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

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48 minutes ago, discountbrains said:

Since﻿ 0 is ﻿the greatest lower bound of S

Why?

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23 hours ago, uncool said:

Why?

Surely u know why. Maybe you're looking for more precision. Precision is required in math. I believe all I need to do is state <* is just like the usual order, <, only its a different ordering.

OK, how are arbitrary orders defined? I need to know this. Is wtf's ordering above acceptable? That is, 1, 1/3, 1/5,...1/2, 1/4, ... etc I've always wondered how u get from numbers in the first sequence to the next sequence. Might this be in the study of ordinals? In a case like his sequences the order might be as f(n), f(n+1), ...

Here's an interesting order: How do they determine all the barcodes, UPC codes, for products in stores like Walmart? Some items in ones home might have to be partially ordered.

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46 minutes ago, discountbrains said:

Surely﻿ u k﻿now﻿ why﻿.

I truly don't. I don't even see a reason why 0 should be a lower bound, let alone the greatest lower bound.

48 minutes ago, discountbrains said:

I﻿ believe all I need to do is state <* is just like th﻿e usual order, <, only its a different ordering.﻿

It's an order, yes. That doesn't mean you can directly copy statements that are true for other orders.

Arbitrary orders are defined as relations: given two numbers x and y, either x < y, or it isn't, with the requirements that either x = y, x < y, or y < x, and that if x < y and y < z, then x < z. That's it. wtf's list clearly defines an order on the reciprocals of positive integers, which is even a well-ordering.

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On 7/8/2019 at 8:17 AM, discountbrains said:

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.

My <* order is as follows:

0 <* 1/2 <*  1/3 <*  pi <*  -sqrt(2) <* 47 <* 1000 <*  -5/2 <* 1, (everything else in its usual order)

Of course this is not a well-order because past 1, it's a dense order. But it  STARTS OUT as a well order.

What is S = {x : 0 <* x <* 1}? It's obviously S = {1/2, 1/3, pi, -sqrt(2), 47, 1000, -5/2}.

In this case S has a greatest lower bound, namely 1/2. It's true that 0 is a lower bound, but 1/2 is the greatest lower bound. Why do you think 0 is the greatest lower bound? It's clear that 1/2 is a lower bound for S and is greater than 0. So 1/2 is the greatest lower bound of the set S. It's

a) A lower bound for S; and

b) Greater than every other lower bound for S.

> Is wtf's ordering above acceptable? That is, 1, 1/3, 1/5,...1/2, 1/4, ... etc I've always wondered how u get from numbers in the first sequence to the next sequence. Might this be in the study of ordinals? In a case like his sequences the order might be as f(n), f(n+1), ...

To see how to get from the first part to the second part, consider the even-odd order on the naturals:

0 <* 2 <* 4 <* 6 <* ... <* 1 <* 3 < 5 <* ...

That's a nice well-order that looks like two copies of the naturals one after the other. Every even number is <* every odd number. You don't have to "get from" one part to the other. You just define your <* any way you like, and show that it's a well-order.

> Might this be in the study of ordinals?

Yes. The even-odd order is the ordinal ω + ω.

ω, lower-case Greek omega, is the ordinal corresponding to the usual order on the natural numbers. ω + ω is two copies of the usual order, one after the other.

But why wouldn't my order be "acceptable?" You can define an order any way you like. It's entirely arbitrary.

4 hours ago, discountbrains said:

Edited by wtf

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"

Why do you think 0 is the greatest lower bound? It's clear that 1/2 is a lower bound for S and is greater than 0. So 1/2 is the greatest lower bound of the set S. It's

a) A lower bound for S; and

b) Greater than every other lower bound for S"

But, I stated at the outset S = (0,1) with the usual order. (S, <*) is strictly made up of numbers from S. 1000, -5/2, etc are not in S.

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As a note, the term "lower bound" refers to a specific order; something that is a lower bound with respect to one order may not be for another order. I have been assuming that in each case, "lower bound" referred to the <* order.

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4 hours ago, discountbrains said:

"

Why do you think 0 is the greatest lower bound? It's clear that 1/2 is a lower bound for S and is greater than 0. So 1/2 is the greatest lower bound of the set S. It's

a) A lower bound for S; and

﻿ b) Greater than every other lower bound for S"

But, I stated at the outset S = (0,1) with the usual order. (S, <*) is strictly made up of numbers from S. 1000, -5/2, etc are not in S.﻿

When you take the lower bound it's with respect to <*. Of course (0,1) has no smallest element in the usual order. That's irrelevant once we're talking about <*. A set is well ordered by <* if every nonempty subset has a smallest element with respect to <*. You keep confusing <* with the usual order < and that has been the source of your mistake since your very first post.

If S = (0,1) with the usual order, then take <* to be 0 <* 1/2 <* 1/3 <* 1/pi <* (everything else) and now 1/2 is the smallest element of {x : 0 <* x <* 1} with respect to <*.

But you keep using the same symbol 'S' to mean the set (0,1) with the usual order and also {x : 0 <* x <* 1} and that is why your argument doesn't work.

But consider this example. Say <* is defined as: 1 <* 0 <* 1/2 <* (everything else).

Then {x : 0 <* x <* 1} is empty because it makes no sense. We have 1 <* 0.

Edited by wtf

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Whoops, I hoped I would correct my mistake before someone else did. L might say S=(0,1) with usual order and (S,<*)={x: a<*x<*b where a,b ∈ (0,1) so I don't restrict myself too much where a counter example can be made. Lets suppose you exhibit an ordering that produces a min for my set I believe I can show it doesn't work for all sets. This may lead to a new theorem that for any set a ordering exists that produces a min for that set, but it won't for all sets. This is leading to a lot to think about.

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2 hours ago, discountbrains said:

L might say S=(0,1) with usual order and (S,<*)={x: a<*x<*b where a,b ∈ (0,1)﻿

Then {x : 0 <* x <* 1} is no longer of interest in that notation, is that correct? What exactly is (S,<*)?

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On 7/10/2019 at 1:41 PM, wtf said:

Then {x : 0 <* x <* 1} is no longer of interest in that notation, is that correct? What exactly is (S,<*)?

Yes, I am just getting back to this. What I have done is use 'S" for both sets and defining (S,<*) like I did I actually made a and b the min and max of (S,<*). I don't want to do this of course. There are other straight forward ways of defining it.

I'm thinking of moving on. Because as I realized long ago if you try to reorder a line segment [0,1] on the x axis you end up with a plane [0,1]x[0,1] of an infinite number of points in XxY or if you draw a squiggly curve you pass through the same numbers repeatedly which is not an order relation. So, what value is this anyway? This gave me another thought which is by drawing a curve f(x) on the y axis of x and if d(xn,xm) is a metric you would get d(f(xn), f(xm)) being greater or less than  d(xn,xm). This might lead to another way to finding the max or min of a curve. That is where d(f(xn), f(xm))=0 with  d(xn,xm)  not =0 for some m and n. I wonder if this has ever been looked at.

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Edit: I might say (S,<*) is S with a and b removed and = {x: a<*x<*b}.

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4 hours ago, discountbrains said:

Because as I realized long ago if you try to reorder a line segment [0,1] on the x axis you end up with a plane [0,1]x[0,1] of an infinite number of points in XxY

Can you explain what you mean? For example suppose I take [0,1] and reorder it in the reverse standard order. So that a <* b iff b < a. Then the reordered set looks like this:

1-----3/4----1/2----1/4----0

I don't see how you get a square from this.

> Edit: I might say (S,<*) is S with a and b removed and = {x: a<*x<*b}.

What exactly is S? What are a and b? It would help if you clearly define your notation.

If S is some set, then S = S throughout your discussion. And if S is an ordered set with standard order <, then (S, <*) must be the EXACT SAME SET with a different order. You can't change the meaning of S in mid-argument.

You could say that T is S with a and b removed, but you can't say that S loses elements by virtue of reordering. That's bad notation and confusing exposition.

What is a reordering of a set? A reordering is just a bijection from the set to itself. The set is exactly the same before and after. A reordering is just a choice of some bijection from the set to itself. The elements remain the same.

Edited by wtf

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OK, what I said is not acceptable. There other ways to go about this like u said.

What I said later about the functions is simply how derivatives have been calculated for centuries-nothing new here. It is interesting though to think of a function expanding or contracting instead of a specific value  y=f(x).

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On 7/19/2019 at 4:34 PM, wtf said:

I agree with what you say. Defining S=(0,1) is not needed or helpful. Let S be a subset of R defined as S={x: a<*x<*b} . Both of you here agree S exists and is nonempty. Here we have a is the greatest lower bound of S with respect to <*. You say all subsets of must have a minimum, M. Now M is in S and a<*M. Therefore the set {x: a<*x<*M} is empty. There can be no elements between a and M. Carrying this further this implies there are sequences in S, a<*b<*c<*... which have no numbers x, y, z,...such that a<*x<*b<*y<*c<*... This doesn't really constitute a proof of my assertion at the start above, though. Whether we believe math is discovered or invented do we want to restrict our notion of it like this?

You talked about ω + ω. This is the problem I'm having with this ordinals: How do you determine when one sequence ends and the next begins?

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6 hours ago, discountbrains said:

You talked about ω + ω. This is the problem I'm having with this ordinals: How do you determine when one sequence ends and the next begins?﻿

The passage you quoted and attributed to me is not mine. In fact I couldn't find it in the thread at all. Is it new text from you that you wanted me to respond to? Let me just talk about $\omega + \omega$.

When you ask, " How do you determine when one sequence ends and the next begins?" you're thinking about it the wrong way. In fact I don't even know what you mean. It's just an order relation where the even numbers, say, precede all the odds.

An ordinal number represents a well-ordered set. Consider the following alternative order $<^*$ defined by:

* If $n, m$ have the same parity (evenness or oddness) then $n <^* m$ just in case $n < m$ in the usual order; and

* If they have opposite parity, then the even one is $<^*$ the odd one.

You can verify that this is a well-order. It's two copies of the standard order $\omega$, one after the other. We denote it $\omega + \omega$.

So if I give you 5 and 46, then we know that $46 <^* 5$. It's not any more complicated than that. You don't have to "find the point" where the first $\omega/math] ends or anything like that. It's just an alternative order where all the evens precede all the odds, and within each group everything's in standard order. And you have to verify it's a well-order. Having done that, you can name it. Another way to look at it is that just as \ The passage you quoted and attributed to me is not mine. In fact I couldn't find it in the thread at all. Is it new text from you that you wanted me to respond to? Let me just talk about [math]\omega + \omega$.

When you ask, " How do you determine when one sequence ends and the next begins?" you're thinking about it the wrong way. In fact I don't even know what you mean. It's just an order relation where the even numbers, say, precede all the odds.

An ordinal number represents a well-ordered set. Consider the following alternative order $<^*$ defined by:

* If $n, m$ have the same parity (evenness or oddness) then $n <^* m$ just in case $n < m$ in the usual order; and

* If they have opposite parity, then the even one is $<^*$ the odd one.

You can verify that this is a well-order. It's two copies of the standard order $\omega$, one after the other. We denote it $\omega + \omega$.

So if I give you 5 and 46, then we know that $46 <^* 5$. It's not any more complicated than that. You don't have to "find the point" where the first $\omega$ ends or anything like that.

Another way to look at it is that we get to $\omega$ by taking the "limit" of the ordinals 0, 1, 2, 3, ... and we get to $\omega + \omega$ by taking the limit of $\omega + 1, \omega + 2, \omega + 3, \dots$.

If we want to we can formalize the definition of limit in this context. It's just the set-theoretic union of all the preceding ordinals, which turns out to be the exact same thing as the set of all the preceding ordinals. How can those two things be intuitively understood as the same? Because ordinals are transitive sets.  That means that every element is a subset and vice versa. That's ordinal numbers in a nutshell.

Edited by wtf

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I'll respond to the quoted text portion of your post, which I think was written by you as a question.

Also my previous post has a copy/paste error, the second half where it starts again is the correct post.

> Let S be a subset of R defined as S={x: a<*x<*b} . Both of you here agree S exists and is nonempty.

I agree to no such thing. You haven't said what a and b are.

* Suppose $a \geq^* b$. Then $S$ is empty.

* Suppose $<^*$ happens to be a well-order. Then perhaps $a <^* b$ and there is nothing between them. Perhaps $b$ is the successor of $a$.

* But maybe $<^*$ isn't a well-order. It doesn't have to be. What if the order is $a <^* b$ followed by all the other real numbers in their usual order? For example let's order the reals as 0, 1, everything else. Then your $S$ is empty.

So I do not agree to what you say unless you tell me what  and$b$are, and enough about the order to know whether there is anything between them.

> Here we have a is the greatest lower bound of S with respect to <^*.

Ok.

> You say all subsets of R must have a minimum, M.

Under the assumption that $<^*$ is a well-order, all nonempty subsets have a minimum. Note well. Nonempty subsets.

> Now M is in S and a<*M.

Not necessarily. Let $<^*$ be the order on the reals:

$0 <^* 1 <^*$ everything else

Let $S = \{ x \in \mathbb R : 0 <^* x <^* 1 \}$

Then $S$ is empty and does not have a minimum element

> Therefore the set {x: a<*x<*M} is empty.

There is no such M. It doesn't exist. There's nothing between 0 and 1. It's like looking for an integer between 0 and 1. There isn't one.

> There can be no elements between a and M.

There is no M.

> Carrying this further ...

is pointless. M doesn't exist. Your set $S$ is empty whenever there's nothing strictly between $a$ and $b$ in the given order.

Edited by wtf

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Yes, I quoted you, but deleted your quote and then gave my response which made my reply confusing.

What you are saying then is that you already assume S is WO and hence u can get the result it is WO. I'll have to think about your claim that the set we are arguing about is empty. If u can give an example set where u say its empty then u can deny my claim. you are giving an example of a set you concocted; I'm presenting a different set.

Yes, that's what I have read about ordinals before: that each number is really represents the set of its predecessors. What puzzles me is what is the purpose of all this.

Anyway, I'll think about all this.

Edited by discountbrains

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2 hours ago, discountbrains said:

Yes, I quoted you, but deleted your quote and then gave my response which made my reply confusing.

What you are saying then is that you already assume S is WO and hence u can get the result it is WO. I'll have to think about your claim that the set we are arguing about is empty. If u can give an example set where u say its empty then u can deny my claim. you are giving an example of a set you concocted; I'm presenting a different set.

Yes, that's what I have read about ordinals before: that each number is really represents the set of its predecessors. What puzzles me is what is the purpose of all this.

Anyway, I'll think about all this.

> What you are saying then is that you already assume S is WO and hence u can get the result it is WO.

No that is not true. My order is 0 <* 1 <* everything else. Then the set of reals strictly between 0 and 1 is empty and has no minimum. It's not a well-order but it falsifies your argument. I'm not using a well-order, I'm not assuming anything is a well-order

But I'm not assuming S is well-ordered. You claimed S is nonempty. I showed an example where S is empty.

> I'll have to think about your claim that the set we are arguing about is empty.

What is the set of integers strictly between 0 and 1?

> If u can give an example set where u say its empty then u can deny my claim.

I gave the example. Twice in the same post. A third time just above. You have the order 0 <* 1 <* everything else. Then the set of reals strictly between 0 and 1 is empty. That's the example that falsifies your argument.

> I'm presenting a different set.

No matter what a and b you pick, I'll just give you the order a <* b < everything else, and then the set of reals strictly between a and b is empty.

> What puzzles me is what is the purpose of all this.

Ordinals are connected with proof strength in logic. They're used to develop the von Neumann hierarchy of sets. They're used to study well-orders. They're an interesting source of counterexamples in topology. For example the first uncountable ordinal, with the order topology, is compact but not sequentially compact. That's because sequences are "too short" to reach their limit. They're the basis of transfinite induction. Lots of other uses in math. The ordinals were discovered by Cantor in his investigations of the zeros of Fourier series. That's an interesting historical point. Set theory came directly out of the study of the physics of heat transfer.

> Anyway, I'll think about all this.

Do you understand the 0 <* 1 <* everything else order, and how there are no reals strictly between 0 and 1 in this order?

Edited by wtf

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I keep telling u over and over that the set I'm presenting is a set that cannot be WO. You keep presenting a different set. This is not my set.

Do u think I can use the induction theorem implies your "everything else" does not include all x in S? I think I can; I'm putting my argument together to post.

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1 hour ago, discountbrains said:

I keep telling u over and over that the set I'm presenting is a set that cannot be WO. You keep presenting a different set. This is not my set.

Do u think I can use the induction theorem implies your "everything else" does not include all x in S? I think I can; I'm putting my argument together to post.

How do you know your set can't be well ordered? That's something you are required to prove. My example falsifies your proof and any similar proof. I have taken YOUR EXAMPLE and shown that the set S might well be empty, making your proof fail.

> I'm putting my argument together to post.

Edited by wtf

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Using the induction theorem might be a little tricky for a lot of proofs. One has to be careful.

Here's what I think should be sufficient to show your example doesn't fit my set definition: Lets pick any arbitrary z in your 'E, everything else'.  So, there could be some x<*z in E and clearly there are x's in E such that z<*x and we generate the set T={x: z<*x}. You must have a min, m, in T such that m <* or = all x in T and it has to be that z<*m and a set {x: z<*x<*m} must be an empty set (you maintain). Keep in mind T is all x in S such that z<*x. S might be R or (a,b)-with usual order-a subset of R, a and b are real numbers....... I think I know what you are saying. That is, how do I know my set (S,<*) can be constructed for any order relation <* or that {x: z<*x<*m} has to contain numbers? I can still build such sets from your 'everything else'. This has always been my premise from the beginning that for any possible order relation sets of the above form can always be constructed.

I need to look up and see if the union of any collection of any countable sets is still countable and maybe use this result to help prove my claim.

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On 7/31/2019 at 1:36 PM, discountbrains said:

Defining S=(0,1) is not needed or helpful. Let S be a subset of R defined as S={x: a<*x<*b} . Both of you here agree S exists and is nonempty.

My "0 < 1 < everything else" example was for the purpose of falsifying this particular point. You said "Both of you" agree S is nonempty. I don't know who both of you is, but I certainly don't agree S is nonempty. I gave an example in which S is empty.

Now of course you can run your argument on the "everything else" part by picking a and b in the everything else portion of the ordering. BUT I AM NOT USING THAT EXAMPLE except to falsify your original claim that S is nonempty.

Suppose the order is 0 < 1 < 2 < everything else. Then you might have picked 1 and 2 as your a and b and your S is still empty.

Suppose you pick a and b in the everything else part of this new order. Then I'll give the example 0 < 1 < 2 < 3 < everything else and you might have picked 2 and 3 as a and b and S is empty.

So what you did is take a specific counterexample I gave to your specific claim, and pretend that I gave the example as some general refutation of your argument. It's not. It's a specific refutation of a specific claim.

But the 0 < 1 < everything else order is an example of a CLASS of counterexamples that will refute your argument no matter how you clip an initial segment.

Your argument fails because no matter what a and b you choose, there might not be any other reals between them in the <* order. Your order might be:

Everything less than pi <* pi <* pi + 1 < everything else.

Then your a and b might be pi and pi+1, and your interval is empty. So I don't even need initial segments. The discrete portion of the order might happen in the middle.

You can never be certain that your (a,b) interval in the <* order isn't empty.

> I need to look up and see if the union of any collection of any countable sets is still countable and maybe use this result to help prove my claim.

You need the axiom of countable choice to proof that. In the absence of countable choice it's not true in general. Although it's true for specific cases like the snaking diagonals proof of the countability of the rationals.

Edited by wtf

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z zz2

OK, here's my proof for your perusal:   z zz2

The proof is by contradiction. Let S =(a,b) where a,b R, 0< a <b. Let T= {x: a<*x<*b} where <* is a total order relation other than, <. I define S this way so all z∈S are in T. S is an denumerable set so T is also. If <* well orders T then ∃z∈T such that z0 ≤*x ∀x∈T. It must be that a<*z and {x: a<*x <*z} is an empty set

z0 ≤*x ∀x∈T T..... I'm going to have to write this out on paper and scan it as an image. This is just too hard. I was nearing the end and it erased my work. I am proving by induction we can generate a denumerable string of minimums of every subset of T arising from after the previous minimum was deleted. This is a denumerable set and actually contains all elements of T. If we choose a z in T it is the min of any subset of T of numbers greater or equal to it. Therefore its also a member of a string of minimums of T. So, T is = to this denumerable string of minimums. T thus is denumerable. Contradiction.

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The proof is by contradiction. Let S =(a,b) where a,b R, 0< a <b. Let T= {x: a<*x<*b} where <* is a total order relation other than, <. I define S this way so all z∈S are in T.

But that is TOTALLY FALSE.

You have S = (a,b) in the standard order. But in the <* order, T may be empty. I've already shown you the counterexample. In the orders 0 <* 1 <* everything else, we have S = (0,1), the USUAL open unit interval. But T is the set of reals strictly between 0 and 1 in the <* order, and T is empty. We've had this same conversation many times already in the past few weeks.

In your example with a = 0 and b = 1, S is the usual unit interval, and T is empty. Do you understand that?

I define S this way so all z∈S are in T.﻿

But that is nonsense. S is the usual standard open unit interval, and T is empty exactly because of the way I defined <*.

Which part of this is confusing you? I have given the same example to the same argument maybe six or eight times already over the past couple of months. If the <* order is: "0 <* 1 <* everything else" then T = {x : 0 <* x <* 1} is empty.

Try to name a real that's in T under the order "0 <* 1 <* everything else". That's a perfectly legitimate total order and T is empty. There's nothing between 0 and 1 in this order.

S is an denumerable set so T is also.

No, S is uncountable as is perfectly obvious, it's the open unit interval in the standard order. And T is empty because there's nothing between 0 and 1 in the <* order.

But your eternal error is that you think T is S or even related to S. It's not. T is an open interval in the <* order. It's empty by the way I defined <*. The fact that S has lots of reals in it means nothing. <* is an entirely different order.

Don't you see that there are NO REALS AT ALL between 0 and 1 in the <* order?

I am proving by induction we can generate a denumerable string of minimums of every subset of T arising from after the previous minimum was deleted.

You could never do that because T MAY BE EMPTY.

Edited by wtf

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