are left with all numbers x from S, z <*x. For your order how do you find the min of these numbers? Correction last line first paragraph: Let this 'set' be S than the statement is correct.
  3. I hope you understand what I mean by this remark. I believe I was saying we should be able to apply the same general concept we use for '<" to the use of '<*'. On 2nd thought I might have to prove denseness in the reals for <*. And, what do u mean by z1 and z2? Is z1 <* z2? Once you identify these numbers how do find the min of the set with these deleted? You keep saying I have proved nothing. Just what is wrong with my latest attempts? You never have said exactly what is wrong. It sounds like you already have a notion of what a proof-if possible-should look like.
  4. f by "might", you mean it happens in some orders, then yes, I agree. But you seem to be claiming this must be true for all orders, and that's something you'll have to prove (in fact, it's false), so no, I do not agree. Don't think I have a problem with this.
  5. "As we might always have an x such that x < z for any z we should have an x <* z for any z" That is if this z is supposed to be a min for a set. Some confusion here. I clearly know some sets can be well ordered by some order. But, my original claim still is always there are some subsets that can't be well ordered by any order.
  6. Disregard my last post. I get complaints that I say something and then go back and say I made an error. The problem here is that in an hour or two someone replies which prevents me from going back and correcting my error. I'd say most msg boards don't get a response in less than a day. What I said in my last post doesn't imply there exists some element below any number in the sets I exhibit. Saying all elements are less or greater than each other doesn't do it. I think I'm getting close, but we know proofs in math depend on the tiniest detail. Wanted to delete the upper part of this. Now I can't.
  7. I think I've been aware all along what you are getting at. And, its a question I've raised for myself. You want me to prove there are numbers between the greatest lower bound for the set and any proposed min for the set. I wrote my last version to address this. I will write out in words some of my key points. Keep in mind we have a total order on the reals so for all u,v in R u<*v or v<*u. Now if I have the a set S={x: z ≤*x} this means this is the set of all numbers greater or equal to z. for all u,v in S not equal to z, u<*v or v<*u. Now take away the z and what do you have left? These are all numbers greater than z. If I pick any number greater than z its less or greater than any other number greater than z. Why can't this be true just like it is for the < relation? After all every number in R is comparable by my definition of <*. This is what this means to me. I wonder if it was suspected I was fairly well respected mathematician I would be taken more seriously.
  8. Sorry, I forgot to add there are several ways to write a set. Of course {a, b, c,...}, {x: z<* x}, and {x: z ≤*x} among them. for a set to have a min it looks like the 3rd one which is a totally different set than the second one. I'm sure you know that.
  9. I need to look at this wiki ordinal stuff but I suspect it won't change what I'm saying. When I 1st studied it I thought I might be over thinking it. I told you 3 times already "Ignore what I said before my last assertion". And yes, I should have included the '<* includes numbers from a set T'. Most all theorems and definitions always include 'nonempty sets or a similar condition. My proof amounts to no more than a simple exercise in a textbook. Given any order, <*, a set S= {x: a<*x} can be constructed which has no min for <*. I don't have to demonstrate this for any order. You know what I said is true-its very fundamental. Quit trying to play like u don't understand my meaning. Think about it. No elaborate proof is needed. Maybe I'm wrong. Maybe there is some arcane theory I don't know about. If u think u are debunking me lets see a counterexample. This is getting tiresome; you keep throwing obstacles in my path at every turn. I hope you dropped that meaningless null set stuff too.
  10. In my last message for any order relation one selects I displayed a set that can't be well ordered by that specific order. This is the only thing I really have to say about it. I believe this is all that is needed. I got off track in the many posts before when asked for proofs. I'm afraid this is the only proof I have that R cannot be well ordered. This realization was what motivated me to assert this some time ago. If I somehow can think of why this needs further proof I'll make an attempt at it here. For now I guess this is all I have to say. But, is what I say any way a proof? All I can say is try constructing my (unspectacular) set for any order you choose on the reals. In order to test it one, of course, needs to know how their particular order relation orders the numbers. I don't know why such a set construction can't be justified based on any order. Yet, for that order a min of the set can't be found.
  11. This is easy. Let S = all real numbers x such that x > 0. There is no least element in S. Of course this would be one of my sets.
  12. Allow me to ask for your take on my debunking of a portion of Einstein's relativity I posted on the physics forum a year ago. It involves one result of his theory and simple arithmetic. I believe uncool commented on my post there at the time: 1) Suppose someone is sitting at a table with a wire going across it perpendicular to him. He measures the length of the wire and finds it 3ft long. A current flows through the wire and he adds up the incremental distances between the near light speed moving charges. Because Einstein's length contraction formula says all these incremental lengths become much smaller the sum of all the incremental lengths becomes much smaller than the original length of the wire.The observer measures it one way and gets one answer; measures another way and gets much shorter. Which is it? Optical illusion? 2) Two rocket ships are traveling on the same path from one planet to another at the same speed of 0.9999c. One is just 100mi from the 2nd planet; the trailing ship is 100mi from the 1st planet. With relativity length contraction adding up the distances the planets, to an observer off to the side, are much, much closer than thought. If the ships are closer together the total distance becomes much larger. Wrote this hurriedly and edited it. I'll simply state: For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed. So we don't go around and around again on this we assume there are numbers in the set which are comparable by <*.
  13. There is no ONE subset that has no min for all order relations. Its just, given any <*, there is always a set of the type {x: a <*x} that can be constructed.
  14. Just because I labeled a number, m, doesn't mean its a min of the set. I'm only saying we're picking a number and testing it to see if its the min of the set. But, we know and I agree there are some sets (natural numbers etc)that can easily be well ordered. And, we can come up with a lot of well orders for a lot of sets. BUT, we cannot come up with an order that well orders all sets. My little one line above should really be my proof.
  15. If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though. If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though. For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too.
  16. Yes, its true I only showed there exists a countable number of z0, z1, z2, ... and the only way I could get more z's from T in my sequence is do wtf's ω+ω thing . I still believe there is a way to show any other z in T is a member of this sequence, but its a mute point now. I went back to my original thought. wtf just commented. Will see what he says. By the way I write this set. It is written in a way that it has no min for <*. Can I not consider a set of this sort?
  17. Maybe my statement requires more detail. But, the answer to both of you is how can a set described like this with this <* order have a minimum no matter what the order is? wtf is already giving me that <* is a WO. I know nothing about the order relation as I ask this.
  18. Uh oh, disregard the last few lines of my argument. At the outset I wanted to consider a set T={x: a<*x<*b, a,b∈R }. In fact we can forget about the 'b' . I say we also understand for all our chosen <* there are u,v∈T such that u<*v. WHAT I DID then was go ahead and try to prove my assertion while almost ignoring this notion that we can construct such an open set for any order relation. So hence for any order we have at least one set which has no minimum element. Its that simple. This seems fundamental and I might even call it an axiom. I actually could even describe subsets of this set which have no min for said order relation. Most everyone should agree that for any given set one can construct at least one order-maybe even unique-that would give it a min. My assertion is the opposite of this notion.....So, I'm saying for any order relation there always exists a set like T on the first line above that it can't well order. In the final part of my previous version I should have taken a different path, but I'm still puzzled on how to deal with wtf's ω1 + ω2 +... I've tried to write orderings like this myself before and they seem to work until they eventually fail.
  19. My new copy of my proof for your perusal: Proof is by contradiction. Let S=(0,1)⊂ R with 0<a<b<1. For any total order relation, <*, let T={x: a<*x<*b} and let S\{a,b}⊂T. Now, if we believe <* well orders T ∃z0∈T such that {x:a<*x<*z0}= ∅. Choose any zn∈T where z0<*zn then by our hypothesis there must be a z(n+1)∈T\{zn} which is the minimum of T\{zn}. So, ∀x∈T\{zn}, zn<*x and {x: zn<*x<* z(n+1)}= ∅ also. Hence, by math induction we have generated a string of minimal elements z0, z1, z2,..., zn,... ∀n∈N of T. Moreover, Z={z0, z1, z2,...,zn,...} =T. To see this choose an arbitrary z in T. z is the min of the subset, {x: z ≤*x} of T. Is z an element of Z? If z≠zn for a zn∈Z zn<*z or z<*zn. If <*zn there is a finite number of steps to reach a zi<*z. If zn<*z, z(n+1)<*z or zn<*z<*z(n+1). If z(n+1)<*z is it >*z(n+2)? Continuing, there is only a finite number of steps to get to zk<*z<*z(k+1). Any of these zi<*z<*z(i+1) are impossible because of the way these minimums were derived above. Therefore, any z in T is in Z and we can create a mapping f(T) to N which is a bijection. So, T contains no more numbers than N and we know this is not true. Can't turn bold off for some of this.
  20. Oh no, no, no, No. You are Wrong, Where do u get that any WO set must contain a minimal set? That's not the definition of WO. Also, I think u added a word to my statement about a set of numbers >* than z. You put in 'every' I believe and here's where u get the argument about null sets. U'r as bad as wtf who tried to redefine my T set. Many months ago it has crossed my mind that these forums (especially stackexchange) might be populated with Russian trolls or operatives. Something suspicious about this picture and I'm the one being duped. Yes, I know the font thing is why I wrote my thing by hand. I'm getting better with typing symbols; maybe next time it won't erase my work when getting near the end. I haven't dealt with this stuff for over 40 years so I'm a little rusty. Today I looked at examples of the math induction theorem use. I will write up a better version of this. But, as long as I show there is a 1 to 1, onto mapping and inverse mapping between T and N they have same cardinality regardless of order.
  21. I take the condition the set is nonempty in the established definition of WO to mean we just don't consider empty sets. For the last few lines this might be a good point which I'm not aware of and need explanation of. Is there a 'work around' of this? I think what you mean about not isomorphically ordered is I didn't show a map from T to N such that f(xi)=i is 1 to 1 and onto for all xi in T . What I say I'm showing is if zi is the min for A⊂T we must have a min, z(i+1), in A\{zi} which is the very next number in the set of minimums of T since there are NO numbers in T between these two. This is part of what I said in my hand written proof above implies there is a 1 to 1, onto correspondence between this set of z's and N and since for any z you choose in T it is the min for every subset of T, {x: z ≤* x, z and x in T}, so its also in this set of minimums. Therefore there is a bijection between T and N which leads to a contradiction. I think you mean my mapping must produce the same order. This is entirely unnecessary. ALSO, after my 'written' proof I defined T so S\{a,b}⊂T so it would be nonempty and I'd have something to work with; this my not really be required. The essence of the proof is the deleting of elements, z's, one by one. I just can't wrap my head around a null set being (trivially) well ordered, If its empty how can it have a min element? This can't be established convention in the math world.
  22. I take the condition the set is nonempty in the established definition of WO to mean we just don't consider empty sets. For the last few lines this might be a good point which I'm not aware of and need explanation of. Is there a 'work around' of this?
  23. "Why? Well-ordering says that any nonempty set has a minimum. If you are using the definition of well-order to say that T has a minimum, you must show T is nonempty." OK, you made this more clear. I take this to mean the set must be nonempty for the particular order which is proposed to be the WO, Oh! I forgot what I wanted to say just above. I wanted to tell wtf that if T is empty WE ARE THROUGH. there is no WO for an empty set.
  24. I could simply sate for my S above S\{a,b} ⊂ T and <* is understood to be an order for T. Look back several posts and see where it shows as an image where I wrote it out by hand. There were too many problems with changing font size when using subscripts.
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