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

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1 hour ago, discountbrains said:

I need to generalize my <* ordering. It needs to be an ordering on the whole set R because it has to be the possibility of any ordering. And, it certainly could be that a<*r<*c<* b<*....In fact, any number of elements not in T could be mixed up in there, but that doesnt mean there is a z in T with z≤*x for all x in T. I might write a typical ordering of mine as

....<* a<*r<*c<* b<*p<*... where r and p are not in R.

wtf, do you mean that Cantor rational number thing? 1    2      3      4 ...

1/2 2/2 3/2 ....

1/3 2/3 3/3....

.       .         .

This, of course, is where you start in the upper left corner and count back and forth diagonally showing thereś a one to one onto relationship to the natural numbers and doesnt apply to what im saying. Clearly this is a WO set. I dont know why u asked this. I was going to show what I think is a way to reorder any set of real numbers in a dense or çontinuous way, but will do that later. Im going to take a break from this now.This kinda gets to me after a while. If anyone has any comments or questions I will still read them.

> I need to generalize my <* ordering. It needs to be an ordering on the whole set R because it has to be the possibility of any ordering.

What does that mean, it has to be the "possibility of any ordering?" It's not even clear what you are trying to do. In order to make progress my suggestion to you would be to force yourself to write more clearly, one line at a time, and make sure each line makes sense. You don't have to write here, do it for yourself. Your ideas are jumbled because your prose is jumbled. That's why they make you write proofs in math class. Clear exposition leads to clear thinking.

> wtf, do you mean that Cantor rational number thing?

Yes, the fact that the natural numbers can be placed into bijection with the rationals.

>  I dont know why u asked this.

Because you have to walk before you can run. You are trying to investigate or understand the relation between the usual dense order on the reals, with the well-order on the reals. Now the reals are a very complicated set. There is a much simpler set we have lying around that has the same phenomenon. You should study it to try to gain insight into the analogous problem in the reals

Specifically, the usual order on the rationals is dense, just like the usual order on the reals is dense. Now you want to find a well-order. We know we can re-order the rationals so that they are well-ordered. This is a perfect analogy for what you are trying to do. That's why I suggested that you study it. It's so you can gain insight into the analogous problem for the reals: How to relate a well-order to a dense order on the same set.

It's a standard technique in math, when you are working on a hard problem, to look at simpler examples.

> I was going to show what I think is a way to reorder any set of real numbers in a dense or çontinuous way

But why? Isn't the usual order already dense or continuous? You don't have to work very hard to find a dense order on the reals, the usual order is dense.

> Im going to take a break from this now.This kinda gets to me after a while.

I hope you got your money's worth from the convo. You should definitely take another run at learning about the ordinals. The ordinals are all about well-ordered sets. If you are interested in well-orders you are by definition interested in the ordinal numbers, even if you don't know it.

I'll check this site from time to time in case you have more questions or comments. All the best, nice chatting with you.

By the way I have a really cool book recommendation for you. Infinity and the Mind by Rudy Rucker. https://www.amazon.com/Infinity-Mind-Philosophy-Infinite-Princeton/dp/0691121273

Edited by wtf

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I'm now revisiting this thread. Like I said several times before to prove my assertion all you have to do is exhibitjust one set that can't be WO by <* assuming you actually found a WO , <*, for a set that contains it.

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Can't find an 'edit' button. Ran out of time yesterday. wtf kept correcting me saying look at an example like the natural numbers and observe these clearly can be WO. I was very surprised at his answer since it sounds like he is a professor or something. This is clearly incorrect: the thing is like what I said above which is 'not finding one example of a set WO'ed by some.<*'. The example he gives is not dense anyway.

Very interesting I get 29 negative points while he and everyone else gets many positive points. I get tons of negative points on every board on the net I post. People just don't like anyone who doesn't go with the crowd-its really, really ridiculous. Its people like me who are the REAL pioneers in everything. But, if I'm wrong please write a valid proof.

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11 minutes ago, discountbrains said:

Can't find an 'edit' button. Ran out of time yesterday. wtf kept correcting me saying look at an example like the natural numbers and observe these clearly can be WO. I was very surprised at his answer since it sounds like he is a professor or something. This is clearly incorrect: the thing is like what I said above which is 'not finding one example of a set WO'ed by some.<*'. The example he gives is not dense anyway.

Very interesting I get 29 negative points while he and everyone else gets many positive points. I get tons of negative points on every board on the net I post. People just don't like anyone who doesn't go with the crowd-its really, really ridiculous. Its people like me who are the REAL pioneers in everything. But, if I'm wrong please write a valid proof.

Feeling like Galileo much?

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4 hours ago, discountbrains said:

Can't find an 'edit' button. Ran out of time yesterday. wtf kept correcting me saying look at an example like the natural numbers and observe these clearly can be WO. I was very surprised at his answer since it sounds like he is a professor or something. This is clearly incorrect: the thing is like what I said above which is 'not finding one example of a set WO'ed by some.<*'. The example he gives is not dense anyway.

Astonishing.

* I told you at least three times that I invited you to look at the example of the naturals since they would help you gain insight into the problem you're attacking. That you continually fail to comprehend this point is mystifying to me. I say again: The way you attack a hard problem is to start by analyzing simpler instances of similar problems. If you don't want to get this, so be it.

* You say I am "incorrect" that the real numbers can be well ordered; but that they can be bijected to the rationals, which are NOT well-ordered. You say that is incorrect. Please provide a proof or explanation.

* Being well-ordered and being dense are mutually exclusive. I thought I pointed this out several times. I thought you UNDERSTOOD this several times. As an exercise -- again, to help you build intuition -- I suggested that you write down a formal proof. I reiterate that suggestion.

Since you reactivated this thread you haven't provided a proof or even any additional argument.

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16 minutes ago, wtf said:

The way you attack a hard problem is to start by analyzing simpler instances of similar problems. If you don't want to get this, so be it.

Where's fun in doing the easy bits first?

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I quite well understand this. Of course they are mutually exclusive. Also, its hard to imagine how any set that is dense with respect to < or about any ordering could be ordered in such a way  that it can be enumerated by 1,2,3,... There are just too many numbers that still pop up in between. I was going to add in my post above that intuitively it seems you can pick one number out of any set in a collection of sets. But, actually doing this is something else. For one thing, which number are you going to choose? Maybe I should ask another question: Can one show that any set dense with respect to one ordering is therefore dense for all possible orderings?

And, yes studiot, I also think I've found the best way to understand a new concept is to see a few examples of it. Its maddening sometimes to try to figure out what a theorem is saying by just looking at it alone.

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21 hours ago, wtf said:

* You say I am "incorrect" that the real numbers can be well ordered; but that they can be bijected to the rationals, which are NOT well-ordered. You say that is incorrect. Please provide a proof or explanation.

32 minutes ago, discountbrains said:

I quite well understand this. Of course they are mutually exclusive. Also, its hard to imagine how any set that is dense with respect to < or about any ordering could be ordered in such a way  that it can be enumerated by 1,2,3,... There are just too many numbers that still pop up in between. I was going to add in my post above that intuitively it seems you can pick one number out of any set in a collection of sets. But, actually doing this is something else. For one thing, which number are you going to choose? Maybe I should ask another question: Can one show that any set dense with respect to one ordering is therefore dense for all possible orderings?

And, yes studiot, I also think I've found the best way to understand a new concept is to see a few examples of it. Its maddening sometimes to try to figure out what a theorem is saying by just looking at it alone.

Be aware that wtf was refering to the rationals (symbol Q) not the integers or natural numbers, (symbols Z and N respectively).

Although the rationals are countable they are also dense, which the integers and naturals are not.

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9 minutes ago, studiot said:

Be aware that wtf was refering to the rationals (symbol Q) not the integers or natural numbers, (symbols Z and N respectively).

Although the rationals are countable they are also dense, which the integers and naturals are not.

In this discussion it's important to note that the rationals are dense in their usual order; and that the naturals are well-ordered in their usual order.

Consider that we may use any bijection between the naturals and the rationals to induce the order of one onto the other. In that way we may impose a dense order on the naturals; or a well-order on the rationals.

Another way to look at this is that the naturals and the rationals are actuall the same exact set. Their order properties depend only on the order we impose on them. If we impose the rational order, our set looks like the rationals. If we impose the naturals, our set looks like the naturals. But it's the same underlying set.

I'm not claiming this as a "fact," as if there could even be any facts about abstract entities such as sets. Rather, I'm presenting this as a point of view that gives insight into the nature of ordered sets.

It's also worth noting (again) that the properties of density and well-order are mutually exclusive. An ordered set may be one or the other, or neither. But it may not be both.

Edited by wtf

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Nice to know that those with greater depth of knowledge of these matters than mine have my back.

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1 hour ago, discountbrains said:

I quite well understand this. Of course they are mutually exclusive. Also, its hard to imagine how any set that is dense with respect to < or about any ordering could be ordered in such a way  that it can be enumerated by 1,2,3,... There are just too many numbers that still pop up in between. I was going to add in my post above that intuitively it seems you can pick one number out of any set in a collection of sets. But, actually doing this is something else. For one thing, which number are you going to choose? Maybe I should ask another question: Can one show that any set dense with respect to one ordering is therefore dense for all possible orderings?

And, yes studiot, I also think I've found the best way to understand a new concept is to see a few examples of it. Its maddening sometimes to try to figure out what a theorem is saying by just looking at it alone.

> I quite well understand this. Of course they are mutually exclusive. Also, its hard to imagine how any set that is dense with respect to < or about any ordering could be ordered in such a way  that it can be enumerated by 1,2,3,... There are just too many numbers that still pop up in between.

Ah! But I just explained that. We know there's a bijection between the naturals and the rationals. So to put a well-order on the rationals, we just enumerate them q1, s2, q3, ... where qn is f(n) where f is the bijection.

Going the other way, we may put a dense order on the natural numbers by simply using the inverse of that bijection, call ig g. So g(q) is some natural number. So order the naturals n, m according to whether g-inverse(n) and g-inverse(m) are < or > to each other in the rationals.

That's why this example is so important. It shows how we may reorder a given set to have strikingly different properties.

>Can one show that any set dense with respect to one ordering is therefore dense for all possible orderings?

On the contrary we can prove that false. Any set may be well-ordered. That's the well-ordering theorem. Given any dense set you just well-order it. That's the point of the entire thread, right?

And of course a MUCH SIMPLER case is that the rationals, which are densely ordered, may nevertheless be well-ordered using any bijection with the naturals. And we do not need to rely on the well-ordering theorem for this fact. That's important. The well-ordering theorem requires the axiom of choice. But establishing a bijection between the naturals and the rationals does not. So your claim is falsified by THE MOST OBVIOUS EXAMPLE.

That's another reason this example is so helpful. It's like the well-ordering theorem but without having to invoke the awesome power of the well-ordering theorem! That's why it's of such great interest.

Edited by wtf

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Ok, the question just popped in my head. I didn't give it any thought. There are, I think, many ways to construct a bijection between N and Q.

"Nice to know that those with greater depth of knowledge of these matters than mine have my back." Yes, I know its universally true its always safest to be on the side of conventional wisdom. I ought to try it sometime, but if everyone did this there would be no advances in anything.

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1 hour ago, discountbrains said:

Ok, the question just popped in my head. I didn't give it any thought. There are, I think, many ways to construct a bijection between N and Q.

"Nice to know that those with greater depth of knowledge of these matters than mine have my back." Yes, I know its universally true its always safest to be on the side of conventional wisdom. I ought to try it sometime, but if everyone did this there would be no advances in anything.

I actually have your back on that last point. Einstein was told that hundreds of physicists disagreed with his theory. He said: "If my theory is wrong, it would only take one."

I agree with you that it doesn't matter who agrees with whom. If the whole of established mathematics believes one thing, and you believe another, and you have a proof, then you are right and they are all wrong.

So ... have you got a proof?

Edited by wtf

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OK, here we go. I'm just answering wtf at this point. Let A = (0,1) with the ususal ordering. Yes, it seems obvious like he says if a set A is WO by some other ordering  then it can be able to be written as A + {a1, a2, a3,...}. This means it can be arranged in one of Cantor's array like things. Now we can do like he did and contruct a diagonal can be chosen to this arrangement and ask if this element is in A. At first I thought 'of course'. Now I see this needs proof. Such a diagonal would show a contradiction.

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11 minutes ago, discountbrains said:

it seems obvious like he says if a set A is WO by some other ordering  then it can be able to be written as A + {a1, a2, a3,...}.

For somebody who is not sure about the precise definitions it may seem plausible that well-ordering implies ability to count all the elements by going up the order from first element to next, and so on. That however is not true about well-orderings. You can well-order the natural numbers as 1,3,5,7,...,2,4,6,8,..., i.e. the odd numbers are in their usual order, and the even numbers as well, but every even number is larger than every odd number in this ordering. You can check that it is a well-order. And if you start from the left and go up infinitely long, you only get to count the odd numbers. Any well-ordering of the reals has similar properties, though much more complicated, well, actually impossible, to describe.

Edited by taeto

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1 hour ago, discountbrains said:

OK, here we go. I'm just answering wtf at this point. Let A = (0,1) with the ususal ordering. Yes, it seems obvious like he says if a set A is WO by some other ordering  then it can be able to be written as A + {a1, a2, a3,...}. This means it can be arranged in one of Cantor's array like things.

No. When you well-order the reals, the reals are still uncountable.

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Oh my goodness. No, no, no. I'm not trying to prove anything about countability. I mean use the same method Cantor used and construct a numer that's not a member of ypur list. If the number is not in the list how can we find its place in the list? How can we compare it to any of the numbers?

Typo above 'A + {a1, a2, a3,...}' should be A = {a1, a2, a3,...}. I have vision problems. And, yes, 1,3,5,... is also clearly a well ordering and so are many other sequences due to the definition of WO..

Edited by discountbrains

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In the version of Cantor's proof in my book if a digit ann=1 then the corresponding digit of the constructed number is 2 and if its not equal to 1 then the the digit=1. So, a typical constructed number might look like 0.1211121... Suppose there is a number a=0.3472000000.... then clearly 0.1211121...< aand its in (0,1). But who knows how to compare the two numbers with the new ordering?

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35 minutes ago, discountbrains said:

In the version of Cantor's proof in my book if a digit ann=1 then the corresponding digit of the constructed number is 2 and if its not equal to 1 then the the digit=1. So, a typical constructed number might look like 0.1211121... Suppose there is a number a=0.3472000000.... then clearly 0.1211121...< aand its in (0,1). But who knows how to compare the two numbers with the new ordering?

You do not have to show that one number is greater than the other in the ordering. You can compare them by just saying that they are different, by the construction.

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22 hours ago, discountbrains said:

Oh my goodness. No, no, no. I'm not trying to prove anything about countability. I mean use the same method Cantor used and construct a numer that's not a member of ypur list. If the number is not in the list how can we find its place in the list? How can we compare it to any of the numbers?

Typo above 'A + {a1, a2, a3,...}' should be A = {a1, a2, a3,...}. I have vision problems. And, yes, 1,3,5,... is also clearly a well ordering and so are many other sequences due to the definition of WO..

You are missing the point. Just because a set is well-ordered does NOT mean it looks like a1, a2, a3, ... First, the a1, a2, a3, ... order is the usual order type of the natural numbers. There are a lot of other well-ordered set. Secondly, enumerating a1, a2, ... implies that your ordered set is countable. But a well-order of the reals is an uncountable well order. It starts as a1, a2, a3 ... but after the dots there's a lot more stuff.

You said a while back that you don't want to learn about ordinals, but that's exactly the study that would clarify many of your misconceptions. A well order of an uncountable set can't be notated a1, a2 ... It BEGINS that way, but after the dots there are many more reals.

If you diagonalize any countable list of reals, you'll find a real not on the list.

Edited by wtf

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40 minutes ago, wtf said:

You are missing the point. Just because a set is well-ordered does NOT mean it looks like a1, a2, a3, ... First, the a1, a2, a3, ... order is the usual order type of the natural numbers. There are a lot of other well-ordered set. Secondly, enumerating a1, a2, ... implies that your ordered set is countable. But a well-order of the reals is an uncountable well order. It starts as a1, a2, a3 ... but after the dots there's a lot more stuff.

You said a while back that you don't want to learn about ordinals, but that's exactly the study that would clarify many of your misconceptions. A well order of an uncountable set can't be notated a1, a2 ... It BEGINS that way, but after the dots there are many more reals.

If you diagonalize any countable list of reals, you'll find a real not on the list.

So many good points packed into one post. +1

I particularly liked the pointer at the ordinals. Where do you (DB) think they got their name from?

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I know this post is directed at me. Let me be clear, the a1, a2, a3 ... sequence was asserted by wtf in one of his posts! Don't accuse me of saying it. I wanted above to state this seems like a fantasy, but forgot to say it.

You can compare them by just saying that they are different, by the construction".

I was just saying the obvious about the usual order.

Edited by discountbrains

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43 minutes ago, discountbrains said:

I know this post is directed at me. Let me be clear, the a1, a2, a3 ... sequence was asserted by wtf in one of his posts! Don't accuse me of saying it. I wanted above to state this seems like a fantasy, but forgot to say it.

In that case I surely misunderstood your notation. Can you please clarify what you meant to say? It certainly seemed to me that you were saying that if you well-order the reals, you get an ordered set that looks like $\{a_1, a_2, a_3, \dots \}$.

That can't be true, because that is the definition of a particular countable well-order; and we know that the reals may only be given an uncountable well-order. That's because the reals are uncountable. A set may be reordered in many ways, but all orders have the same cardinality.

Now if you did not mean to imply that meaning, just tell me what you meant.

Edited by wtf

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I'm sorry, I misunderstood what wtf was saying. I was accusing him as saying the same thing I was being accused of. No, we all know the reals or (0,1) can't be simply ordered like a1, a2, a3, ... Like someone said, its way more complicated than that. We know there are an infinite number of disjoint and intersecting subsets of these sets. I kinda want to think of them as multidimensional which I don't have a definition of.

I would like someone to focus more directly on my original claim at the start and show its wrong.

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On 8/5/2018 at 8:43 PM, wtf said:

let <* and ≤* be total order relations on ℝ. Let S be any complete (in relation to <*) subset of ℝ , let z ε S such that z ≤* x ∀x ε S. Then ∃T = S - {z}. And, T is complete also. So, ∀x ε T, z <* x. Hence any such z is not a least element of T. This implies there is no way to Well Order ℝ

This shows for any possible well ordering of ℝ there still exists subsets of ℝ such that they have no least element in relation to the ordering.

I assume this is the post?

The problem is that you seem to have misunderstood the well ordering principle by accidentally rearranging quantifiers. It doesn't say that there is a minimum that works for all sets; it says each set has its own minimum; that is, while multiple sets can have the same minimum, they are not all required to. z may not be a minimum of T, but another thing can be.

Your proof uses nothing about the real numbers. If it worked, it would apply equally well to the natural numbers, which are well-ordered. Try it out, and see if you can understand why it doesn't work there. Alternatively, try writing your proof formally - you should be able to see where you have wrongly exchanged quantifiers.

Edited by uncool

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