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

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36 minutes ago, discountbrains said:

OK, I see what I did. That a set can be WO depends on the AC. The two are equivelent statements. What I showed is WO does not exist naturally on its own. Its like saying if we can live on air we do not have to grow food and we know the first part is impossible. What I did was just an exercise. Of course something like this could never overturn 100 years of history. Of course it cannot be that simple. If a set is WO then there you have your set of numbers for the AC.

Yes, its clear the natural numbers can be WO, but my thesis was that not every set can be WO. If you accept the AC then of course you say they can.

> OK, I see what I did. That a set can be WO depends on the AC. The two are equivelent statements.

The natural numbers are well-ordered in the absence of AC. And as I noted, even the real numbers might be well-ordered even in the absence of AC. What AC says is that EVERY set may be well ordered.

> What I showed is WO does not exist naturally on its own. Its like saying if we can live on air we do not have to grow food and we know the first part is impossible.

That doesn't make much sense. Even in the absence of AC, there's a model of set theory in which the reals are well-ordered. And without AC, plenty of sets are "naturally" well-ordered, like the natural numbers, or any finite set, or any of the transfinite ordinals. You certainly did NOT show that "WO does not exist naturally on its own." You didn't show anything like that.

> What I did was just an exercise.

You don't seem to have understood the flaws in your own thinking. You think you proved that "a WO does not naturally exist on its own." First that doesn't make sense, since you haven't defined your terms. And second, under any reasonable interpretation, it's false.

> Of course something like this could never overturn 100 years of history. Of course it cannot be that simple. If a set is WO then there you have your set of numbers for the AC.

No no no no no no no. Even in the absence of AC, all the ordinal numbers are well-ordered. And even in the absence of AC, the real numbers might be well-ordered.

> Yes, its clear the natural numbers can be WO, but my thesis was that not every set can be WO. If you accept the AC then of course you say they can.

"I say" they can? What do you say? You deny that AC implies that every set can be well-ordered? Your phrasings are so imprecise that they are causing your thinking to become confused.

Edited by wtf

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No.no, no. Yes you showed there are other sets of reals that can be WO even without AC just like the natural numbers can. But, I said above that the AC guarantees every set can be WO. Of course, we are in agreement in what you said. It looked like taeto said exactly what I had said above.

Let me take this opportunity to show my little math construction-may be unrelated and of little value:

Consider the interval I= [0,1] and draw the plane XxY. Let each point in the plane (x,y) be constructed from a number  a in I such that x=the 1st digit and every other digit after that of a and y=the 2nd digit and every other digit after that of a. For example if a=0.297051... then x=0.275... and y=0.901...You could plot these numbers on XxY and draw any kind of line or any shape in this area and it would be a subset of I. Also, this is a way to make a bijection from I to this plane. I wonder if anyone has done this before.

Edited by discountbrains

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7 minutes ago, discountbrains said:

No.no, no. Yes you showed there are other sets of reals that can be WO even without AC just like the natural numbers can. But, I said above that the AC guarantees every set can be WO. Of course, we are in agreement there. It looked like taeto said exactly what I had said above.

Let me take this opportunity to show my little math construction-may be unrelated and of little value:

Consider the interval I= [0,1] and draw the plane XxY. Let each point in the plane (x,y) be constructed from a number  a in I such that x=the 1st digit and every other digit after that of a and y=the 2nd digit and every other digit after that of a. For example if a=0.297051... then x-0.275... and y=0.901...You could plot these numbers on XxY and draw any kind of line or any shape in this area and it would be a subset of I. Also, this is a way to make a bijection from I to this plane. I wonder if anyone has done this before.

Yes Cantor did this. See the section titled "Squared Infinities" here ... https://plus.maths.org/content/glimpse-cantors-paradise

Edited by wtf

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OK then. I know of his very clever diagonal proof thing which was a bit different than this.

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34 minutes ago, discountbrains said:

OK then. I know of his very clever diagonal proof thing which was a bit different than this.

Are you talking about the proof that the reals are uncountable? Or that the unit interval can be placed into bijection with the unit square? It's the proof of the latter that you outlined.

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Iḿ not proving anything by this. I just thought it was kind of a fun thing to do. Don know if its good for much. I thought one could define different kinds of addition. Using the usual addition you get interesting results. Or you might say 0.1265+ 0.0018=0.1373... or all kinds of ideas.

Interesting. I didn know this. We proved that a line has the same number of points of this square in topology class a different way. I see it was  Gödel who I should have said was associated with the CH Actually, I quickly realised my construction could prove this Cantor thing.

i

Edited by discountbrains

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wtf made numerous reemarks that my thinking is seriously flawed. No problem there. Then wtf tells me I proved my assertion for dense sets and nothing else. If my proof is valid for these why does it not prove my assertiion? This makes no sense. The essence of the proof is if you can show there is no <* l.e. for some subset in the collection you cant say this works for ALL subsets in the collection which is what WO requires-if Im not mistaken. Then he talks about the set of natural numbers. Well, of course. This is obvious Its like the very definition of a WO set. Then he mentions ordinals. I don know what ordinals are. Actually, I know a little more about them than I let on. Something I dont understand is they seem to all be ordered the same way and must all be well ordered so whatś the distinction? And, whatś the use? Iĺl get pushback from wtf for this. wtf mentioned other models of set theory that allow well ordering without the need of the AC. I truly want to see these. I hope he will post something for me. I guess the natural numbers being well ordered is supposed to contradicts my proof.

Edited by discountbrains

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1 hour ago, discountbrains said:

wtf made numerous reemarks that my thinking is seriously flawed. No problem there. Then wtf tells me I proved my assertion for dense sets and nothing else. If my proof is valid for these why does it not prove my assertiion? This makes no sense. The essence of the proof is if you can show there is no <* l.e. for some subset in the collection you cant say this works for ALL subsets in the collection which is what WO requires-if Im not mistaken. Then he talks about the set of natural numbers. Well, of course. This is obvious Its like the very definition of a WO set. Then he mentions ordinals. I don know what ordinals are. Actually, I know a little more about them than I let on. Something I dont understand is they seem to all be ordered the same way and must all be well ordered so whatś the distinction? And, whatś the use? Iĺl get pushback from wtf for this. wtf mentioned other models of set theory that allow well ordering without the need of the AC. I truly want to see these. I hope he will post something for me. I guess the natural numbers being well ordered is supposed to contradicts my proof.

> I hope he will post something for me.

That's a very kind thing to say. I agree that I deliver criticisms that could be taken personally, but they are not intended like that. When I say your argument is a trainwreck, that's a statement of objective fact based on considerable experience with proofs in this particular area of math. If I'm your TA I'd write "trainwreck" in big red letters.

My intention is to be of assistance to you in understanding these ideas and learning to express your mathematical ideas more effectively, if that's your desire. To do that a certain level of precision is needed, both in concept and exposition. And I'll call you on lack of same.

> Then wtf tells me I proved my assertion for dense sets and nothing else. If my proof is valid for these why does it not prove my assertiion? This makes no sense.

Oh how funny. Yes I can see that you misunderstood this. I have a perfect explanation for you.

I told you to carefully write out your proof that a dense linear order can't be a well-order, for the purpose of you getting practice in expressing your mathematical ideas and writing formal proofs. I reiterate that suggestion. I did not mean to imply that this would help your cause. We agree that it's not sufficient, because you still have all those OTHER linear orders that aren't dense but that might be well-orders.

But you would learn a lot from struggling -- and yes it is a neverending struggle for ALL of us -- to force yourself to write math clearly. That is the only way to get better at it. I suggested this as an exercise. I hope that's not presumptuous in a message board format. I've been through this material and I've had proof-writing beaten into me by distinguished professors at some of the nation's finest universities. It's simply the only way to learn this stuff. If there were a pill we could take, we'd all take it. As Euclid said: There is no royal road to geometry. Even the King has to "do the math" if he wants to learn.

> I guess the natural numbers being well ordered is supposed to contradicts my proof.

You are asking why I talk about N. The reason is that they are the SIMPLEST example of an infinite well-ordered set. So when we're trying to get insight into a well-ordering of the reals, one place to look for intuition is in the much more familiar well-ordering of the naturals.

And this is a math trick! Whenever you have convinced yourself that you proved something about a complicated object, run the proof on a simpler object that the proof still applies to. If you end up proving the naturals aren't well-ordered, then you know your proof's wrong.

Re ordinals: Your interest in well-orders has led you back to the ordinal numbers. They are hard for everyone at first but they are one of the very coolest things I know about in math. They're well worth another run at that part of the book.

By the way a well-order of the reals is an uncountable ordinal. That's a very strange beast to get one's mind around, it takes some work and study. So if you want to think about well-ordering the reals, you need to learn about ordinals and work your way up to the wild idea of an uncountable ordinal.

The existence of an uncountable ordinal can be proved WITHOUT the axiom of choice. That is a very interesting fact I think. I can walk through the proof but you do need to believe in the ordinals first.

> Iĺl get pushback from wtf for this.

Didn't you initially ask for someone who knows this material? Be careful what you wish for

> wtf mentioned other models of set theory that allow well ordering without the need of the AC. I truly want to see these.

Yes this is not a hard reference. It's a mention of a fact that, once I thought about it, must be true. I found it in a very unlikely place, in the Talk page for the Wiki article on the Banach-Tarski paradox. Whenever I look at a Wiki page I often look at its Talk page, where the editors argue about what the article should say. You get a lot of insight from these sometimes.

And way down the page there is this conversation about how much Choice is needed. (Sometimes you don't need the full axiom of choice, you can get by with weaker principles like countable choice, dependent choice, and some other intermediate axioms.) Without going into this in detail -- readers can click if desired -- someone at some point said this:

And it must have been known since the sixties that it's consistent with ZF that the reals can be wellordered but some larger set (say, the powerset of the reals) cannot.

The moment I thought about this I saw it must be true. The well-ordering theorem says that ANY set can be well-ordered. Its negation says that SOME set can't be well-ordered. But it doesn't say anything about which set that is. So it must be logically consistent that the real numbers are well-ordered, yet some larger set isn't. In this particular model of set theory, well-ordering is false and the axiom of choice is false yet the reals are well-ordered.

These are deep waters.

Edited by wtf

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Thank you for taking the time and effort to help me with this. I have alittle book called Ëquivalents of the Axiom of Choice which mentions weaker forms and numerous forms. Clearly many people have thought about this stuff a long time. The natural numbers is a subset of the reals. All I need to do is show there is at least one subset of the reals that doesnt have a <* l.e. to prove my point. Yeah, I am not approaching this with the precision I would have been doing if I were taking a class for a grade. I probably wouldn get by with some of this.....Back in the 1980s I showed what I thought was a WO of [0,1] to a math prof at the Univ of Denver. He walked over to his blackboard, picked up a piece of chalk, thought for a few seconds and wrote out two infinite strings of numbers and asked me which one was greater with my order. I said uh oh and left.

I will look at your suggestions. A possible flaw in my hypothetical ordering is that I might be thinking of it in a usual sense. But, there just seems to be no other way. Any way you look at it you run into infinity somewhere. Again, I say all I need to do is show there is one subset that no l.e. can be found with the assumed WO to prove my assertion.

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1 hour ago, discountbrains said:

Thank you for taking the time and effort to help me with this. I have alittle book called Ëquivalents of the Axiom of Choice which mentions weaker forms and numerous forms. Clearly many people have thought about this stuff a long time. The natural numbers is a subset of the reals. All I need to do is show there is at least one subset of the reals that doesnt have a <* l.e. to prove my point. Yeah, I am not approaching this with the precision I would have been doing if I were taking a class for a grade. I probably wouldn get by with some of this.....Back in the 1980s I showed what I thought was a WO of [0,1] to a math prof at the Univ of Denver. He walked over to his blackboard, picked up a piece of chalk, thought for a few seconds and wrote out two infinite strings of numbers and asked me which one was greater with my order. I said uh oh and left.

I will look at your suggestions. A possible flaw in my hypothetical ordering is that I might be thinking of it in a usual sense. But, there just seems to be no other way. Any way you look at it you run into infinity somewhere. Again, I say all I need to do is show there is one subset that no l.e. can be found with the assumed WO to prove my assertion.

> Back in the 1980s I showed what I thought was a WO of [0,1] to a math prof at the Univ of Denver. He walked over to his blackboard, picked up a piece of chalk, thought for a few seconds and wrote out two infinite strings of numbers and asked me which one was greater with my order. I said uh oh and left.

At least he didn't call your idea a trainwreck! Much nicer individual than me

> A possible flaw in my hypothetical ordering is that I might be thinking of it in a usual sense. But, there just seems to be no other way. Any way you look at it you run into infinity somewhere. Again, I say all I need to do is show there is one subset that no l.e. can be found with the assumed WO to prove my assertion.

I don't understand your assertion. Can you write it down clearly in one sentence? Are you saying no well-order of the reals exists even in in the presence of AC? That's false. Are you saying no well-order of the reals exists in the absence of AC? That might be false too

You seem to understand both these points. So I don't understand what you are trying to prove or what you are claiming.

Edited by wtf

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Sorry for wasting all this time insisting Im correct. You are ABSOLUTELY right. My assertion before had several fatal flaws. To save face I never put the word Proof at the top of my text. Like I said this was submitted for a critique. To begin with I asserted that <* is a supposed WO for ℝ. Intuitively we see a WO ,<', seperates the elements of each set so there already can't be some x,y,a∈S , with x<'a<'y.

Now, let me try this: Suppose <* is any total order relation on ℝ. And <* is not the usual order <. There are nonempty subsets S of ℝ such that for any a,b∈S ∃x∈S, a<*x<*b (right?). So, if S has a min z then z≤*x for all x∈S. Let T=S\{z}, for any u∈T there exists another x∈T such that x<*u. Which implies there is no min in T for <*......Correction: I don't need this set S above. I had been trying to prove this by contradiction assuming <* is a WO. BUT, if <* is a WO I cannot make the "dense" set claim above. I think I just proved it directly-I hope. That is, ∃T⊂ℝ such that for any a,b∈T ∃x∈T , a<*x<*b. This is all of course based on the assumption there is no knowledge of well ordering relations or the AC.
In this iteration am I already assuming no <* can be a WO at the start? I don't think so.
To negate my claim one would also have to prove if <* is a WO you can't construct sets S and T above.

Im going to try to think of a real world analogy to the idea of my proof.

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I say again: Please state in one sentence what you are trying to claim or show.

* You agree what AC implies the well-ordering theorem.

* You seem to understand that even in the absence of AC, the reals might still be well-ordered.

You keep diving into your argument without saying in one clear sentence what it is you are trying to show.

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1 hour ago, discountbrains said:

Now, let me try this: Suppose <* is any total order relation on ℝ. And <* is not the usual order <. There are nonempty subsets S of ℝ such that for any a,b∈S ∃x∈S, a<*x<*b (right?).

You have not found a good way to explain what you want to say. First because there actually never are subsets S like that: to say "for any a,b" includes the case a=b, and then you do not have x. But that slip aside, and otherwise, the property that you ask about does hold for the usual order $$<$$ in place of the given different order $$<^*.$$ But how will you argue if you get asked why it also holds for $$<^*?$$ Like, as wtf points out, if you replace $$\mathbb{R}$$ by $$\mathbb{N}$$ and $$<^*$$ by the usual (well-)order, then it does not quite look right.

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How is this for an anology in my previous post: If I have a pet tiger and it bites me(=the existance of WO). If I feed it (= sets S or T are possible) it wont bite me. If it bites me anyway Iḿ in the hospital and cannot feed it. So now what? This is kind of a puzzler.

I know my previous post about running into infinity is a bit vague.It was something going on in my head: I messed around with trying to wo stuff for years and it seems  that u always end up dealing with something going to infinity. BTW, I thought there was nogreater number than c. Ive forgotten much of this stuff. No, I dont necessarily believe you can have wo without ac. You have told me you could.  looked at your references and find them interesting. Iĺl try harder.

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But how will you argue if you get asked why it also holds for <This might be a good question. I think that the fact this needs to work for all subsets of R means this is true. Do we have to ban such sets? You are right; how exactly do I know what you asked? I think I can pick any subsets of R I want and pick any order. Will this fly?

Edited by discountbrains
mistake

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2 hours ago, discountbrains said:

But how will you argue if you get asked why it also holds for <This might be a good question. I think that the fact this needs to work for all subsets of R means this is true. Do we have to ban such sets? You are right; how exactly do I know what you asked? I think I can pick any subsets of R I want and pick any order. Will this fly?

Run your argument on the naturals ordered by the usual <. Is there such a subset S?

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2 hours ago, discountbrains said:

But how will you argue if you get asked why it also holds for <This might be a good question. I think that the fact this needs to work for all subsets of R means this is true. Do we have to ban such sets? You are right; how exactly do I know what you asked? I think I can pick any subsets of R I want and pick any order. Will this fly?

Your basic problem might lie in the use of formal logic. It is not about you personally, but I see this quite a lot. It is possible to use formal logic language to, in a kind of silly way, to express ordinary occurrences. Like you would say in ordinary language "every person has a mother", and for fun express the same in a formal logic language "for every person x there exists a person y, such that y is the mother of x". This kind of translation is routine for anyone mathematically trained, and in fact it is necessary to be able to parse such statements in order to function professionally. But you would be amazed at how many I have met who freely take the latter statement phrased in formal logic language to mean the same as "there exists a person y such that for every person x, it is true that y is the mother of x." I leave it to you to translate this back into a statement of ordinary language and notice that it does not mean the exact same.

My point is that we are entertaining the idea of one particular ordering $$<^*$$ for which something is true for every subset $$S \subseteq \mathbb{R}$$. It is a statement of the form "there exists ... so that for all ... something is true".

Edited by taeto

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I will now show that there are subsets ofsuch that there are no order relations which have a min for iit. Let T=(t,s). Now for the usual ´<'and a string of elements a,b,c,d,... all in T, we might have t<a<b<c<d<...<s. We might reorder these and still get t<*b<*d<*a<*c<*....<*s.  Notice there is no way to produce a number z in T such that z≤*x for all x in T. Each letter has a position in the order or a slot and another letter is put in that position by the reordering, <*, so any new element in that slot has another element <* less than it just as with the original ordering <. Often I drift into little daydreaming sessions and think about how I might do something. Ive played with trying to reorder sets before. Now I might have found a use for this.

On 8/10/2018 at 9:51 AM, discountbrains said:

Edited by discountbrains
spelling

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No numeric values need considering for the elements of T. Just the relative positions of the elements to each other.

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1 hour ago, discountbrains said:

I will now show that there are subsets ofsuch that there are no order relations which have a min for iit. Let T=(t,s). Now for the usual ´<'and a string of elements a,b,c,d,... all in T, we might have t<a<b<c<d<...<s. We might reorder these and still get t<*b<*d<*a<*c<*....<*s.  Notice there is no way to produce a number z in T such that z≤*x for all x in T. Each letter has a position in the order or a slot and another letter is put in that position by the reordering, <*, so any new element in that slot has another element <* less than it just as with the original ordering <. Often I drift into little daydreaming sessions and think about how I might do something. Ive played with trying to reorder sets before. Now I might have found a use for this.

Just because t < a, why should we still have t <* a?

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Run your argument on the naturals ordered by the usual <. Is there such a subset S?  Did I not say <* is not the same ording as <? Again, I say there are subsets of R with this property. Thats all I need to show; clearly the natural numbers are not going to work. Are u saying if one can find a <* such that even the natural numbers satisfy my claim that negates it?

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10 minutes ago, discountbrains said:

Run your argument on the naturals ordered by the usual <. Is there such a subset S?  Did I not say <* is not the same ording as <? Again, I say there are subsets of R with this property. Thats all I need to show; clearly the natural numbers are not going to work. Are u saying if one can find a <* such that even the natural numbers satisfy my claim that negates it?

Just because t < a, why should we still have t <* a?

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t is not in T therefore the reordering is not being applied to t. But, thats a good question. I believe I came up with 3 different ways to prove my claim. I will have to think about this. Now, just where is t in relation to the others? Yes, you are right; I will have to define <* so it works for the whole set R.

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Here's another example you should be thinking of. We know that there's a bijection between the naturals N and the rationals Q. The usual order on Q is dense and the usual order on N is a well-order. You should try to run your ideas through this example since it's a countable and well-understood parallel to the case of a well-order on the reals versus the usual dense order.

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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.

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