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

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59 minutes ago, wtf said:

Secondly, suppose you do find a nonempty interval. Then the argument as I understand it says that there's a smaller element, ad infinitum, giving  a countably infinite sequence of ever smaller reals.

I don't see that in the proof being used.

The proof seems to be based on the claim that a well-ordered set must be exhausted by the denumerable sequence of its minimum, the successor of the minimum, the successor of that, etc. This is distinct from the "halfway between" proof that discountbrains was using earlier, and is on surer footing notationally, but fails in that it cannot prove the above key claim (and in fact, the claim is false, as you have pointed out with omega + omega). 

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17 hours ago, wtf said:

I said IF <* happens to be a well order, your proof would fail. How do you know it isn't?

If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though.

17 hours ago, wtf said:

I said IF <* happens to be a well order, your proof would fail. How do you know it isn't?

If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though.

For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too.

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Posted (edited)
3 hours ago, discountbrains said:

If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though.

If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though.

For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too.

If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'.

Ok. Take the natural numbers. Then 0 is the smallest element. And S' might be all the numbers greater than or equal to 14. Ok. What does that prove? Of course there are always UPWARD chains, even infinite ones depending on the order.

And can someone please point me to the definitive version of the proof? I seem to have missed the new proof. 

For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too.

Wait what? A moment ago m was the minimum of S, now you have some a that's smaller. What is the argument you are making?

Edited by wtf

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1 hour ago, wtf said:

>Wait what? A moment ago m was the minimum of S, now you have some a that's smaller. What is the argument you are making?

Just because I labeled a number, m, doesn't mean its a min of the set. I'm only saying we're picking a number and testing it to see if its the min of the set. But, we know and I agree there are some sets (natural numbers etc)that can easily be well ordered. And, we can come up with a lot of well orders for a lot of sets. BUT, we cannot come up with an order that well orders all sets. My little one line above should really be my proof.

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24 minutes ago, discountbrains said:

Just because I labeled a number, m, doesn't mean its a min of the set. I'm only saying we're picking a number and testing it to see if its the min of the set. But, we know and I agree there are some sets (natural numbers etc)that can easily be well ordered. And, we can come up with a lot of well orders for a lot of sets. BUT, we cannot come up with an order that well orders all sets. My little one line above should really be my proof.

I didn't see an argument. Can you repeat it? All you said is that 0, say, is less than or equal to all natural numbers; and 14 is less than or equal to all of them except for some. But so what? There's no argument there.

> BUT, we cannot come up with an order that well orders all sets.

We can't "come up" with one, but we can easily prove that one exists in the presence of the axiom of choice. And even without choice, the reals might be well ordered even if some other set isn't. You haven't got a proof.

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3 hours ago, wtf said:

And even without choice, the reals might be well ordered even if some other set isn't.

There is no ONE subset that has no min for all order relations. Its just, given any <*, there is always a set of the type {x: a <*x} that can be constructed.

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13 minutes ago, discountbrains said:

There is no ONE subset that has no min for all order relations. Its just, given any <*, there is always a set of the type {x: a <*x} that can be constructed.

That's just not a coherent argument. I can't understand what you are trying to say. 

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Posted (edited)

Allow me to ask for your take on my debunking of a portion of Einstein's relativity I posted on the physics forum a year ago. It involves one result of his theory and simple arithmetic. I believe uncool commented on my post there at the time:

1)  Suppose someone is sitting at a table with a wire going across it perpendicular to him. He measures the length of the wire and finds it 3ft long. A current flows through the wire and he adds up the incremental distances between the near light speed moving charges. Because Einstein's length contraction formula says all these incremental lengths become much smaller the sum of all the incremental lengths becomes much smaller than the original length of the wire.The observer measures it one way and gets one answer; measures another way and gets much shorter. Which is it? Optical illusion?

2) Two rocket ships are traveling on the same path from one planet to another at the same speed of 0.9999c. One is just 100mi  from the 2nd planet; the trailing ship is 100mi from the 1st planet. With relativity length contraction adding up the distances the planets, to an observer off to the side, are much, much closer than thought. If the ships are closer together the total distance becomes much larger.

28 minutes ago, wtf said:

That's just not a coherent argument. I can't understand what you are trying to say. 

Wrote this hurriedly and edited it. I'll simply state: For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed. So we don't go around and around again on this we assume there are numbers in the set which are comparable by <*.

Edited by discountbrains
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Posted (edited)
50 minutes ago, discountbrains said:

Allow me to ask for your take on my debunking of a portion of Einstein's relativity I posted on the physics forum a year ago.

 

...

 

Wrote this hurriedly and edited it. I'll simply state: For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed. So we don't go around and around again on this we assume there are numbers in the set which are comparable by <*.

I have no comments on matters of physics.

> For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed.

Ok. The usual order on the reals is a total order. We can form the set of all reals > 0 for example. What of it? Where is your argument? Premises, reasoning, conclusion. I see no argument.

Edited by wtf

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This is easy. Let S = all real numbers x such that x > 0. There is no least element in S. Of course this would be one of my sets.

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Posted (edited)
13 minutes ago, discountbrains said:

This is easy. Let S = all real numbers x such that x > 0. There is no least element in S. Of course this would be one of my sets.

Correct. You just proved that the usual order on the reals isn't a well order. 

However you have NOT shown that NO linear order on the reals can be a well order. Your proof fails.

I looked back a couple of pages but could not find the supposed new version of the argument. I see no argument at all being made. Of course it is true that you can find a countably infinite ascending sequence of reals in a given linear order. For example let [math]<^*[/math] be defined as:

all natural numbers [math]<^*[/math] everything else

Now you have a countably infinite well-ordered subset. 

But we could extend it up through all the countable ordinals and you still wouldn't have a proof. You have not presented a proof that there is NO possible well order on the reals.

But our current problem is that you have not presented a complete argument in quite some time; and I do not understand the purpose of the things you're saying. You take the trouble to point out that the usual order on the reals isn't a well-order? Why? We all agree that it isn't. Why bring it up now? 

Please give a complete argument. Give me something to work with. Nothing you say makes sense. In YOUR mind you think you're filling in details to something you've already explained. From where I sit there is no official version of your argument that I can refer to. I have no idea what you are talking about.

Please:

a) Point me to the post in this thread that expresses the coherent argument you're assuming I know; or

b) Give such a coherent argument here. Write a complete argument so I can understand what you're thinking so I can respond. What is the point of your one-liners? I have no idea WHY you are telling me these things.

 

Edited by wtf

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In my last message for any order relation one selects I displayed a set that can't be well ordered by that specific order. This is the only thing I really have to say about it. I believe this is all that is needed. I got off track in the many posts before when asked for proofs. I'm afraid this is the only proof I have that cannot be well ordered. This realization was what motivated me to assert this some time ago. If I somehow can think of why this needs further proof I'll make an attempt at it here. For now I guess this is all I have to say. But, is what I say any way a proof? All I can say is try constructing my (unspectacular) set for any order you choose on the reals. In order to test it one, of course, needs to know how their particular order relation orders the numbers. I don't know why such a set construction can't be justified based on any order. Yet, for that order a min of the set can't be found.

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Posted (edited)
1 hour ago, discountbrains said:

In my last message for

Can you please point me to the exact post that contains your proof? There are so many different versions. The proofs I've seen, I've debunked.

I do see you often posting a proof then later saying you made a mistake, but not going back and reposting a complete proof. It's impossible for me to have any idea what you are referring to when you refer to "your last message," which most definitely did not contain a proof. You haven't got a proof unless you post one.

You did say that you thought the [math]\omega + \omega[/math] example gives you problems. But then what about [math]\omega + \omega + \omega[/math] and so forth. The upward limit of that process is [math]\omega^\omega[/math]. You can keep going with that idea to get a countable tower of [math]\omega[/math] exponentiated with itself, which is a countable ordinal called [math]\varepsilon_0[/math]. And you can keep on going. These are all countable ordinals. The upward limit of all the countable ordinals is the first uncountable ordinal [math]\omega_1[/math], whose existence can be proved WITHOUT using the axiom of choice.

Nothing you've written handles any of these cases. You have no proof.

https://en.wikipedia.org/wiki/Epsilon_numbers_(mathematics)

https://en.wikipedia.org/wiki/First_uncountable_ordinal

Edited by wtf

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4 hours ago, wtf said:
4 hours ago, wtf said:

debunked

I need to look at this wiki ordinal stuff but I suspect it won't change what I'm saying. When I 1st studied it I thought I might be over thinking it. I told you 3 times already "Ignore what I said before my last assertion". And yes, I should have included the '<* includes numbers from a set T'. Most all theorems and definitions always include 'nonempty sets or a similar condition. My proof amounts to no more than a simple exercise in a textbook. Given any order, <*, a set S= {x: a<*x} can be constructed which has no min for <*. I don't have to demonstrate this for any order. You know what I said is true-its very fundamental. Quit trying to play like u don't understand my meaning. Think about it. No elaborate proof is needed. Maybe I'm wrong. Maybe there is some arcane theory I don't know about. If u think u are debunking me lets see a counterexample. This is getting tiresome; you keep throwing obstacles in my path at every turn. I hope you dropped that meaningless null set stuff too.

 

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45 minutes ago, discountbrains said:

Given any order, <*, a set S= {x: a<*x} can be constructed which has no min for <*. I don't have to demonstrate this for any order.

Um. Yes, you do, if you want claim you have proved that the reals cannot be well-ordered.

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Posted (edited)
2 hours ago, discountbrains said:

I need to look at this wiki ordinal stuff but I suspect it won't change what I'm saying. When I 1st studied it I thought I might be over thinking it. I told you 3 times already "Ignore what I said before my last assertion". And yes, I should have included the '<* includes numbers from a set T'. Most all theorems and definitions always include 'nonempty sets or a similar condition. My proof amounts to no more than a simple exercise in a textbook. Given any order, <*, a set S= {x: a<*x} can be constructed which has no min for <*. I don't have to demonstrate this for any order. You know what I said is true-its very fundamental. Quit trying to play like u don't understand my meaning. Think about it. No elaborate proof is needed. Maybe I'm wrong. Maybe there is some arcane theory I don't know about. If u think u are debunking me lets see a counterexample. This is getting tiresome; you keep throwing obstacles in my path at every turn. I hope you dropped that meaningless null set stuff too

> I need to look at this wiki ordinal stuff but I suspect it won't change what I'm saying.

I have in mind to write a bit of exposition that develops the ordinals in a very simple and natural way. My aim is to bring to life as best I can, a visualization of the space of the countable ordinals that might or might not be interesting to anyone who reads it, now or in the future. I'll give myself a couple of days to get this together, and meanwhile I'll fight the urge to post in this thread. Not saying I'll succeed. It's like flicking your tongue at a bad tooth.

 

> I told you 3 times already "Ignore what I said before my last assertion".

Yes and this is exactly the problem. Please read carefully what I'm about to write.

Your pattern is to post a "definitive" version of the proof. Then someone finds an error, and you post, "You're right, that was an error."

Now what NEVER happens is that you post a repaired version of your post. So what we have is a post containing a proof; followed by one or more posts saying, "Yes I see there's an error." So in YOUR mind there's some canonical version of the proof, which YOU think I see, but I assure you, I DON'T. I have no idea what the proof in your mind is; because you have not written it down. It's all in your head and frankly the ideas in your head about ordinals are fuzzy and malformed. You asked a long time ago for someone who "knows this material" to jump in. I am not a math prof as you asked once, just another forum denizen who loves math more than math loved me. But I know ordinals and I am qualified to tell you that you (a) Have not got a proof in general; and (b) Have not told me what your latest version of the proof is. That is my statement to you.  

> Quit trying to play like u don't understand my meaning.

I am not doing that and I WOULD NOT do that. I'm earnestly trying to understand your idea. If I tell you I don't follow your argument and/or don't even know what your latest argument is; it's with the intention of giving you feedback about your mathematical exposition

If you would recognize that I'm putting my mathematical experience at your service, instead of being someone who is deliberately playing dumb to make you feel bad; all this would go a lot better. Of course nothing I could ever say would definitively prove that my intentions are of the sincere and hopefully helpful kind. If you prefer to think the worst of me, well ... actually you might not be alone around here! LOL. 

I will agree that I can speak sharply. If an argument is garbage I say so. That is not ever meant as a personal judgment. But I am qualified to judge proofs in this particular area. If I say your argument is garbage, you should try to do better. You should ask me WHY it's garbage and I'll try to explain it.

 

Why do I insist that you write down a coherent proof?  It's so that your ideas will get sharper. Your intuitions will get sharper. Writing down a proof makes you think right about the mathematical abstractions. You don't write proofs for others. You write them for yourself. Your ideas are muddled and the reason I know that is that you haven't written down a coherent proof. 

Ok I'm going to go work up a little writeup on the countable ordinals as they pertain to the question of well-ordering the reals without the axiom of choice.

I should add that I"m trying not to show umbrage, but you should be aware that I was startled by your remark and did not respond to it kindly.  You are wrong about my playing dumb. You are wrong about imagining you've expressed an argument. Those two things are related.

And -- on the chance that I am wrong and you have in fact presented a perfectly coherent argument, which I'm too dumb to see -- then you should just repeat it for my dimwitted benefit; and in so doing, you would show that you appreciate my effort to understand you. I know, that's a lot to ask. I'm a dreamer.

Edited by wtf

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Posted (edited)
9 hours ago, wtf said:

>

 

 

 It's so that your ideas will get sharper.

I agree. This helps me have a clearer picture of what I'm up against. Uncool also says I produced no real proof. 

I'm going to make a new attempt: Like I said before there always exists a subset of like {x: a <*x} for some a for any total, linear order, <*, which has no min for <*. I think you and uncool want absolute evidence there is such a set. My proof is as follows: So, now lets assume for any <* there are at least a few subsets of the reals  that contain a minimum element. Otherwise, I am done: there are NO subsets with a min. Suppose we have a subset that has a min, z, and its S={x: z ≤*x}. Lets consider S\{z}. This set is {x: z <*x} and is the same form as the set on the first line and has no min with respect to <*. Thus, I have constructed an actual set with no min with respect to <*. Such a set actually exists. There exists at least one subset that contains no min. Since such an argument can be made for any order relation not all subsets of can be made to contain a min. So cannot be well ordered. I believe this is a proof.

 

Edited by discountbrains
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35 minutes ago, discountbrains said:

 Lets consider S\{z}. This set is {x: z <*x} and is the same form as the set on the first line and has no min with respect to <*.

Why not?

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1 hour ago, uncool said:

Why not?

Sorry, I forgot to add there are several ways to write a set. Of course {a, b, c,...}, {x: z<* x}, and {x: z ≤*x} among themfor a set to have a min it looks like the 3rd one which is a totally different set than the second one. I'm sure you know that. 

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Posted (edited)
45 minutes ago, discountbrains said:

Sorry, I forgot to add there are several ways to write a set. Of course {a, b, c,...}, {x: z<* x}, and {x: z ≤*x} among themfor a set to have a min it looks like the 3rd one which is a totally different set than the second one. I'm sure you know that. 

I know that if you are given z, then {x: z<* x}, and {x: z ≤*x} are different; I don't know that for all orders <*, {x: z1<* x} can't be written as {x: z2 ≤*x}, no - in fact, I sincerely doubt it. And "I'm sure you know that" is another example of this:

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

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.

Stop saying things like "I'm sure you know that", and start examining those parts closely.

Edited by uncool

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I think I've been aware all along what you are getting at. And, its a question I've raised for myself. You want me to prove there are numbers between the greatest lower bound for the set and any proposed min for the set. I wrote my last version to address this. I will write out in words some of my key points. Keep in mind we have a total order on the reals so for all u,v in u<*v or v<*u. Now if I have the a set            S={x: z ≤*x} this means this is the set of all numbers greater or equal to z. for all u,v in S not equal to z, u<*v or v<*u. Now take away the z and what do you have left? These are all numbers greater than z. If I pick any number greater than z its less or greater than any other number greater than z. Why can't this be true just like it is for the < relation? After all every number in R is  comparable by my definition of <*.  This is what this means to me. I wonder if it was suspected I was fairly well respected mathematician I would be taken more seriously.

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Posted (edited)
44 minutes ago, discountbrains said:

Why can't this be true just like it is for the < relation?

What does "this" refer to?

44 minutes ago, discountbrains said:

I wonder if it was suspected I was fairly well respected mathematician I would be taken more seriously.

A "fairly well respected mathematician" writing the same things you have would not be taken seriously, and would rather quickly become a disrespected mathematician. You should be working on explaining your proof clearly and precisely (which you still have not done); being taken seriously is a result of that.

Edited by uncool

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Disregard my last post. I get complaints that I say something and then go back and say I made an error. The problem here is that in an hour or two someone replies which prevents me from going back and correcting my error. I'd say most msg boards don't get a response in less than a day. What I said in my last post doesn't imply there exists some element below any number in the sets I exhibit. Saying all elements are less or greater than each other doesn't do it. I think I'm getting close, but we know proofs in math depend on the tiniest detail.

1 hour ago, uncool said:

What does "this" refer to?

I am applying the same meaning to a <*x <* b as one would to a< x< b-just a different order relation.  As we might always have an x such that x < z for any z we should have an x <* z for any z. I'm saying this should be allowed in our meta math system. Do you not agree?

 

Wanted to delete the upper part of this. Now I can't.

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18 minutes ago, discountbrains said:

As we might always have an x such that x < z for any z we should have an x <* z for any z. I'm saying this should be allowed in our meta math system. Do you not agree?

If by "might", you mean it happens in some orders, then yes, I agree. But you seem to be claiming this must be true for all orders, and that's something you'll have to prove (in fact, it's false), so no, I do not agree.

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"As we might always have an x such that x < z for any z we should have an x <* z for any z" That is if this z is supposed to be a min for a set.

2 minutes ago, uncool said:

If by "might", you mean it happens in some orders, then yes, I agree. But you seem to be claiming this must be true for all orders, and that's something you'll have to prove (in fact, it's false), so no, I do not agree.

Some confusion here. I clearly know some sets can be well ordered by some order. But, my original claim still is always there are some subsets that can't be well ordered by any order.

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