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.