discountbrains

My first of several proofs: Consider any total linear ordering, <*, of the reals. To make it simpler consider <* for S={x: 0<x}. At this point we don't know if <* is a well ordering or not. I will show by math induction that a well ordering of S must produce a countable number of minimums for a particular collection of subsets of S. Then I'll show all numbers, z, must be in this collection or set of minimums. Thus, the conclusion must be that if R can be well ordered it must be a countable set and we know this is not true. The above is a preliminary test before going further to make sure my topic does not get closed

All good answers. An overwhelming over 90% of climate scientists believe we are experiencing changes in the global climate. There is no 'one size fits all' answer to any of your questions. Also, If someone made these dire predictions in the 70s they were wrong. After all the Chicago School of Economics has been so wrong about so many things they shouldn't be considered worthy of funding. Doctors have tools that are very limited in measuring conditions. A prof Modi at Columbia Eng school says if we installed heat pumps we could reduce our carbon emissions by 90%. I say if we install solar panels and zinc air batteries we could generate sufficient electrical power in Summer to use all Winter. Even crude zinc air batteries without the sophisticated catalysts etc could do it because so much solar energy is otherwise goes to waste in Summer. The zinc air batteries would be simple tubs containing zinc plates, felt soaked in an electrolyte, and some ss steel wool cathodes. They would be like a plating bath in Summer and then stored for Winter. In Winter they would be reassembled or reactivated for use. They would have only one cycle per year so dendrites should not be an issue.

The "bare assertion" must be wrong. Really? Wow! Whatever. No sense in repeating myself from above.

Yes, I guess that should work. I'm trying my best as quickly as I can to thoroughly, thoroughly examine what I'm saying. If u can give an example where I'm saying nonsense please tell me. The one u just quoted above you're question.

The use of 'property P' is to assign a property to or define the relation, <*, by giving it some arbitrary property, With x<*y both x and y have this property; this may also mean x has one characteristic any y does not. And, then we clearly see we can exhibit a set whose elements have none of these properties. AS I TOLD U MY LAST TIME, its reasonable to say that, but I said its not consistent with my derivation. You refute my derivation? Is it wrong? What does it imply?

Allow me to examine the concern in your last paragraph more carefully. But, yes I do believe that's what I mean. I believe the example I wrote illustrates this. Maybe you've never seen this 'property P' idea. I had an instructor for 2 classes who I think used this at least twice. I don't recall seeing it in textbooks. I think I saw it in my Axiomatic Set Theory . I essentially said if we can identify a z in S, and then consider S\{z}. This set has no min for any order. That is with S containing ALL elements of the interval (0,1). You can't refute this! You know what I'm saying; both of u quit acting like u don't understand. This is not hard. My stuff is simple yet profound-very profound. I don't know what else I can do for you 2 to make this more straightforward. You 2 keep saying its incoherent and badly stated to keep from admitting I'm right.

In reply to both of u please state specifically about what the point is in your questions. I think I may need to include a paragraph or two before my statement or axiom or whatever so u will know exactly my train of thought. It would actually be pretty hard to try to figure out what a definition or theorem means in a textbook without a background. I'll give an example of a property P: Let x,y ∈ (0,1) with x=0.xxxxx..., y=0.yyy.... and if no digit of y is 5 and no digit of x is 3 or otherwise if x<y. Then we would say x<*y iff it has property P. Note, there is no claim here at all that <* is or is not a WO.- I believe I've covered all cases here. Written logically property P could be that 'A and B and C are true.' If any of these are untrue then we would have ~P. I guess my argument I kept repeating-to some people-doesn't mean the reals can't be WO. I concluded to back off this claim and simply say my perfectly derived results are inconsistent with the assertion there is a WO for the reals..... Now,I constructed an entirely new argument with the property P thing. So, I have actually 3 arguments for my claim. I think I have enough now to write a paper to submit to a journal. I could also include some of you people's responses.

I guess especially in my case where I have two people to answer I should always quote the message I was answering. It was a response to wtf's last msg. I have no idea what you're talking about.

I implied this order is perfectly conceivable; however, its incompatible with my perfectly derived set manipulation. The use of property P is meant to refer to any arbitrary property. Could it be a collection of all properties? Then the collection would have to include ~P and they would cancel each other.

First let me specify S is a nonempty set. xRy iff x and y have property P. -this is how I was going to say it when I went to bed last night; I forgot it this morning. Property P might be x > y AND x and y are rational numbers. Property P might include many things.

"What is your precise true-or-false statement?" ......Yeah, that's a common way of expressing the AC. Sounds reasonable doesn't it? On 2nd thought, 3rd thought, or 4th thought I take back saying I was wrong concerning my repeated set manipulation. Its true one would expect at least one order in a collection of orderings would be such that it would produce a min in any set. But, this is incompatible with what my set manipulations above dictate. That's simply the way it is. Can u dispute this? Let me state my so called axiom: If R is a linear order relation on the reals and for all (x,y) in the relation, R, x and y are related to each other because they share a certain property P then a set S exists such that for all u,v in S. u and v have property ~P. This sounds reasonable doesn't it? m ≤*x for all x in S could be a property of R wtf kept saying given any set defined by me he could produce an order that would make my set empty. He has it backwards; this should be 'for any order there can be a set for which it can't be a well order'. And, his last post adds nothing. Clearly I'm no John Nash. I;m a John Nash wannabee. I have a couple other minor little mathematical observations. One I showed to a math prof. He directed me to a couple of books where I found my thing was a sort of 'fixed point theorem'. I liked mine better because it was more pure or simpler.

Well then its no worse than the axiom of choice. I mean how is anyone going to evaluate that? You see even though examples can vary very widely there is a common issue to all of them. Its the simple fact that for every rule there is not-rule. I say this is not bad meta math at all. Who appointed u the judge?

Some examples of this are: What if we had an ordering for any set beginning with the midpoint of the set? Some sets don't have midpoints. What if a property of an ordering was everything is ordered by increasing rational numbers? Many sets have no rational numbers. What if we had √3 < 2√3 < 3√3 <....? Of course many sets don't have these numbers. Of course there is an infinite number of examples. What conditions apply to an order relation can also apply to sets. The ordinary '<' for the real numbers already has its built in failing case because there are many sets that can't be WO by <.

Yes, there might be room for more clarification etc. I should and plan to give a number of examples to show how this works. I have pondered at times over some years how one might WO the real numbers and concluded this is why there is a problem. Just as u can define an ordering of numbers based on certain properties u can likewise construct a set whose membership is precisely numbers that have none of these properties.

You guys are Right; I was wrong. Right about almost everything. I still go by my opening statement at the very beginning though. I had a thought in my head and allowed myself to get off track. I am not to be doing this. I will make my original statement and say because: Every order relation has certain properties that determine the order of elements in some set or sets. If the order relation's properties produce a min in any set in a collection of subsets of reals than another subset of the reals always exists that contains elements with none of these properties. This might be more generally stated that for any ordering based on certain properties a subset of the reals exist containing only elements with none of these properties. I have had this thought for several years. It seems pretty obvious-at least as obvious as the AC. I might want to call this an axiom. Do u think it needs proof? I can provide a number of examples of this statement. For any operation that identifies elements in a set by certain rules there exists a set entirely of elements satisfying none of these rules. The null set has none of the rules. There is no min for the null set.

OK good, I'll do a proof after some fine tuning of my previous iterations. I'll bet what u see in textbooks are quite often the result of a number of revisions.....You know I double check myself often to see if I'm in error. Last evening I found myself succumbing to your way of thinking. I thought given a set S=(0,1), the usual interval, why couldn't it be that 1/2 ≤* x for all x in S and 1/2 hence is the min for S. I delete 1/2 and get S\{1/2}. Maybe 1/4 is the min for this set etc. But this just can't be because for all x in S\{1/2}, 1/2 <* x so 1/2 is the greatest lower bound for S\{1/2}. If 1/4 is the min for S\{1/2}.than its also the greatest lower bound for S\{1/2}.. 1/2 <* 1/4 because 1/4 is in S\{1/2} and for all x in S\{1/2}, 1/2 <* x. So, this is a contradiction. This is the central issue of my claim. I've been over this a half dozen times. This sort of greatest lower bound stuff is very common in undergraduate analysis. I just can't help sticking to my original argument.

I really, really think both of u are not making an effort to visualize what I'm saying.Actually my statements require some deep thought. I know many theorems I have come across in a textbook on first reading I can't make any sense of. Because what I say is not in a textbook u immediately dismiss it. Maybe I'm wrong, but I think everything needed is there in my presentations . You call it handwaving and refuse to visualize what I'm saying. Oh, and the numbering I gave above: I believe I was careful to not talk about ONE reordering; I mean ALL reorderings. All possible orderings.

"So yet again, I have to ask: what, precisely, do you mean?" An order relation must satisfy 3 properties. Sets and manipulations of sets are understood to take a certain form and meaning. I only used these basic ideas and ended up with my result. That's all I'm reduced by critics to saying at this point. Your question to me makes me ask just precisely is your issue with my train of thought? Point to exactly what I said that u can say is not true and correct me. Wtf's examples seem vague to me; might actually be more imprecise than he accuses me of. My sense now of ordinals is that its almost obvious they are well ordered. How did u like my ordering of numbers in (0,1)?

Don't know how to answer u. Your position is there is an established well accepted theory of ordinals that contradicts my claims. My argument stands on itself; everything follows logically from each proceeding statement in the argument. It is what it is without your fancy argument. If u didn't already had this bag of tricks back in your 1st few responses to me why didn't u use them then? You keep saying "everything else". How are u going to find a min let alone a least element of that? If this is a well order for the reals than u have ACTUALLY exhibited one. I will now give yet another argument that's probably non-rigorous nor intuitive: If you reorder a set S=(0,1) you realize all numbers in S are infinite strings of digits (countable number of digits). Some end in an infinite string of 0s of course. A string might look like 0.045xxxx0....xx. To produce ANY reordering in the most general way so no one can say you skipped any all digits need to be replaced by another not = to it or of course some or all digits could stay the same. Now consider any possible reprdering of these numbers. You would have to rearrange the numbers one by one. Of course we know even if there were a countable number of arrangements we couldn't do this, but theoretically its conceivable. We know, however, there is an uncountable number of ways to do this. We need the Axiom of Choice to do this. We can't even produce all reorderings without the AC. We can make a few reorders like reversing the order of all numbers etc. I was trying to model my logic after this Goerdel, Continuum Hypothesis stuff. I forget how all that stuff comes out. I mean that if u follow the basic rules of order relations and basic definitions of sets without the concept of WO you must conclude there is a conflict here if you start with the premise every set can be well ordered.

No

You are saying I am making statements about ordinals. They never crossed my mind. After some thought I could be wrong trying to prove a negative. Empirically it often is impossible at least: Many public figures concluded there were no wmd's in Iraq because they never found any. Logically its wrong to say this-not taking sides though. Was trying to find an analogy of my type of statement that no WO existed for R. I know of a theorem due to Von Neumann who you mentioned that states "No two variables can be maximized at the same time". So, your equivalent response to him would be "that YOU know of a function that does this". So, let me end by changing my statement to 'the rules on order relations on the reals lead to inconsistent results with the notion there exists a well ordering' for the reals. This would make my statement similar to some others in set theory, symbolic logic etc. My source of much of my knowledge of set theory is a book-long out of print-called Axiomatic Set Theory by Rubin and Rubin. In the book they discussed what you were saying about ZFC, ZF, and whatever models of set theory. They used a lot of symbolic logic notation and got a little into meta mathematics. I think my statements are a little on the meta math side.

The wiki site is saying the ordinals are well ordered in respect to set inclusion. "ω1 is a well-ordered set, with set membership ("∈") serving as the order relation. ω1 is a limit ordinal, i.e. there is no ordinal α with α + 1 = ω1." Don't know quite what this means. I guess α + 1 is still a countable number or set and thus can't equal ω1. That these topologies are not metrizable I'm a bit curious about. If I say A and B and C ⇒ ¬D and you say D ⇒D you are inserting information which is not there. You're saying a WO relation is a relation so I must automatically include it in my set of relations on my set S even though I don't know it exists. I can make all kinds of manipulations with order relations without even exhibiting them and the WO fails.

Is this your counterexample? Let me say the set for consideration is the usual (0,1). My <* applies to this set and for more numbers outside this set if someone want's to use them in an argument. You can't be bringing up your special set of 1, 2, 3,... I really, really, really don't think u understand this subject. My <* applies to each and every number in (0,1). If we delete a number from (0,1) we still have a very large uncountable number of elements. Maybe uncool can explain your error to you.... I keep telling u the only thing of concern is finding ONE set u can't find a min of, geez! ....No one answered my opening question. I'll look up various real math societies or maybe see if I can get a real prof to answer me. These groups have got to be some sort of scam.

Those are not counterexamples to my sets. I'll reread what u said; maybe I missed something. It looks like the same obvious old thing over and over again.