Where to submit my proof that the set of real numbers can't be well ordered

[discountbrains]

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.

[uncool]

20 minutes ago, discountbrains said:

Maybe I'm wrong, but I think everything needed is there in my presentations .

Then assemble them in a readable fashion. 

As wtf has observed, you have taken back and reiterated and modified your claimed proof to the point that it is impossible to tell what form it takes now. So please:

Write your proof out, in one post, with no handwaving. 

We have asked you to do this repeatedly. You have not done so. 

29 minutes ago, discountbrains said:

Because what I say is not in a textbook u immediately dismiss it.

No. We dismiss what you are saying because the claimed proofs you use to support your claims are wrong. They have fatal errors. These fatal errors cannot be remedied by "visualizing them". 

[discountbrains]

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. 

[uncool]

44 minutes ago, discountbrains said:

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}.

That only proves that 1/2 is a lower bound for S\{1/2}. Why must it be the greatest lower bound? (Hint: it doesn't; there are many orders where your claim is false)

45 minutes ago, discountbrains said:

This is the central issue of my claim. I've been over this a half dozen times.

And we've been over why it's wrong a half dozen times.

46 minutes ago, discountbrains said:

This sort of greatest lower bound stuff is very common in undergraduate analysis.

Yes, it is, but it only works well because of the common order on reals. When you lose that order, your intuitions all go wrong, too.

[uncool]

nm, tried to do something and forum software refused

Edited August 27, 2019 by uncool

[wtf]

8 hours ago, discountbrains said:

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. 

Suppose I give you the following linear order: 1/2 <* 1/4 <* 1/8 <* 1/16 <* ... <* everything else.

What happens to your argument?

[wtf]

> 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. 

You know I read through that and as far as I can see IT'S EXACTLY RIGHT. 1/2 is the min, then when you delete that 1/4 is the next min, and when you delete that 1/8 is the new min. But it's NOT a contradiction. It's exactly the way it might be ... which refutes your argument.

Also you're confusing your intuitions of greatest lower bounds in the real numbers in the STANDARD order, with  what's going on here. Again, think of the naturals 0, 1, 2, 3, 4, 5. What's the greatest lower bound of {x >= 4}? It's 4. What's the GLB of {x >= 5}? It's 5. It's a discrete order. Thinking about limit points in the real numbers is what's throwing you off. 

Edited August 28, 2019 by wtf

[discountbrains]

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. 

Edited August 28, 2019 by discountbrains
added words

[uncool]

35 minutes ago, discountbrains said:

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 no idea what this is supposed to mean. 

[discountbrains]

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. 

[discountbrains]

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 <.

 

 

 

[uncool]

This is beginning to sound like philosophy of math - and, to be blunt, really bad philosophy of math - rather than math itself. I don't see an "axiom" that can be stated based on what you've said, because what you've said doesn't seem coherent enough to evaluate whether it's true or false.

[discountbrains]

32 minutes ago, uncool said:

This is beginning to sound like philosophy of math - and, to be blunt, really bad philosophy of math - rather than math itself. I don't see an "axiom" that can be stated based on what you've said, because what you've said doesn't seem coherent enough to evaluate whether it's true or false.

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?

[uncool]

18 minutes ago, discountbrains said:

Well then its no worse than the axiom of choice. I mean how is anyone going to evaluate that?

By whether or not it is true that for every collection-set of sets J_i indexed by I, there is a function f: I -> union of J_i such that for all i, f(i) is an element of J_i. This is a simple true-or-false statement based entirely around set inclusion, which is the basis for set theory. If you want your "axiom" to be used, you will have to be able to specify it to that level (and possibly further). What is your precise true-or-false statement?

21 minutes ago, discountbrains said:

You see even though examples can vary very widely there is a common issue to all of them.

I really, truly cannot see that there is anything beyond the trivial, essentially that "Different sets have different elements".

22 minutes ago, discountbrains said:

I say this is not bad meta math at all. Who appointed u the judge?

I am the judge of how things sound to me. 

[wtf]

1 hour ago, discountbrains said:

Well then its no worse than the axiom of choice. I mean how is anyone going to evaluate that?

The axiom of choice says that you can choose a legislature. If you have a country with a bunch of states, you can pick a resident from each state and form them into a lawmaking body. Suppose in the future they prove the universe is infinite and there is some country with infinitely many states. Why can't they form a legislature to organize their affairs? The axiom of choice is more natural than its negation.

Now the next time you hear about the US Senate, just think of two applications of the axiom of choice. Two representatives from each state. That's what the axiom of choice says. If you have a collection of nonempty sets, then Mitch McConnell is involved. 

Edited August 30, 2019 by wtf

[discountbrains]

"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.

Edited August 31, 2019 by discountbrains
add word

[uncool]

1 hour ago, discountbrains said:

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?

It sounds like confused thinking. Here's why:

First, you say x and y are related "because" they share a property. Since you are talking about an arbitrary relation R, there is no "because". Either they are related, xRy, or they are not. So how are you defining P?

Second, when you say they "share" a property, I infer that you mean that P is a property of a single element. But you keep talking about P as if it were a property of pairs of elements. 

Third, no matter how you define P, such a set does exist: if S is the empty set, then every universal statement is true on S, trivially. 

 

Edited August 31, 2019 by uncool

[discountbrains]

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.

[uncool]

So to be clear, it sounds like you are saying P is a property of pairs of elements. Is that correct?

Also, are you entirely sure you want to say that xRy if and only if  x and y have property P?

Edited August 31, 2019 by uncool

[wtf]

3 hours ago, discountbrains said:

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.

This business with P is half baked because you haven't yet incorporated it into a complete argument about anything. 

Secondly if you are retracting your admission that you're wrong, please respond to the question I asked earlier. How does your argument handle the case of the order 1/2 <* 1/4 * <* 1/8 <* ... <* everything else?

Edited August 31, 2019 by wtf

[discountbrains]

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.

[uncool]

38 minutes ago, discountbrains said:

I implied this order is perfectly conceivable; however, its incompatible with my perfectly derived set manipulation.

I have no idea what this is supposed to mean.

[wtf]

48 minutes ago, discountbrains said:

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.

Your earlier arguments that there's no well-order of the reals are wrong. This paragraph is "not even wrong." It's incoherent. It says nothing and means nothing. Is Property P like Preparation H?

Edited September 1, 2019 by wtf

[discountbrains]

18 hours ago, uncool said:

I have no idea what this is supposed to mean.

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.

18 hours ago, wtf said:

Your earlier arguments that there's no well-order of the reals are wrong. This paragraph is "not even wrong." It's incoherent. It says nothing and means nothing. Is Property P like Preparation H?

I have no idea what you're talking about.

[uncool]

5 minutes ago, discountbrains said:

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 realized that you were trying to refute wtf's implicit claim; it didn't help me figure out how it was supposed to address it.