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

[uncool]

Expand on that "clearly". Why can't {x: a <* x <* b} have a least element?

As I have asked you before: What step in your proof fails if you try to apply it to the natural numbers instead of the reals?

[discountbrains]

6 hours ago, uncool said:

Expand on that "clearly". Why can't {x: a <* x <* b} have a least element?

As I have asked you before: What step in your proof fails if you try to apply it to the natural numbers instead of the reals?

I keep telling u over and over again I am exhibiting such a set: this set. Clearly this is not the natural numbers. Only one set needs to be exhibited that has no least element with <* for WO to fail.

[uncool]

And I keep asking you: why is your statement true for "this set"? Why can't {x: a <* x <* b} have a least element?

You're relying on "clearly" for a statement that isn't at all clear. 

Edited May 24, 2019 by uncool

[discountbrains]

First I see I need to add more detail. Suppose S is a subset of R  S =(0,1) or S={x: 0 < x < 1} (usual ordering). <* is an order relation on S so x <*y <*z or y<*x, z<*y or any order holds for any x,y,z in S. 'a' above is strictly <* less than any element in S and all elements in S are greater than a. How are you going to find a min for <* in S? If you want to show me one I will be glad to look at it.

[uncool]

That's not how proofs work. You have made the claim that it can't have a least element; it's your job to demonstrate it, not for me to provide a method to find a minimum.

You can try a proof by contradiction, to show that any method will result in a problem, but you don't get to look at my method when your statement should be true independent of which method I choose. 

Alternatively: I choose 1/2 to be the minimum of S. Prove that that's impossible.

Edited May 24, 2019 by uncool

[discountbrains]

Yes, I knew that when I wrote it. Just wanted to see if u could contradict me. 

I'm talking about an ordered field F which has 5 properties anti symmetric, transitive , etc AND for any x in F, 0<*x  if a<*b a+*x<*b+*x and similarly for multiplication for all a and b in F. 

If we have a set S={x: a<*x} then a is the greatest lower bound for S a is not in S.A greatest lower bound is a min for S iff if is in S. Therefore S has no min with <*.

Also by the definition of an ordered field above. for any S subset of F, S\{m} should contain a min, M, by your assumption there should be no elements in between them. But, by above there exists a z such that m+*z+*z=M and m+*z<*M which is a contradiction.

For my 3rd proof I note that no matter what order for elements in F is possible every element can be assigned a a number such as 0.750004xxxx. or 0.xxx7xx....etc Lets suppose we the digits 1 thru 9 are ordered 17542...9 and we also take 0.xxx...7xxx...<*0.xxx2xxx... where the two digits are in the nth place. So, in order to compare all numbers with 7 and 2 in the same place any two unequal numbers in an mth place will cause the number to be between 0.xxx7xxx... and 0.xxx2xxx... depending on what the numbers are. This contradicts like above that a WO must separate the elements.

The +* is a concoction of mine to model the'+' notion.

 

Hopefully, I can fine tune what I said later. I hope you can tell what I'm saying from this presentation. BTW, where's WTF?

 

[uncool]

11 minutes ago, discountbrains said:

I'm talking about an ordered field F which has 5 properties anti symmetric, transitive , etc AND for any x in F, 0<*x  if a<*b a+*x<*b+*x and similarly for multiplication for all a and b in F. 

Then you're adding a restriction to the original problem (eta: as I have told you before). The question is whether the real numbers have some ordering that is well-ordered - no other restrictions. You are adding the restriction that addition and multiplication must be compatible with the ordering. The well-ordering principle does not require compatibility.

Edited May 28, 2019 by uncool

[discountbrains]

On 5/28/2019 at 4:03 PM, uncool said:

Then you're adding a restriction to the original problem (eta: as I have told you before). The question is whether the real numbers have some ordering that is well-ordered - no other restrictions. You are adding the restriction that addition and multiplication must be compatible with the ordering. The well-ordering principle does not require compatibility.

I used this so I could attempt a proof using a notion of '+' to show for any a, b in a subset S of the reals there exists an x in S between them contradicting a necessity for WO in my  post way above that any WO must produce a set of exactly the same elements of any original subset S such that they have no other elements of S between them.....I was going to show a proof in (i), (ii), and (iii) ways.  (the computer I was using was frustrating me with odd typing results). 

For my (iii) proof one can notice that for any element in a set it can be represented by a unique number 0.xxxx.... ,whatever, and the only way you can reorder them is make some digit of one greater than the same digit of the other.

I once thought I could order the reals by writing: 0.1, 0.2,....0.01, .... 0.02,...etc A professor very quickly wrote 0.101010... and 0.010101... on his blackboard and asked which is greater? I knew I was stumped. In fact in any attempt to order any infinite string of numbers like this fails because there is no longer an order relation by definition, ie we must show either a<b or b<a which we can't do.

My (i) statement in my post you reference is all that is really needed to be sufficient.

Yes, AC says nothing about ordering at all, but once we are given any random order of objects we can assign numbers to order it. An example might be items on grocery store shelves or library books.

[uncool]

 

23 hours ago, discountbrains said:

I used this so I could attempt a proof using a notion of '+'

I realize why you want to be able to use it, but you have to demonstrate that it makes sense in the first place. 

To be honest, I think you are attempting to write a proof where you don't have an understanding of the underlying question, or of how a proof works.

[discountbrains]

On 6/1/2019 at 1:08 PM, uncool said:

 

I realize why you want to be able to use it, but you have to demonstrate that it makes sense in the first place. 

To be honest, I think you are attempting to write a proof where you don't have an understanding of the underlying question, or of how a proof works.

Right, I should ask u how to write proofs. I actually majored in math. I have been planning to come back to this. I want to delete everything I might have said about finding numbers between the minimums of S, S\{a}, S\{a,b},... where a, b,... are the minimums of the respective sets. I will go back to my proof using Cantor's method way back in this thread. Before I showed that if (0,1) or [0,1] is well ordered then every element in it must be in some subset of it of the form {a, b, c,...}. Since we can construct a number using Cantor's diagonal method which clearly is in [0,1], but can't be found in any of these subsets and hence it fails the requirement that <* is an order relation. 

I ask again, as I did from the start,  if anyone can tell me given any order relation <* I can't exhibit a set defined as {x: 0<*x<*1}. This set, (0,1), with respect to <* has no least element related to <*. Hence no <* well ordering can be found. This is so obvious I'll call it my axiom. 

For someone who understands how to write a proper proof as u suggest about yourself u sure jump to conclusions about me. It very well may be u, as with wtf, who doesn't quite grasp the  true subtleness of this this subject. 

[discountbrains]

Someone said before all I was doing by using this Cantor diagonal method was proving what he did: that the reals are not countable. But, I am proving much more than this: That not all sets can be well ordered.

[uncool]

On 7/2/2019 at 3:35 PM, discountbrains said:

Right, I should ask u how to write proofs. I actually majored in math.

I said that for a specific reason: you keep trying to use a structure ("+*") that you fail to define, and then worse, you assume properties about it without proving those properties. That's not how a proof can work.

On 7/2/2019 at 3:35 PM, discountbrains said:

I ask again, as I did from the start,  if anyone can tell me given any order relation <* I can't exhibit a set defined as {x: 0<*x<*1}.

You can define such a set. Basic set theory, axiom schema of specification.

On 7/2/2019 at 3:35 PM, discountbrains said:

This set, (0,1), with respect to <* has no least element related to <*.

Prove it. Your attempted proofs have not been sufficient.

[discountbrains]

On 7/4/2019 at 11:41 AM, uncool said:

 

You can define such a set. Basic set theory, axiom schema of specification.

Prove it. Your attempted proofs have not been sufficient.

Let the set (S, <*) = {x: 0<*x<*1} where S=(0,1). Then we have <* is a WO means if m is the min of S , S\{m} has no elements less than m and if n is min of S\{m} we can write S={m ,n, o, ...} where each of these successive elements is the minimum of the set left with the minimum of each previous set deleted. Like I showed before this is a countable set so its elements can all be listed in a table like Cantor did. Using his diagonal method we can produce a number which is clearly in (0,1), but is NOT in the list. Therefore <* is not an order relation on (0,1) because for all x,y  in (0,1) we must have either x<*y or y<*x and the generated number can't be placed anywhere in the list. It remains to show S really contains no elements other than the above generated m, n, o,... numbers which is very similar to using the math induction theorem.  Therefore, no WO for the reals exists. 

Edited July 5, 2019 by discountbrains
no change

[uncool]

4 minutes ago, discountbrains said:

Then we have <* is a WO means if m is the min of S , S\{m} has no elements less than m and if n is min of S\{m} we can write S={m ,n, o, ...}

Why should this exhaust S?

6 minutes ago, discountbrains said:

 It remains to show S really contains no elements other than the above generated m, n, o,... numbers which is very similar to using the math induction theorem.

You keep trying to wave away key parts of your proof. Something you should learn when attempting a proof: the things you most want to handwave are often the most important parts of the proof.

Your attempted proof seems to come down to the claim that any infinite well-ordering is isomorphic to the usual ordering on the positive integers. That is, if a well-ordering is infinite, then it looks like m<n<o<..., and every element must be represented in that sequence. Correct?

[discountbrains]

21 hours ago, discountbrains said:

Let the set (S, <*) = {x: 0<*x<*1} where S=(0,1). Then we have <* is a WO means if m is the min of S , S\{m} has no elements less than m and if n is min of S\{m} we can write S={m ,n, o, ...} where each of these successive elements is the minimum of the set left with the minimum of each previous set deleted. Like I showed before this is a countable set so its elements can all be listed in a table like Cantor did. Using his diagonal method we can produce a number which is clearly in (0,1), but is NOT in the list. Therefore <* is not an order relation on (0,1) because for all x,y  in (0,1) we must have either x<*y or y<*x and the generated number can't be placed anywhere in the list. It remains to show S really contains no elements other than the above generated m, n, o,... numbers which is very similar to using the math induction theorem.  Therefore, no WO for the reals exists. 

THIS ALL NEEDS TO BE DELETED. Many errors!

[Phi for All]

51 minutes ago, discountbrains said:

THIS ALL NEEDS TO BE DELETED. Many errors!

!

Moderator Note

We don't delete anything.

 

[discountbrains]

On 7/5/2019 at 2:31 PM, uncool said:

 

Your attempted proof seems to come down to the claim that any infinite well-ordering is isomorphic to the usual ordering on the positive integers. That is, if a well-ordering is infinite, then it looks like m<n<o<..., and every element must be represented in that sequence. Correct?

I keep getting myself sidetracked. I'll go back to my very first original premise. That is, let (S,<*)={x: 0<*x<*1}. Such a set can be described and by its very nature has no minimum with respect to <*. I know  this is extremely simple and unsophisticated. You admitted such a set existed. This is true for any ordering <* of course.

Edited July 6, 2019 by discountbrains
adding/subtracting words

[uncool]

2 hours ago, discountbrains said:

Such a set can be described and by its very nature has no minimum with respect to <*. 

Why not?

2 hours ago, discountbrains said:

I know  this is extremely simple and unsophisticated. 

Then prove it.

[discountbrains]

14 hours ago, uncool said:

Why not?

Then prove it.

This was my original concept. The idea is given any order relation a set, S, can be described containing a number, a, such that for any number you choose, M, in the set not equal a,  a<*M. The opposite notion to this is given any subset of R you can describe an order relation that produces a minimum in this set. The problem is the order relation might only produce this for this set. I know what you are saying. That I must produce every possible ordering <* and prove it doesn't produce a minimum M for S. I believe all I need to do is state what seems obvious that a set {x: 0<*x<*1} exists for any <*. Set theory allows us to produce any set we can think of.

[uncool]

I'm sorry, I couldn't make heads nor tails of what you were trying to say there. 

What is your proof that for any order relation <*, the set S = {0 <* x <* 1} has no minimum?

I am not asking you to show that S is a set. It is a set. I stipulate that. But your proof depends on your claim that S has no minimum with respect to <*, and you have yet to prove that claim. 

Edited July 7, 2019 by uncool

[wtf]

5 hours ago, discountbrains said:

all I need to do is state what seems obvious that a set {x: 0<*x<*1} exists for any <*. Set theory allows us to produce any set we can think of.

Most def. But how do you know that the smallest element of that set in the <* order isn't 1/2, say?

Does <* well-order the reals? Or just (0,1)? That's unclear to me in your exposition.

If the former, what if the well-order is 0, 1, 1/2, pi, -47, etc. Then  {x :  0 <* x <* 1} is the empty set. 

Either way, if {x :  0 <* x <* 1} is nonempty, can you explain why 1/2 can't be its least element?

Suppose <* well-orders (0,1), where the notation (0,1) means the open unit interval in the usual order. Suppose the <* order starts with 1/2, 1/3, 1/pi, -2/3, etc. Why can't that happen?

And continuing, what if it starts with 1/2, 1/3, 1/4, 1/5, ... That's countably many, that's your m, n, o from earlier. But after all those you have 1/pi, 2/pi, 3/pi, ... And after all those, 1/e, 2/e, 3/e, ...

To visualize the beginning of that well order, imagine the natural numbers well ordered like this:

0, 3, 6, 9, 12, ..., 1,4, 7, 10, 13, ..., 2, 5, 8, 11, 14, ...

That's still a countable set, because it's just a rearrangement of the natural numbers. But it's not the usual order type. It has the order type of three copies of the naturals, one after the other. 

It turns out that there are uncountable well-ordered sets. That can be proved even without the axiom of choice.

You'd do well to study the ordinals. I suggested that a couple of months ago. I reiterate the suggestion. Your m, n, o argument shows that you don't understand the ordinals. If you spent a day reading up on the ordinals all this would become clear to you.

 

 

Edited July 7, 2019 by wtf

[discountbrains]

You meant S={x: 0<*x<*1} right? 0 is the greatest lower bound and 0 is not in S. There is no <* lower bound in S.  To answer wtf if 1/2 is the min of S then 1/2 <* x for all x in S not = to 1/2. I never really wanted to do this, but lets go back to the very primitive notion of arithmetic-back to ancient Egypt or even the cave man. He might think the fish he caught is longer than all the others. He has to cut some notches in a stick to measure and compare each fish. We can order things anyway we like, but we are using math when we assign numbers to, lets say, a and b. We say a<*b if there is some way to get from a to be. We don't care how. We say a<*b iff we make a symbol, +*, such that a+*σ=b. We can continue with this idea and say σ=q+*q. It follows that a+*q <*b. So if 1/2 is in S we can write there is a q such that 0<*0+*q<*1/2 and of course 0+*q is in S by the definition of S. Therefore 1/2 can't be the min of S.

σ is a number -σ is not defined for this. Lets say σ >*0 and σ > 0.

47 minutes ago, wtf said:

51 minutes ago, wtf said:

 

"You'd do well to study the ordinals. I suggested that a couple of months ago. I reiterate the suggestion. Your m, n, o argument shows that you don't understand the ordinals. If you spent a day reading up on the ordinals all this would become clear to you."

 

I tried to discard everything I said there.

 

 

 

[wtf]

48 minutes ago, discountbrains said:

You meant S={x: 0<*x<*1} right? 0 is the greatest lower bound and 0 is not in S. There is no <* lower bound in S.  To answer wtf if 1/2 is the min of S then 1/2 <* x for all x in S not = to 1/2. I never really wanted to do this,

Yes. If <* is a well-order, then S={x: 0<*x<*1} might be empty, or it might be nonempty. If it's nonempty, why can't 1/2 be its least element in the <* order? After all, why can't a well-order (of R or of (0,1)) start with 1/2? It's a well-order, it has to start with something. Might as well start with 1/2.

Also for clarity, can you please tell me if <* well-orders all of R? Or just (0,1)? I can't determine that from your exposition. It would make things more clear if you'd just say which you mean.

But in general, let's pretend we want to at least START a well-order of R. Maybe we can't finish, as you claim. But we can at least start: 0, 1/2, 1/3, pi, -sqrt(2), 47, 1000, -5/2, 1, ... 

Then your set S doesn't contain 0, but it does contain 1/2 as its first element. In fact in this case S has exactly 7 elements, with 1/2 being the smallest.

In this case S = {1/2, 1/3, pi, -sqrt(2), 47, 1000, -5/2}.

Do you agree?

Edited July 7, 2019 by wtf

[uncool]

48 minutes ago, discountbrains said:

We say a<*b if there is some way to get from a to be. We don't care how. We say a<*b iff we make a symbol, +*, such that a+*σ=b. 

That's not how arbitrary orders are defined. 

48 minutes ago, discountbrains said:

We can continue with this idea and say σ=q+*q.

Why must there be such a q?

Your earlier attempt ("S = {m,n,o,...}") was better; at least it dealt with the actual definition and question at hand, and the argument could at least be repurposed to something somewhat meaningful. This "argument" is an attempt to ignore definition.

Edited July 7, 2019 by uncool

[discountbrains]

wtf made an error-a fatal error. He supposes 1/2 is the min of S. To be such 1/2 has to be in S ⇒ 0 <*1/2 <*x for all other x in S. Since 0 is the greatest lower bound of S, 1/2 is no lower bound and hence not a min of S.

To uncool I'll have to think about if the way I defined S is sufficient or do I need all this other stuff which is unconventional and needs some cleaning up.