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

[wtf]

T CAN be nonempty but it MAY be empty. Your proof fails already in the second sentence. 

[discountbrains]

Here it is. If you find any little mistakes let me know. For reasons I gave above how can we imagine the only sets that satisfy my definition above of T have to be empty.

[wtf]

46 minutes ago, discountbrains said:

Here it is. If you find any little mistakes let me know. For reasons I gave above how can we imagine the only sets that satisfy my definition above of T have to be empty.

I found a mistake. I found the same mistake multiple times over the past several months that this thread has been going on.

Your claim is that there is NO well order of the reals. Your proof consists of picking SOME order and showing it's not a well-order.

But you have to prove that EVERY order is not a well-order. So you have to consider the case where T might be empty. And since you never consider that case -- indeed, since you give no appearance of even understanding the point -- you have not proven your claim.

BTW pasting an image makes it impossible for me to reply line by line. But your approach is wrong. You have to prove that <* can NEVER be a well-order. You can't cherry-pick some <* where T is nonempty. Because there are perfectly obvious <* orders, and choices of a and b, for which T is empty. Your proof fails in your second sentence.

Edited August 10, 2019 by wtf

[wtf]

ps -- You define T = {x : a <* x <* b} and then in your next sentence you claim a and b are elements of T. That's ridiculous.

Edited August 11, 2019 by wtf

[wtf]

pps -- By the end you claim T = S. But that's absurd. I already showed you the example of the 0 <* 1 <* everything else order in which S = (0,1), the uncountably infinite open unit interval; and T = {x : 0 <* x <* 1} is empty. 

If you would take the time to understand this example you would find it enlightening. 

Edited August 11, 2019 by wtf

[discountbrains]

 I know what you're trying to say. Since you are starting with its given every set can be well ordered it must follow my sets are empty. It that not correct? Somebody here doesn't understand what's going on. First of all, I defined S=(0,1) with usual order not your set. Do u not understand I am presenting a set, T, which I showed can't be well ordered. You keep saying my sets are empty. Let me put it this way: we can think of any uncountable subset of the reals as an infinite deck of cards. We can shuffle this deck any way u like and get a new order. If u pick any card and claim there's no other card below it I can always find one that is. Or, lets replace any two digits of some decimal number with two others and then replace the two others with another two. We have a reordering. Or we might change all the digits of every number. Whatever number u say is the first I still can find another number before it. This doesn't matter how u order things. What u are trying to say is my set T does not exist by saying its empty.

[discountbrains]

Here's another way to look at it: We don't hesitate writing a set like {x: a<x<b} with natural ordering. Isn't it natural to write {x: a<*x<*b} for a different ordering, <*? ....Lets suppose there is some dude in an alternative universe for which <* is the natural order of things and < is an alternative order? So, any of these sets will have plenty of elements in them.

[uncool]

5 minutes ago, discountbrains said:

Here's another way to look at it: We don't hesitate writing a set like {x: a<x<b} with natural ordering. Isn't it natural to write {x: a<*x<*b} for a different ordering, <*?

It is perfectly natural.

5 minutes ago, discountbrains said:

So, any of these sets will have plenty of elements in them.

This doesn't follow. Naturally defined sets can be empty.

[discountbrains]

19 hours ago, wtf said:

ps -- You define T = {x : a <* x <* b} and then in your next sentence you claim a and b are elements of T. That's ridiculous.

I did no such thing!

[wtf]

1 hour ago, discountbrains said:

I did no such thing!

It's really tedious to work this way, I'm not going to mark up your image post like this again. Please write out your argument line by line in a way that can be quoted in the forum. 

It says, plain as day what T is, and then you say a, x, b ∈ T.

Of course x is a variable used to define a set, it's not an element of anything in this context. 

The larger point here is that you are NOT FREE to choose a and b. If you claim <* is not a well-order, you have to make your argument for arbitrary a and b. But you can't possibly do that since "a <* b <* everything else" always results in a T that is empty.

BTW you can use a site like https://math.typeit.org/ for math markup.

We're now at the point where you have made the same faulty argument repeatedly for weeks, and I keep giving the same counterexample. But now you're posting images that are labor-intensive to quote, and you don't seem to be proofreading your own work. We're going backward.

 

Edited August 11, 2019 by wtf

[wtf]

5 hours ago, discountbrains said:

 I know what you're trying to say. Since you are starting with its given every set can be well ordered it must follow my sets are empty. It that not correct? Somebody here doesn't understand what's going on. First of all, I defined S=(0,1) with usual order not your set. Do u not understand I am presenting a set, T, which I showed can't be well ordered. You keep saying my sets are empty. Let me put it this way: we can think of any uncountable subset of the reals as an infinite deck of cards. We can shuffle this deck any way u like and get a new order. If u pick any card and claim there's no other card below it I can always find one that is. Or, lets replace any two digits of some decimal number with two others and then replace the two others with another two. We have a reordering. Or we might change all the digits of every number. Whatever number u say is the first I still can find another number before it. This doesn't matter how u order things.

>  I know what you're trying to say. Since you are starting with its given every set can be well ordered it must follow my sets are empty. It that not correct?

No it is not correct. I am NOT starting with a well-order. And your T sets may be empty even if <* is not a well order. 

> Somebody here doesn't understand what's going on.

That's right. 

> Do u not understand I am presenting a set, T, which I showed can't be well ordered.

For suitable choice of <*, which need not be a well-order, T is empty and your argument fails.

> You keep saying my sets are empty.

T MAY be empty for suitable choice of <* and a and b. But to make your argument work, you have to show that T is nonempty for any possible <*, a, and b. But you can't do that because there is always a counterexample of the form I gave.

> What u are trying to say is my set T does not exist by saying its empty.

It does exist. It just happens to be empty. 

Suppose we take the order "0 < 1 < everything else". Now this is NOT a well-order because the "everything else" is in its usual dense order. But T = {x : 0 <* x <* 1} is empty. DO YOU OR DO YOU NOT AGREE WITH THIS? Let's not discuss anything else till we get clarity since this example falsifies your argument.

 

2 hours ago, discountbrains said:

Here's another way to look at it: We don't hesitate writing a set like {x: a<x<b} with natural ordering. Isn't it natural to write {x: a<*x<*b} for a different ordering, <*? ....Lets suppose there is some dude in an alternative universe for which <* is the natural order of things and < is an alternative order? So, any of these sets will have plenty of elements in them.

Sure, it's perfectly natural to write that. But let's get clarity on this one thing. 

Suppose <* is defined as: "0 <* 1 <* everything else". Now this is NOT a well order since the "everything else" is all the real numbers in their usual order, with the exception of 0 and 1, which we pulled out and put in front of everything else.

Then T = {x : 0 <* x < 1} is empty. Let's stop right here and drill this down till this is clear to you. Do you see that with this particular <* it must be the case that T is empty?

Edited August 11, 2019 by wtf

[discountbrains]

On 8/10/2019 at 4:24 PM, wtf said:

I found a mistake. I found the same mistake multiple times over the past several months that this thread has been going on.

Your claim is that there is NO well order of the reals. Your proof consists of picking SOME order and showing it's not a well-order.

But you have to prove that EVERY order is not a well-order. So you have to consider the case where T might be empty. And since you never consider that case -- indeed, since you give no appearance of even understanding the point -- you have not proven your claim.

BTW pasting an image makes it impossible for me to reply line by line. But your approach is wrong. You have to prove that <* can NEVER be a well-order. You can't cherry-pick some <* where T is nonempty. Because there are perfectly obvious <* orders, and choices of a and b, for which T is empty. Your proof fails in your second sentence.

That's not what I'm trying to say. Clearly, to me I don't intend to cherry pick any order relation. I believe I clearly stated this applies to any order you choose. I should revisit my proof, but I should stipulate a,b are in S, not in T, and T contains all other elements of S. This leaves no doubt T is nonempty.

Interesting some of my replies are missing. Could it be I used a popular slang word that's not demeaning to anyone?

2 hours ago, wtf said:

>  I know what you're trying to say. Since you are starting with its given every set can be well ordered it must follow my sets are empty. It that not correct?

No it is not correct. I am NOT starting with a well-order. And your T sets may be empty even if <* is not a well order. 

> Somebody here doesn't understand what's going on.

That's right. 

> Do u not understand I am presenting a set, T, which I showed can't be well ordered.

For suitable choice of <*, which need not be a well-order, T is empty and your argument fails.

> You keep saying my sets are empty.

T MAY be empty for suitable choice of <* and a and b. But to make your argument work, you have to show that T is nonempty for any possible <*, a, and b. But you can't do that because there is always a counterexample of the form I gave.

> What u are trying to say is my set T does not exist by saying its empty.

It does exist. It just happens to be empty. 

Suppose we take the order "0 < 1 < everything else". Now this is NOT a well-order because the "everything else" is in its usual dense order. But T = {x : 0 <* x <* 1} is empty. DO YOU OR DO YOU NOT AGREE WITH THIS? Let's not discuss anything else till we get clarity since this example falsifies your argument.

 

Sure, it's perfectly natural to write that. But let's get clarity on this one thing. 

Suppose <* is defined as: "0 <* 1 <* everything else". Now this is NOT a well order since the "everything else" is all the real numbers in their usual order, with the exception of 0 and 1, which we pulled out and put in front of everything else.

Then T = {x : 0 <* x < 1} is empty. Let's stop right here and drill this down till this is clear to you. Do you see that with this particular <* it must be the case that T is empty?

No, no, no, NO! You don't redefine my T!!! What is the issue here is that <* is arbitrary and you can define it any way like. 

Three times u redefined my set, T....... OK, got it. I will need to take a look at your example of <*. Maybe I can find a way around it. Wait a minute, who said the "everything else" is in the usual order? I really think I need not even require x<*b in T. All I'm concerned about is the lower bound or the min .

[wtf]

5 hours ago, discountbrains said:

That's not what I'm trying to say. Clearly, to me I don't intend to cherry pick any order relation. I believe I clearly stated this applies to any order you choose. I should revisit my proof, but I should stipulate a,b are in S, not in T, and T contains all other elements of S. This leaves no doubt T is nonempty.

Interesting some of my replies are missing. Could it be I used a popular slang word that's not demeaning to anyone?

No, no, no, NO! You don't redefine my T!!! What is the issue here is that <* is arbitrary and you can define it any way like. 

 

 

> That's not what I'm trying to say. Clearly, to me I don't intend to cherry pick any order relation. I believe I clearly stated this applies to any order you choose.

That's right. You claim that for ANY linear order, you can prove it's not a well order. Ok, then I get to pick the order. And if you tell me which [math]a[/math] and [math]b[/math] you want to use, I can still modify the order and dare you to make your proof work. That is CORRECT, you do understand that. 

> Interesting some of my replies are missing. Could it be I used a popular slang word that's not demeaning to anyone?

I represent nobody but myself here and have my own issues with the moderators from time to time. I'm the last person who can do anything about your posting issues. 

> No, no, no, NO! You don't redefine my T!!! What is the issue here is that <* is arbitrary and you can define it any way like. 

I'm not redefining your T at all. As you agree, I am allowed to pick the total order <* and then challenge you to prove it's not a well-order. And once you tell me your [math]a[/math] and [math]b[/math], which is an unnecessary complication we could do without, I can just adjust my <* accordingly. 

For example you want [math]0 < a < b < 1[/math] where < is the usual order. Then my <* is just "a < b < everything else" and your T is empty. 

> Three times u redefined my set, T

I never redefined it. I used it exactly as you specified it. You're right that I used 0 and 1 instead of a and b but that's a minor detail. I can use a and b if you like. 

> ....... OK, got it. I will need to take a look at your example of <*.

I would truly ask you to do that. 

> Maybe I can find a way around it.

Give it a try. When you understand my example you'll see why your proof doesn't work.

> Wait a minute, who said the "everything else" is in the usual order?

I said that though I never said it explicitly. My order <* is "0 < 1 < everything else in its usual order". 

So for example 0 <* 1, 1 <* 47, 102 <* 104 and so forth. 

But the numbers in (0,1) are of special interest. We have 1 <* 1/2, 1 <* pi - 3, 1 < sqrt(2) -1, and so forth. 

0 <* 1 <* everything else in its usual order. So this order has a discrete initial segment and a dense tail. 

If you want to call it a and b that's fine but I'll just use a <* b <* everything else in its usual order. 

So in the order 0 < 1 < everything else [can I stop writing "in its usual order" and let that be understood?], can you see that T = {x : 0 <* x <* 1} must be empty? It MUST be since we defined the order exactly so it would be! I am allowed to do that since I get to pick the order and you must supply a proof it's not a well order. 

So to narrow this down to one simply YES or NO question: 

If I define <* as 0 <* 1 <* everything else, would you agree that T = {x : 0 <* x <* 1} must be empty? Just answer this, and then we can move on. We must get clarity on this single point before discussing anything else. 

> I really think I need not even require x<*b in T. All I'm concerned about is the lower bound or the min .

An empty set does not have a lower bound OR a min. And since I can always cook up a <* that makes T empty, your proof fails.

[discountbrains]

I see in my latest definition of T I made a couple minor errors. Since T doesn't contain a and b T can't = S near the bottom of my proof. Since my initial conditions state all other elements in S are in T clearly T is not empty. However, are you really a math prof? You know and you stated your example is not a well ordering. I don't care about this at all. This is entirely not the point. I care only about what is proposed to be a WO and about 6 lines down I conclude such an ordering implies, in fact, various sets I present must be empty. You do realize if u can produce an ordering that well orders my set thus negating my argument u actually have done something pretty profound.  You will have actually produced a WO for the reals. You seem to be stuck the idea of finding anything to show my T is empty.  Yet I get -32 votes here-amazing.

[wtf]

2 hours ago, discountbrains said:

I see in my latest definition of T I made a couple minor errors. Since T doesn't contain a and b T can't = S near the bottom of my proof. Since my initial conditions state all other elements in S are in T clearly T is not empty. However, are you really a math prof? You know and you stated your example is not a well ordering. I don't care about this at all. This is entirely not the point. I care only about what is proposed to be a WO and about 6 lines down I conclude such an ordering implies, in fact, various sets I present must be empty. You do realize if u can produce an ordering that well orders my set thus negating my argument u actually have done something pretty profound.  You will have actually produced a WO for the reals. You seem to be stuck the idea of finding anything to show my T is empty.  Yet I get -32 votes here-amazing.

You seem to have changed the subject and decided not to answer the question I posed to you. Not much for me to say here. Let me know if you want to discuss some math at some point.

>  Since my initial conditions state all other elements in S are in T clearly T is not empty.

You may "state" that T is nonempty but my example shows that it might be. And you're making a point of not responding to that issue. So for the record, I ask again: If we define <* as "0 <* 1 <* everything else in its usual order", is T = {x : 0 <* x <* 1} empty or not?

Edited August 12, 2019 by wtf

[discountbrains]

I admit I should take more time for serious thought on your example, but will need to see if your example is meant to be totally unworkable with my definition of T. Yes, again I can come up with definitions of <* that will make T seem empty right off the bat. So what. If I say T contains all elements of S except a and b  while a and b are also in S then that's what T contains. Again defining <* to make T contain nothing is beside the point because u must have at least one min in T to make T well ordered. You're trying to make T empty so u don't have to answer the question. You could offer to help me define T and <* like suggest I start by saying "if T is a nonempty set".

[wtf]

5 hours ago, discountbrains said:

I admit I should take more time for serious thought on your example, but will need to see if your example is meant to be totally unworkable with my definition of T. Yes, again I can come up with definitions of <* that will make T seem empty right off the bat. So what. If I say T contains all elements of S except a and b  while a and b are also in S then that's what T contains. Again defining <* to make T contain nothing is beside the point because u must have at least one min in T to make T well ordered. You're trying to make T empty so u don't have to answer the question. You could offer to help me define T and <* like suggest I start by saying "if T is a nonempty set".

> I admit I should take more time for serious thought on your example,

That would be a good idea. I'd like you to just say YES or NO to my question: If <* is defined as "0 <* 1 <* everything else", is T = {x : 0 <* x <* 1} empty?

If you agree, then there's something we agree on. If you disagree, we can focus on this example till we agree.

Now having done that, you might argue that I'm wrong to MENTION it. We can argue about its RELEVANCE. But at least we would agree on whether this particular T is empty or not. That is important. 

> but will need to see if your example is meant to be totally unworkable with my definition of T.

We can argue about that LATER. First just say YES or NO as to whether my T is empty or not. Once we do that, you can argue that I had no right to mention it in the first place. But it would be something we agree on. And I can then explain why it's relevant. So please don't let trying to figure out the relevance, get in the way of you just saying YES or NO to this basic question.

> Yes, again I can come up with definitions of <* that will make T seem empty right off the bat. So what. If I say T contains all elements of S except a and b  while a and b are also in S then that's what T contains.

Fine. Let's talk about that LATER. First please answer the question. I for one can not go forward unless we have agreement on that one technical point. I'll justify its relevance later, but first it's important for us to at least agree on the basic fact.

> Again defining <* to make T contain nothing is beside the point because u must have at least one min in T to make T well ordered.

Ah. No I am not trying to show anything is well ordered. I am pointing out that YOUR PROOF FAILS that <* is NOT well-ordered. Do you see that?

> You're trying to make T empty so u don't have to answer the question.

I'm giving an example where T does happen to be empty, because your proof depends on T not being empty. If T is empty even for a SINGLE <*, then your proof fails; because you have to make your proof go through for EVERY linear order <*. Once there exists even a SINGLE <* that makes T empty, your proof fails because the empty set does not have a min. Remember well-order definition is that every NONEMPTY set has a smallest element. This is vital.

> You could offer to help me define T and <* like suggest I start by saying "if T is a nonempty set".

Ah you think I am being deliberately unhelpful to not do so. On the contrary. If I could say T is nonempty I would. But here is the problem:

We are playing the following game. You claim that NO linear order <* can be a well-order. So I am entitled to cherry-pick a <* that makes T empty, which blocks your proof. That's why I'm using that example. Not to be unhelpful. But to be HELPFUL. To show you the error in your proof.

 

Edited August 13, 2019 by wtf

[discountbrains]

Yes, Neither 0 or 1 are in T. 1<*everything else makes any other x and y such that (x,y)∈<* not be in T. So T is empty with your order relation. You are pointing out an imprecision in my argument. That's good. I should restrict any order relation to be if x and y are in T, (x,y)∈<*. Or this could be written in other ways......

No, there are lots of well orderings for a lot of sets. I'm trying to show there are some sets that can't be well ordered.

[wtf]

7 hours ago, discountbrains said:

Yes, Neither 0 or 1 are in T. 1<*everything else makes any other x and y such that (x,y)∈<* not be in T. So T is empty with your order relation. You are pointing out an imprecision in my argument. That's good. I should restrict any order relation to be if x and y are in T, (x,y)∈<*. Or this could be written in other ways......

No, there are lots of well orderings for a lot of sets. I'm trying to show there are some sets that can't be well ordered.

> Yes, Neither 0 or 1 are in T. 1<*everything else makes any other x and y such that (x,y)∈<* not be in T. So T is empty with your order relation. You are pointing out an imprecision in my argument. That's good

Ok!! Good. We agree that if I am allowed to cherry-pick <*, then I can make T be empty. 

Now YOU are claiming that the reals can't be well-ordered. So your argument is: Given ANY <* and you'll show it's not a well-order.

But I showed that for SOME <* orders, T is empty and your proof is blocked, because you can no longer pick the min of T. 

So now, you need a new argument. Yes?

> I'm trying to show there are some sets that can't be well ordered.

Aren't you trying to show explicitly that the reals can't be? It's impossible to do that because even in the absence of the axiom of choice, it's possible that the reals are well-ordered. But nobody can produce an explicit well order.

Since it is logically consistent for the reals to be well-ordered even in the absence of Choice, you can not possibly prove that the reals aren't well-ordered. We covered this point months ago if I recall.

 

[discountbrains]

Yes, I know that's exactly what I'm saying: the reals can't be well ordered. And, you say there is a proof without the AC that says they can. Yet, I'm getting the result I'm getting. I think I can come up with many, many examples of your logic. I might say all my numbers in my set S add up to a whole number like n+m=p (p a whole number). For example 1/3+2/3 =1, 2+1=3... and you say what about 1/2+1/3 etc? And then say my statement is false. This may not be an exact analogy though. Let me reiterate, I believe my last version of <* must include numbers from T. Let me see if this is what u are saying: I say all the car prices on a Chevy dealer's lot are ordered a certain way and u say my statement is invalid because the Ford dealers are not ordered that way.

[wtf]

20 minutes ago, discountbrains said:

Yes, I know that's exactly what I'm saying: the reals can't be well ordered. And, you say there is a proof without the AC that says they can. Yet, I'm getting the result I'm getting. I think I can come up with many, many examples of your logic. I might say all my numbers in my set S add up to a whole number like n+m=p (p a whole number). For example 1/3+2/3 =1, 2+1=3... and you say what about 1/2+1/3 etc? And then say my statement is false. This may not be an exact analogy though. Let me reiterate, I believe my last version of <* must include numbers from T. Let me see if this is what u are saying: I say all the car prices on a Chevy dealer's lot are ordered a certain way and u say my statement is invalid because the Ford dealers are not ordered that way.

> And, you say there is a proof without the AC that says they can.

No it's more subtle than that. There is a model of set theory in which the reals can be w.o. but Choice fails. But this is not relevant to our discussion, except that you are looking for a proof that the reals can't be w.o. and there can be no such proof.

The rest of your post I did not find helpful. I don't know why you're adding numbers. Your car analogy doesn't apply. You are trying to prove that no <* can be a w.o. but your proof is faulty, as you already agreed. 

>  Let me reiterate, I believe my last version of <* must include numbers from T.

Which version of <*? It will be very helpful for you to write out a specific proof. 

But it doesn't matter that SOME <* has a nonempty T, that's perfectly fine. Your proof depends on NO <* having a nonempty T and that is false.

 

 

[discountbrains]

Here's your problem: You have a problem with your counterexample. In order to demonstrate there is a well order relation for R you must have that my set, T, contains an element to be a min for T. Hence for any order you present to be a WO T cannot be empty! Its as simple as that. I actually should have said this a while back. Of course, there is an unimaginable number of examples of order relations which would give no elements in T. You have to present one that produces an element in T to be a min of T and also elements of all of its subsets likewise.

Naturally anyone is on risky ground when they go against conventional wisdom like me. I will revert back to my original title of my post. I think I mentioned contacting the AMS journal and was told my work did not fit their requirements by their secretary-true its no multi page article. Wonder if she might even be another one of these low level gate keepers like are encountered many places. I have another issue with your example; won't bother mentioning it here. I'm comfortable with my argument now. You brought attention to potential flaws. There are other math societies. I haven't had any direct access to any academia for a long time.  I will get someone's attention somewhere. My 'proof(?)' is what it is.

[uncool]

7 minutes ago, discountbrains said:

In order to demonstrate there is a well order relation for R you must have that my set, T, contains an element to be a min for T.

Why? Well-ordering says that any nonempty set has a minimum. If you are using the definition of well-order to say that T has a minimum, you must show T is nonempty.

7 minutes ago, discountbrains said:

My 'proof(?)' is what it is.

What is it, at this point? Please post it in full - with no elisions, no "This is obvious", no "This should be easy to see". 

Edited August 14, 2019 by uncool

[discountbrains]

I could simply sate for my S above S\{a,b} ⊂ T and <* is understood to be an order for T. 

9 minutes ago, uncool said:

Why? Well-ordering says that any nonempty set has a minimum. If you are using the definition of well-order to say that T has a minimum, you must show T is nonempty.

What is it, at this point? Please post it in full - with no elisions, no "This is obvious", no "This should be easy to see". 

Look back several posts and see where it shows as an image where I wrote it out by hand. There were too many problems with changing font size when using subscripts.

[uncool]

I can't quite read this line: "Note T can be nonempty since we can choose any [???] and b elementof S"; however, you have already chosen a and b at this point. Your proof would be better if you chose a and b appropriately to begin with.

"To be sure, suppose z is an arbitrary element of T. z is the min of some subset of T since there exists a subset of T of all elements greater than or equal to z. Hence z is an element of this string [...]"

This doesn't follow.