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

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"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."

OK, you made this more clear. I take this to mean the set must be nonempty for the particular order which is proposed to be the WO, Oh! I forgot what I wanted to say just above. I wanted to tell wtf that if T is empty WE ARE THROUGH. there is no WO for an empty set.

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52 minutes ago, discountbrains 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."

OK, you made this more clear. I take this to mean the set must be nonempty for the particular order which is proposed to be the WO, Oh! I forgot what I wanted to say just above. I wanted to tell wtf that if T is empty WE ARE THROUGH. there is no WO for an empty set.

That is incorrect. The empty set has a unique order, and that order is (trivially) a well-ordering.

I am saying that the definition of a well-ordering is that a set S and order <* is a well-ordering if every nonempty subset T of S has a minimum. It says nothing if T is empty.

Going back to your proof: your proof, between the lines "Now if we assume <* well orders T" and "Hence z is an element of this string of minimums above" can be condensed to the claim that any infinite well-ordered set must be order-isomorphic to the positive integers.

And that's easily shown to be false. Your proof cannot work on those lines.

Edited by uncool

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8 hours ago, discountbrains said:

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've already explained that I"m NOT claiming I have a well-order.

As you ALREADY AGREED, my showing that T might be empty is a refutation of your proof that there is no well-order.

I will get someone's attention somewhere. My 'proof(?)' is what it is.

You haven't got a proof. And you even agreed that I found a flaw in your claimed proof.

Edited by wtf

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38 minutes ago, wtf said:

I've already explained that I"m NOT claiming I have a well-order.

As you ALREADY AGREED, my showing that T might be empty is a refutation of your proof that there is no well-order.

I will get someone's attention somewhere. My 'proof(?)' is what it is.

You haven't got a proof. And you even agreed that I found a flaw in your claimed proof.

I do think that the error you've identified is a relatively small error that can be worked around. The point he's aiming for is that the set T must be denumerable; if he instead uses countable, then should the sequence ever stop because a subset is empty, then T must be finite. The larger error is in his claim that the "mins" must eventually exhaust T.

Edited by uncool

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9 hours ago, uncool said:

That is incorrect. The empty set has a unique order, and that order is (trivially) a well-ordering.

I am saying that the definition of a well-ordering is that a set S and order <* is a well-ordering if every nonempty subset T of S has a minimum. It says nothing if T is empty.

Going back to your proof: your proof, between the lines "Now if we assume <* well orders T" and "Hence z is an element of this string of minimums above" can be condensed to the claim that any infinite well-ordered set must be order-isomorphic to the positive integers.

And that's easily shown to be false. Your proof cannot work on those lines.

I take the condition the set is nonempty in the established definition of WO to mean we just don't consider empty sets. For the last few lines this might be a good point which I'm not aware of and need explanation of. Is there a 'work around' of this?

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2 hours ago, discountbrains said:
12 hours ago, uncool said:

That is incorrect. The empty set has a unique order, and that order is (trivially) a well-ordering.

I am saying that the definition of a well-ordering is that a set S and order <* is a well-ordering if every nonempty subset T of S has a minimum. It says nothing if T is empty.

Going back to your proof: your proof, between the lines "Now if we assume <* well orders T" and "Hence z is an element of this string of minimums above" can be condensed to the claim that any infinite well-ordered set must be order-isomorphic to the positive integers.

And that's easily shown to be false. Your proof cannot work on those lines.

I take the condition the set is nonempty in the established definition of WO to mean we just don't consider empty sets.﻿

We "don't consider" empty subsets when determining whether a set is well-ordered. That's not the same as saying that every subset of a well-ordered set is nonempty, which your reasoning seems to imply. Nor does it mean that a well-ordered set can't itself be empty.

2 hours ago, discountbrains said:

For the last few lines this might be a good point which I'm not aware of and need explanation of. Is there a 'work around' of this?

I sincerely doubt that there is a workaround. I'm surprised you aren't aware of the objection, as it's essentially the same objection I've been making for a while now.

Edited by uncool

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10 hours ago, uncool said:

We "don't consider" empty subsets when determining whether a set is well-ordered. That's not the same as saying that every subset of a well-ordered set is nonempty, which your reasoning seems to imply. Nor does it mean that a well-ordered set can't itself be empty.

I sincerely doubt that there is a workaround. I'm surprised you aren't aware of the objection, as it's essentially the same objection I've been making for a while now.

22 hours ago, uncool said:

I take the condition the set is nonempty in the established definition of WO to mean we just don't consider empty sets. For the last few lines this might be a good point which I'm not aware of and need explanation of. Is there a 'work around' of this?

I think what you mean about not isomorphically ordered is I didn't show a map from T to such that f(xi)=i is 1 to 1 and onto for all xi in T . What I say I'm showing is if zi is the min for A⊂T we must have a min, z(i+1), in A\{zi} which is the very next number in the set of minimums of T since there are NO numbers in T between these two. This is part of what I said in my hand written proof above implies there is a 1 to 1, onto correspondence between this set of z's and N and since for any z you choose in T it is the min for every subset of T, {x: z ≤* x, z and x in T}, so its also in this set of minimums. Therefore there is a bijection between T and N which leads to a contradiction. I think you mean my mapping must produce the same order. This is entirely unnecessary.

ALSO, after my 'written' proof I defined T so S\{a,b}⊂T so it would be nonempty and I'd have something to work with; this my not really be required. The essence of the proof is the deleting of elements, z's, one by one.

I just can't wrap my head around a null set being (trivially) well ordered, If its empty how can it have a min element?

This can't be established convention in the math world.

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26 minutes ago, discountbrains said:

I think what you mean about not isomorphically ordered is I didn't show a map from T to such that f(xi)=i is 1 to 1 and onto for all xi in T . What I say I'm showing is if zi is the min for A⊂T we must have a min, z(i+1), in A\{zi} which is the very next number in the set of minimums of T since there are NO numbers in T between these two. This is part of what I said in my hand written proof above implies there is a 1 to 1, onto correspondence between this set of z's and N and since for any z you choose in T it is the min for every subset of T, {x: z ≤* x, z and x in T}, so its also in this set of minimums. Therefore there is a bijection between T and N which leads to a contradiction. I think you mean my mapping must produce the same order. This is entirely unnecessary.

It producing the same order is an immediate implication of the way you tried to prove it. If z_i <* z_j, then i <* j, by the definition of the z_i. And this conclusion can't be true, as shown by the existence of countable well-ordered sets that aren't isomorphic to N. The problem is in this line: "since for any z you choose in T it is the min for every subset of T, {x: z ≤* x, z and x in T}, so its also in this set of minimums". There is no reason to think that the list z0, z1, z2, ... will include the minimum of every subset of T.

Side note: this reformatting of text to get larger and larger is distracting.

26 minutes ago, discountbrains said:

I just can't wrap my head around a null set being (trivially) well ordered, If its empty how can it have a min element?

It doesn't. A set being well-ordered isn't based on that set having a minimal element. It's based on every nonempty subset of that set having a minimal set. The empty set has no nonempty subsets, so trivially, every nonempty subset of the empty set has a minimum.

Edited by uncool

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Oh no, no, no, No. You are Wrong, Where do u get that any WO set must contain  a minimal set? That's not the definition of WO. Also, I think u added a word to my statement about a set of numbers >* than z. You put in 'every' I believe and here's where u get the argument about null sets. U'r as bad as wtf who tried to redefine my T set. Many months ago it has crossed my mind that these forums (especially stackexchange) might be populated with Russian trolls or operatives. Something suspicious about this picture and I'm the one being duped.

Yes, I know the font thing is why I wrote my thing by hand. I'm getting better with typing symbols; maybe next time it won't erase my work when getting near the end.

I haven't dealt with this stuff for over 40 years so I'm a little rusty. Today I looked at examples of the math induction theorem use. I will write up a better version of this. But, as long as I show there is a 1 to 1, onto mapping and inverse mapping between T and N they have same cardinality  regardless of order.

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26 minutes ago, discountbrains said:

Oh no, no, no, No. You are Wrong, Where do u get that any WO set must contain  a minimal set? That's not the definition of WO.

Sorry, my typo. I meant to say "A set being well-ordered isn't based on that set having a minimal element. It's based on every nonempty subset of that set having a minimal element."

26 minutes ago, discountbrains said:

You put in 'every' I believe and here's where u get the argument about null sets.

I did not change what you said. Check your post. However, I took your meaning correctly and my post remains accurate even if the word "every" is removed.

26 minutes ago, discountbrains said:

But, as long as I show there is a 1 to 1, onto mapping and inverse mapping between T and N they have same cardinality  regardless of order.  ﻿

While that is true, the method you have used proves that they would have to have the same order. Your proof says there are z0 < z1 < z2 < ..., and that every element of T must be listed as some z.

Edited by uncool

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1 hour ago, discountbrains said:

maybe nex﻿t time it won't erase my work when getting near the end.

What I do is compose in a separate text file, then paste into a forum window when I'm done. That way I reduce the chance of some hungry message board eating my post.

I for one would like to see a new, clean, latest version so that we're all sure we're referencing the same proof.

Perhaps as you write you might consider writing down numbered sentences or paragraphs instead of running everything together. Just for readability. If nothing else, just put some horizontal space every sentence or two.

Edited by wtf

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My new copy of my proof for your perusal:

Proof is by contradiction. Let S=(0,1)R with 0<a<b<1. For any total order relation, <*, let T={x: a<*x<*b} and let S\{a,b}⊂T. Now, if we believe <* well orders T ∃z0∈T such that {x:a<*x<*z0}= ∅. Choose any zn∈T where z0<*zn then by our hypothesis there must be a z(n+1)∈T\{zn} which is the minimum of T\{zn}. So, ∀x∈T\{zn}, zn<*x and

{x: zn<*x<* z(n+1)}= ∅ also. Hence, by math induction we have generated a string of minimal elements z0, z1, z2,..., zn,... ∀nN of T. Moreover, Z={z0, z1, z2,...,zn,...} =T. To see this choose an arbitrary z in T. z is the min of the subset, {x: z ≤*x} of T. Is z an element of Z? If z≠zn for a zn∈Z zn<*z or z<*zn. If <*zn there is a finite number of steps to reach a zi<*z. If zn<*z, z(n+1)<*z or zn<*z<*z(n+1). If z(n+1)<*z is it >*z(n+2)? Continuing, there is only a finite number of steps to get to zk<*z<*z(k+1). Any of these zi<*z<*z(i+1) are impossible because of the way these minimums were derived above. Therefore, any z in T is in Z and we can create a mapping f(T) to N which is a bijection. So, T contains no more numbers than  N and we know this is not true.

Can't turn bold off for some of this.

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1 hour ago, discountbrains said:

Proof is by contradiction. Let S=(0,1)R with 0<a<b<1. For any total order relation, <*, let T={x: a<*x<*b} and let S\{a,b}⊂T. Now, if we believe <* well orders T ∃z0∈T ﻿such that {x:a<*x<*z0}= ∅. Choose any zn∈T where z0<*zn then by our hypothesis there must be a z(n+1)∈T\{zn} which is the minimum of T\{zn}. So, ∀x∈T\{zn}, zn<*x and

Let S=(0,1) R with 0<a<b<1. For any total order relation, <*, let T={x: a<*x<*b} and let S\{a,b}⊂T. Now, if we believe <* well orders T ∃z0∈T

This is the same damn error again. Why won't you take the trouble to understand it?

Let <* be the total order: "a <* b <* everything else in its usual order". Then T is empty. Your proof fails on the first line. There is no such z0. You keep making this same error and I keep giving the same counterexample over and over. This is the Groundhog Day thread. Not a shred of progress has been made since the very first post from August 2, 2018. This has been going on over a year now.

Edited by wtf

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1 hour ago, discountbrains said:

Continuing, there is only a finite number of steps to get to zk<*z<*z(k+1)

Why?

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On 8/14/2019 at 5:28 PM, uncool said:

I do think that the error you've identified is a relatively small error that can be worked around. The point he's aiming for is that the set T must be denumerable; if he instead uses countable, then should the sequence ever stop because a subset is empty, then T must be finite. The larger error is in his claim that the "mins" must eventually exhaust T.

> I do think that the error you've identified is a relatively small error that can be worked around.

It's interesting that you think that. I think the error is absolutely decisive and not repairable. Perhaps we are understanding the argument of @discountbrains differently.

I take things to be thus: We play a game. I give discount (for short) a proposed linear order (= total order, same thing) $<^*$; and discount proves that it can not possibly be a well-order. In this way discount would show that there is no well-order.

Whatever  $a$ and  $b$ he picks, I choose the linear order:

$a <^* b <^* \text{everything else in its usual order}$

In this order it's clear that  $T = \{x \in \mathbb R : a <^* x <^* b\}$ is empty. Therefore there is no  $z_0$ to start the "induction" (which of course isn't really an induction and doesn't prove what discount says it does).

Now I think this is decisive. The compiler in my mind has detected a fatal error and I generally don't bother to read the rest except at a very cursory level. It doesn't matter that discount might have picked  $c$ and  $d$, I'd give the corresponding example. Remember discount's claim is that NO linear order can be a well order. I only have to produce a single example that breaks the proof.

I think you might be thinking something like this. Correct me if I'm misunderstanding your point. You think you can fix this by saying that ok,  $T$ is empty, so we don't care about it. We'll take some other interval. But I can give you the order

$a <^* b <^* c <^* d <^* \text{everything}$

So ok you say never mind THAT interval. But I can do this all day. I could put countably many natural numbers, say all of them, in front of the "everything else" part.

Or I could put all the even natural numbers, then the odds, and then everything else. So this would still be a countable initial segment, but it would be $\omega + \omega$, two copies of the naturals. Or three. I can just walk these examples right up through the countable ordinals.

You might NEVER find a nonempty interval!

And what if  $<^*$ happens to actually be a well-order of the reals? It's some uncountable ordinal. You might find some nonempty  $T$'s but it wouldn't mean anything. It is a fact that descending chains of ordinals are always finite. Otherwise since ordinals are transitive sets, you'd have an infinite descending chain of element-hood, contradicting the axiom of foundation. You can't have infinite epsilon-chains. So at best you can only iterate the construction finitely many times and it doesn't prove anything.

But the mental image here is that if the reals are well-ordered, it does in fact look like a long ordinal-indexed "sequence" of discrete elements. That's what ordinals look like. They are literally "one thing after another" like bowling balls. So it's discount's conceptualization that's still incomplete. He still conflates the dense order with the possible well-order.

I better stop writing and let you tell me if you think I'm understanding this correctly. I could be misunderstanding everything, I would not be surprised.

My point is that my first example actually implies all the rest of that! It's a refutation of not only the proof, but of all possible attempted rescues of the proof by trying to choose a different interval.

Edited by wtf

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What I mean is that he could basically say "OK, maybe some intervals are empty", and continue with his proof. The existence of empty intervals doesn't affect the rest of his proof.

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Uh oh, disregard the last few lines of my argument. At the outset I wanted to consider a set T={x: a<*x<*b, a,b}. In fact we can forget about the 'b' . I say we also understand for all our chosen <* there are u,v∈T such that u<*v. WHAT I DID then was go ahead and try to prove my assertion while almost ignoring this notion that we can construct such an open set for any order relation. So hence for any order we have at least one set which has no minimum element. Its that simple. This seems fundamental and I might even call it an axiom. I actually could even describe subsets of this set which have no min for said order relation. Most everyone should agree that for any given set one can construct  at least one order-maybe even unique-that would give it a min. My assertion is the opposite of this notion.....So, I'm saying for any order relation there always exists a set like T on the first line above that it can't well order.

In the final part of my previous version I should have taken a different path, but I'm still puzzled on how to deal with wtf's ω1 + ω2 +...

I've tried to write orderings like this myself before and they seem to work until they eventually fail.

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54 minutes ago, discountbrains said:

So hence for any order we have at least one set which has no minimum element.

Why? Please write out the proof of this statement.

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2 hours ago, discountbrains said:

So hence for any order we have at least one set which has no minimum element. Its that simple.

What about a well-order? Then there is no nonempty set that has no minimum element. How do you know <* doesn't happen to be a well order?

You're claiming as a fact the very thing you are trying to prove.

Edited by wtf

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3 hours ago, uncool said:

Why? Please write out the proof of this statement.

Maybe my statement requires more detail. But, the answer to both of you is how can a set described like this with this <* order have a minimum no matter what the order is? wtf is already giving me that <* is a WO. I know nothing about the order relation as I ask this.

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23 minutes ago, discountbrains said:

Maybe my statement requires more detail. But, the answer to both of you is how can a set described like this with this <* order have a minimum no matter what the order is? wtf is already giving me that <* is a WO. I know nothing about the order relation as I ask this.

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

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Yes, its true I only showed there exists a countable number of z0, z1, z2, ... and the only way I could get more z's from T in my sequence is do wtf's ω+ω thing . I still believe there is a way to show any other z in T is a member of this sequence, but its a mute point now. I went back to my original thought. wtf just commented. Will see what he says.

6 minutes ago, wtf said:

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

By the way I write this set. It is written in a way that it has no min for <*. Can I not consider a set of this sort?

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1 hour ago, discountbrains said:

Maybe my statement requires more detail. But, the answer to both of you is how can a set described like this with this <* order have a minimum no matter what the order is? wtf is already giving me that <* is a WO. I know nothing about the order relation as I ask this.

I don't believe either of us said that any order relation has a minimum. You are the one who made a positive statement; it's your job to support it.

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1 hour ago, discountbrains said:

By the way I write this set. It is written in a way that it has no min for <*. Can I not consider a set of this sort?

Which set? You keep claiming T has a min when in fact I keep showing T might be empty. But now you say "this set" has no min. It would be helpful to me if you could repeat which set you are talking about at any given time.

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The idea of the proof seems to be that given an arbitrary linear order, you define $T$. If it's not empty, you do your construction. If it is empty, you pick another interval.I have two points to make.

* First, it has not been proven that this procedure will eventually find a nonempty interval. A lot depends on how the intervals are chosen. There are some details to be filled in if someone wants to claim they must eventually be able to choose a nonempty interval. I believe it's trickier than it looks.

* 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. This is impossible if the linear order happens to be a well order. Every downward chain of elements of a well-ordered set must be finite. That's because (as I noted earlier) there can be no infinite downward chains of set membership; and when it comes to ordinals, their order relationship is set membership. That's what it means that ordinals are inductive sets.

Now discount you keep objecting that I'm "assuming" $<^*[math] is a well-order. No I'm not. I'm using it as follows. YOU claim that ANY linear order can NOT be a well order. I am pointing out that if ht DOES happen to be a well order, your proof doesn't work! So the only way for your proof to work is to assume that the linear order is not a well order. In other words you have to assume the thing you're trying to prove. The idea seems to be that given an arbitrary linear order, you define [math]T$. If it's not empty, you do your construction. If it is empty, you pick another interval.I have two points to make.

* First, it has not been proven that this procedure will eventually find a nonempty interval. A lot depends on how the intervals are chosen. There are some details to be filled in if someone wants to claim they must eventually be able to choose a nonempty interval. I believe it's trickier than it looks.

* 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. This is impossible if the linear order happens to be a well order. Every downward chain of elements of a well-ordered set must be finite. That's because (as I noted earlier) there can be no infinite downward chains of set membership; and when it comes to ordinals, their order relationship is set membership. That's what it means that ordinals are inductive sets.

Now discount you keep objecting that I'm "assuming" $<^*$ is a well-order. No I'm not. I'm using it as follows. YOU claim that ANY linear order can NOT be a well order. I am pointing out that if $<^*$ DOES happen to be a well order, your proof doesn't work!

So the only way for your proof to work is to assume that the linear order is not a well order. In other words you have to assume the thing you're trying to prove.

Edited by wtf

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