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

[wtf]

1 hour ago, discountbrains said:

"YOU CLAIM there is no well-order on the reals".

You say I begin by stating something as fact and then use the statement to prove itself; thus, making a circular argument.  I don't know how I;m doing that. I just went through a simple illustration to show that  there exists at least one set that doesn't have a min  with any order relation.

This is like its established that x is defined to have certain properties if its divisible by 3 and I show x is not divisible by 3.

FWIW, I would like u to look at a post I put on Physics Forum on this scienceforum board. I submitted a flaw in Einstein's special relativity. I got no responses contradicting me; all I got were personal attacks. Haven't looked at the board for months though. I used very simple straightforward arithmetic. I don't say the whole theory is wrong. Fatal flaw?

Your argument seems to be: Assume we can't well-order the reals. Then there is a nonempty subset of the reals that can't be well-ordered.

That's circular. If you claim some subset of the reals can't be well-ordered, I can refute that argument by well-ordering the reals. So the burden is on you to show that can't be done. You have not done so.

 

> FWIW, I would like u to look at a post I put on Physics Forum on this scienceforum board. I submitted a flaw in Einstein's special relativity. I got no responses contradicting me; all I got were personal attacks. Haven't looked at the board for months though. I used very simple straightforward arithmetic. I don't say the whole theory is wrong. Fatal flaw? 

I can't address different claims you made on different forums, especially when you didn't bother to link the thread. I will say that PhysicsForums is a much more strictly moderated forum than this one, and they have zero tolerance for alternative points of view not backed up by a rigorous argument. I can't comment further since I have no idea what post you might be talking about or what you said. Nor am I qualified to discuss alternative physics. Or standard physics for that matter. I do know that bowling balls fall down. My understanding of the latest theory is that we happen to live in a universe where bowling balls fall down. In some other universe, they fall up. This is what passes for scientific thinking in our postmodern age.

Edited November 14, 2018 by wtf

[discountbrains]

No, no, I didn't say that at all. I think u'r the one confusing things. Since u take everything can be WO'd as dogma u think any statement not consistent with that must be negating it. I'm only testing the theory.

I'll look up the link. Yes, I see what goes on on these boards. They DO NOT like challenges to conventional wisdom. They like questions from people less well versed in the subject so they can appear smart in answering. That stack exchange I don't even visit anymore. Oddly, supposedly intelligent people act literally like 10 year olds- so frustrating.. 

[wtf]

33 minutes ago, discountbrains said:

No, no, I didn't say that at all. I think u'r the one confusing things. Since u take everything can be WO'd as dogma u think any statement not consistent with that must be negating it. I'm only testing the theory.

I'll look up the link. Yes, I see what goes on on these boards. They DO NOT like challenges to conventional wisdom. They like questions from people less well versed in the subject so they can appear smart in answering. That stack exchange I don't even visit anymore. Oddly, supposedly intelligent people act literally like 10 year olds- so frustrating.. 

> No, no, I didn't say that at all. I think u'r the one confusing things. Since u take everything can be WO'd as dogma u think any statement not consistent with that must be negating it. I'm only testing the theory.

I take the well-ordering of the reals as a logical consequence of the ZFC axioms; or alternately, a logical consequence of ZF + CH. That's the opposite of dogma. That's logical proof from axioms using standard principles of reasoning.

You're not testing any theories. You're presenting convoluted arguments that, when carefully unpacked, don't hold water.

 

[discountbrains]

wtf:  

I had a book called "Axiomatic Set Theory" I copied because it was out of print. I know where it is. I believe it contains a lot of the technical stuff u'r talking about such as ZFC set theory models etc. I used to study it. I take it it was the gold standard on set theory and meta mathematics.

 Why didn't u tell me this 50 posts ago?

Any nonusual order <* might order the interval S = (0,1) like for example, <0.5, 0.0000002, 0.46, a, b,...> where a ∈ (0.2, 0.4), b ∈ (0.75, 0.54)-reverse order. If σ is the min for S then there is no min for the set I mentioned several times above for <*. This is true  no matter how the new ordered set looks.  That's just the way it is even though its not consistent with pretty well established set theory for years. I have to add u at one time claimed I declared the reals to be well ordered and another time u maintain I say the opposite. I just can't win.

Edited November 16, 2018 by discountbrains

[wtf]

13 hours ago, discountbrains said:

wtf:  

I had a book called "Axiomatic Set Theory" I copied because it was out of print. I know where it is. I believe it contains a lot of the technical stuff u'r talking about such as ZFC set theory models etc. I used to study it. I take it it was the gold standard on set theory and meta mathematics.

 Why didn't u tell me this 50 posts ago?

Any nonusual order <* might order the interval S = (0,1) like for example, <0.5, 0.0000002, 0.46, a, b,...> where a ∈ (0.2, 0.4), b ∈ (0.75, 0.54)-reverse order. If σ is the min for S then there is no min for the set I mentioned several times above for <*. This is true  no matter how the new ordered set looks.  That's just the way it is even though its not consistent with pretty well established set theory for years. I have to add u at one time claimed I declared the reals to be well ordered and another time u maintain I say the opposite. I just can't win.

 

> If σ is the min for S then there is no min for the set I mentioned several times above for <*.

You keep "mentioning" it but your claim is false.

> u maintain I say the opposite.

Isn't that exactly what you've been arguing? That the reals can't be well ordered? If that's not your claim, what is?

Isn't "Where to submit my proof that the set of real numbers can't be well ordered?" a clue?

Edited November 16, 2018 by wtf

[discountbrains]

You are determined to find fault with my assertion. You say I'm starting off with one claim and then u accuse me of starting with the opposite claim.

For answer to taeto, I believe I left going back to my original conclusion on physics forum. I'll have to go back and see what I said.

[wtf]

1 hour ago, discountbrains said:

You are determined to find fault with my assertion. You say I'm starting off with one claim and then u accuse me of starting with the opposite claim.

For answer to taeto, I believe I left going back to my original conclusion on physics forum. I'll have to go back and see what I said.

If you would clearly state a thesis and a claimed proof, I can point out your errors. I've been doing so regularly for weeks.

I'm not "determined to find fault," I'm responding to your initial request for critique by someone who understands the material. Do you revoke your request? If so I'm happy to comply.

ps -- Here is your direct quote from your initial post in this thread:

On 8/2/2018 at 8:32 PM, discountbrains said:

I want answers from people who really know this stuff.

Do you still want that? If not, I'll happily stop responding.

Edited November 16, 2018 by wtf

[discountbrains]

OK, I see what's happening here. There is a subtle disagreement in our logic and my approach is unorthodox. At least I've never seen anything like it. My thought is: Given any nonempty, noncountable, subset of  R that is dense in R ( (0,1) is an example) for any function or algorithem or whatever that produces an order relation, <*, that identifies a min for this subset another subset with this minimum deleted has no minimum for this <*. You say pf course it has a min because if the <* is a well ordering it will order all subsets like {a, b, c,...} and each min can easily be found. I believe the power set for R contains any kind of even unimaginable subsets of R and such a subset exist which implies there is at least one subset with no min no matter what the order. I defined this subset in a previous post.

Maybe I need to go ahead and try to prove your discrete order can't exist. I'm still trying to modify one of my failed attempts to prove this. Or should it suffice to say clearly a set of the form I described exists?

[studiot]

Let us work this in terms of your example subset.

I assume the interval (0,1) is meant to include the endpoints 0 and 1.

Thus 0 is a minimum and 1 a maximum for this set, using your terminology.

(I prefer smallest member and largest member)

 

So if you wish to then consider the ordering of the rest of R, you now have two subsets 

minus infinity to 0

and

1 to plus infinity

Both as open sets

The first has no maximum (although it has an upper bound) and no minimum or lower bound.

The second has no minimum though it has a lower bound and no maximum or upper bound.

 

(Sorry I can't do math symbols tonight.)

Edited November 18, 2018 by studiot

[wtf]

11 minutes ago, discountbrains said:

Given any nonempty, noncountable, subset of  R that is dense in R ( (0,1) is an example) ...

If you think (0,1) is dense in the reals you need to dig out that Real Analysis text of yours and review the definition of dense in R. It's true that (0,1) is dense; but it's not dense in the reals. Do you know the defs? An ordered set is dense if there's a third distinct element between any other two. A subset of the reals is dense in the reals if every real is the limit of a sequence of elements from the subset. Easy to confuse the two. (0,1) in the usual order is dense, but it's certainly not dense in the reals. 2, for example, is not the limit of any sequence of elements of (0,1). 

More later, but you seem to have a fundamental misunderstanding here.

[discountbrains]

I made a mistake. You're right. When one considers the power set they're talking about distinct subsets. Here we are talking about the same subset. The only difference is the order of the elements is different. Yes, I know, I was a bit loose with my terminology about dense sets. I didn't consult my book. I really meant any two elements of the set has anther element between them.

I'm left with needing to prove your ordering of my original set is impossible....If I can.

23 minutes ago, studiot said:

 

24 minutes ago, studiot said:

Let us work this in terms of your example subset.

I assume the interval (0,1) is meant to include the endpoints 0 and 1.

 

No, not at all. (0,1) is an open set or open interval as usual.

 

 

[wtf]

1 hour ago, discountbrains said:

I made a mistake. You're right. When one considers the power set they're talking about distinct subsets. Here we are talking about the same subset. The only difference is the order of the elements is different. Yes, I know, I was a bit loose with my terminology about dense sets. I didn't consult my book. I really meant any two elements of the set has anther element between them.

 

 

No prob, everyone confuses dense with dense in the reals. They literally mean two different things. Bad terminology, like "open set." Just made up to confuse people.

I haven't read your earlier post past the business with (0,1). I need to review that. But meanwhile, just consider the following ordering on (0,1):

1/2, 1/3, 1/pi, 1/e, 1/sqrt(2), and everything else stays exactly where it was. And in the reals at large, everything stays were it is. So it's the usual order on the reals; but within (0,1), the first few elements of the order are discrete, and the rest are in their usual dense order. And clearly you can do this trick with as many discrete elements at the beginning as you like. Finitely many discrete values. But also infinitely many. For example suppose we order (0,1) such that we start with an enumeration of the rationals: q1, q2, q3, q4, ... Then after all those, everything else stays the same. So we have an order that's not a well-order, and isn't dense, and starts with a countably long infinite sequence of discrete elements. 

So now with these examples in mind, we can see that even if the reals aren't well-ordered; your argument STILL does not work. You want to remove the first element of some proposed well-order of S and claim that what's left doesn't have a first element. But my examples show that this claim is simply false. Your proof fails EVEN IF the reals can not be well-ordered.

I've shown that the assumption that the reals may be well-ordered is no longer needed. I thank you for challenging me on this point. I don't actually need that claim. Even if the reals could not be well-ordered, your argument would fail because we can always order (0,1) in such a way that it has as many initial discrete elements as we like. Your idea of continually removing first elements can always be defeated by some counterexample analogous to the ones I showed.

1 hour ago, studiot said:

Let us work this in terms of your example subset.

I assume the interval (0,1) is meant to include the endpoints 0 and 1.

 

The open brackets are universally agreed on to mean the open interval, excluding its endpoints. I've seen ]0,1[ used for this (in France, perhaps elsewhere in Europe) to avoid ambiguity with the ordered pair (0,1). But when its meaning as an interval is certain, (0,1) is the open unit intervals. It does not include its endpoints. 

 

Edited November 18, 2018 by wtf

[wtf]

2 hours ago, discountbrains said:

Given any nonempty, noncountable, subset of  R that is dense in R ( (0,1) is an example) for any function or algorithem or whatever that produces an order relation, <*, that identifies a min for this subset another subset with this minimum deleted has no minimum for this <*.

I have refuted this claim even without the assumption that I can well-order the reals. 

I assume "another subset" refers to the same order with its first element deleted. My examples show that no matter how finitely many times you delete the first element, what remains may still have a least element. And you might even remove countably many initial elements and what's left still has a least element. 

 

Edited November 18, 2018 by wtf

[studiot]

1 hour ago, wtf said:

The open brackets are universally agreed on to mean the open interval, excluding its endpoints. I've seen ]0,1[ used for this (in France, perhaps elsewhere in Europe) to avoid ambiguity with the ordered pair (0,1). But when its meaning as an interval is certain, (0,1) is the open unit intervals. It does not include its endpoints. 

 

And I have seen several different notations.

So what?

So it is far from universal, especially in British practice.

So I asked what the OP means and stated the consequences of my assumption.

Clearly if DB's intention was as you say then I need to revise my list of consequences, but they will leave us with essentially the same problem to overcome.

I also said I can't do special characters from this keyboard.

Edit I just saw that DB added words into the quote box I did not say, despite recent moderator requests and requirements not to do this.

Thank you for clarifying the meaning of your brackets.

So it leaves a set (0,1) which is bounded above and below but has neither a least member nor a greatest member.

Do you agree with this now?

[wtf]

10 minutes ago, studiot said:

 

And I have seen several different notations.

So what?

So it is far from universal, especially in British practice.

So I asked what the OP means and stated the consequences of my assumption.

Clearly if DB's intention was as you say then I need to revise my list of consequences, but they will leave us with essentially the same problem to overcome.

I also said I can't do special characters from this keyboard.

Edit I just saw that DB added words into the quote box I did not say, despite recent moderator requests and requirements not to do this.

Thank you for clarifying the meaning of your brackets.

So it leaves a set (0,1) which is bounded above and below but has neither a least member nor a greatest member.

Do you agree with this now?

Not really, I didn't follow your point. Might just be me.

(0,1) as an open interval is universal except among those who use ]0.1[ If you have a reference using (0,1) as a closed interval I'd be interested in seeing it. 

Edited November 18, 2018 by wtf

[studiot]

1 minute ago, wtf said:

Not really, I didn't follow your point. Might just be me.

I don't know what you are referring to, in my post you quoted.

I made several points and asked the OP one question.

[wtf]

34 minutes ago, studiot said:

I don't know what you are referring to, in my post you quoted.

I made several points and asked the OP one question.

I was referring to your most recent post. I'm thinking there's enough vagueness in @discountbrains's exposition that we might be reading different things into it. I didn't think that your comments were relevant to my interpretation of the argument, which is why I said I didn't understand your point(s).

Edited November 19, 2018 by wtf

[studiot]

7 minutes ago, wtf said:

I was referring to your most recent post. I'm thinking there's enough vagueness in @discountbrains's exposition that we might be reading different things into it. I didn't think that your comments were relevant to my interpretation of the argument, which is why I said I didn't understand your point(s).

I don't think my comments were directly relevant to yours or you long sweated painstaking explanations (well done for keeping going)

I was beginning an alternative viewpoint that DB might find helpful (though then again he might not but we can't know till I try).

FYI It relies on the self similarity property of a part to the whole of the Reals.

[wtf]

2 hours ago, studiot said:

I don't think my comments were directly relevant to yours or you long sweated painstaking explanations (well done for keeping going)

I was beginning an alternative viewpoint that DB might find helpful (though then again he might not but we can't know till I try).

FYI It relies on the self similarity property of a part to the whole of the Reals.

Well that's sort of my point. I don't think OP's argument relates to self-similarity. However, that does not mean I think I'm right and you're wrong.

On the contrary, I think OP's presentation is so convoluted and ambiguous, that my interpretation is not the only possible interpretation. For all I know you are understanding OP better than I am. 

But self-similarity is important in the sense that the open unit interval (0,1) is homeomorphic to the entire real line. That's kind of a cool thing about the real numbers. You can take a bounded interval like (0,1) and stretch it infinitely like idealized taffy, until it coveres the entire real line to infinity in both directions. And in this transformation virtually nothing is lost. You lose boundedness but the cardinality and the topology are identical. Whether any of this bears on the OP's interests, I can't say.

Did you happen to read my earlier post showing that OP's argument doesn't work? If that makes sense to you, that's what I think OP is getting at. But if not, maybe I'm misunderstanding the OP. It's certainly possible. But basically OP takes a subset S of the reals (under any order). If S has a smallest element, take that element away. The remaining set, OP says, doesn't have a least element. I've decisively refuted that idea. That's my take on this whole business at this point.

I stay in because I love the ordinals. There's a theorem that every countable ordinal can be order-embedded in the rationals in their usual order. That's a really cool fact. The proof looks complicated but not beyond a little diligence.

 

Edited November 19, 2018 by wtf

[discountbrains]

"1/2, 1/3, 1/pi, 1/e, 1/sqrt(2), and everything else stays exactly where it was. And in the reals at large, everything stays were it is. So it's the usual order on the reals; but within (0,1), the first few elements of the order are discrete, and the rest are in their usual dense order. And clearly you can do this trick with as many discrete elements at the beginning as you like. Finitely many discrete values. But also infinitely many. For example suppose we order (0,1) such that we start with an enumeration of the rationals: q1, q2, q3, q4, ... Then after all those, everything else stays the same. So we have an order that's not a well-order, and isn't dense, and starts with a countably long infinite sequence of discrete elements. 

So now with these examples in mind, we can see that even if the reals aren't well-ordered; your argument STILL does not work."

Yes, your assumption is right, it would always be whatever the chosen order relation is. And, after all your nth number in the list we see we can generate an n+1th number (and of course we would have a 1st one) so this would be by math induction at least a countable number of sets. I'm wondering here where do you start with the first real number after you run out of your initial list? 

To add to what I said earlier if u claim <* happens to be in fact a well ordering then it therefore has the property of creating a discrete set of numbers which negates my claim this is analogous  of saying the Bible says this is the way it is therefore other things can't be true.

[wtf]

23 minutes ago, discountbrains said:

 

Yes, your assumption is right, it would always be whatever the chosen order relation is. And, after all your nth number in the list we see we can generate an n+1th number (and of course we would have a 1st one) so this would be by math induction at least a countable number of sets. I'm wondering here where do you start with the first real number after you run out of your initial list? 

To add to what I said earlier if u claim <* happens to be in fact a well ordering then it therefore has the property of creating a discrete set of numbers which negates my claim this is analogous  of saying the Bible says this is the way it is therefore other things can't be true.

@dcb  I did not understand anything you wrote here.

I reiterate that I have refuted your claim that you can always remove enough initial elements to leave a set with no smallest element. On the contrary that is false and I showed you why very clearly. 

> I'm wondering here where do you start with the first real number after you run out of your initial list?

Anywhere! You just pick a next number you haven't picked yet. 

Edited November 19, 2018 by wtf

[discountbrains]

So, if I am getting it correctly u are separating out the rational numbers and placing them 1st. Then u have the irrationals. Where do u start with them? Maybe u could say 2^1/2, 3^1/2, ... Then what? I tried this stuff years ago and don't think it works. So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC. So, we don't need an axiom like this at all. If u could do this u ought to get the Clay Prize. Don't think it is offered for this because its considered unprovable. 

I wonder if some theorems that depend on the AC could work for just the sets you need and you need only talk about a finite number of sets. I need to try to find theorems that depend on the AC on the internet. I know I've seen one or two in my studies, but don't remember them.

[studiot]

14 hours ago, wtf said:

I don't think OP's argument relates to self-similarity

No I don't think he has considered this, which I why I wanted to raise it. +1

In any event since he has not had the courtesy to answer me I am out of here.

I wish you well in your struggle.

Edited November 19, 2018 by studiot

[wtf]

2 hours ago, discountbrains said:

So, if I am getting it correctly u are separating out the rational numbers and placing them 1st. Then u have the irrationals. Where do u start with them? Maybe u could say 2^1/2, 3^1/2, ... Then what? I tried this stuff years ago and don't think it works. So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC. So, we don't need an axiom like this at all. If u could do this u ought to get the Clay Prize. Don't think it is offered for this because its considered unprovable. 

I wonder if some theorems that depend on the AC could work for just the sets you need and you need only talk about a finite number of sets. I need to try to find theorems that depend on the AC on the internet. I know I've seen one or two in my studies, but don't remember them.

> So, if I am getting it correctly u are separating out the rational numbers and placing them 1st. Then u have the irrationals. Where do u start with them?

I don't need to. I don't need to well-order the reals to show that your argument is wrong.

You say to pick a nonempty subset S of the reals. Say it has a first element a. Then you say that S\{a} has no first element. I showed the counterexample to that claim. So your argument fails. It also fails no matter how many elements you take away, since what's left is always nonempty and you can simply pick an element from it as the least.

> So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC.

I said no such thing. I simply pointed out that even if the reals can't be well ordered, your argument fails.

You complained that I was using well-ordering of the reals to defeat your argument. So I turned it around and defeated your argument WITHOUT assuming AC or the well-ordering of the reals. 

I don't need to well order anything to prove that your argument fails.

> So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC.

No. This type of argument requires AC. It says that we end up with an endless sequence of nonempty sets; and that we can keep picking an element from each one. That's AC. This is the intuition behind why well-ordering requires AC. But I have not made that argument. All I need is to put two discrete elements at the beginning of a set to show that your argument is false. S\{a} has a least element. That falsifies your claim.

Edited November 19, 2018 by wtf

[discountbrains]

On 11/19/2018 at 12:47 PM, wtf said:

> So, if I am getting it correctly u are separating out the rational numbers and placing them 1st. Then u have the irrationals. Where do u start with them?

I don't need to. I don't need to well-order the reals to show that your argument is wrong.

You say to pick a nonempty subset S of the reals. Say it has a first element a. Then you say that S\{a} has no first element. I showed the counterexample to that claim. So your argument fails. It also fails no matter how many elements you take away, since what's left is always nonempty and you can simply pick an element from it as the least.

> So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC.

I said no such thing. I simply pointed out that even if the reals can't be well ordered, your argument fails.

You complained that I was using well-ordering of the reals to defeat your argument. So I turned it around and defeated your argument WITHOUT assuming AC or the well-ordering of the reals. 

I don't need to well order anything to prove that your argument fails.

> So, what u seem to be telling me is that by doing this u can identify a minimum for any set and u thus well ordered the reals without the AC.

No. This type of argument requires AC. It says that we end up with an endless sequence of nonempty sets; and that we can keep picking an element from each one. That's AC. This is the intuition behind why well-ordering requires AC. But I have not made that argument. All I need is to put two discrete elements at the beginning of a set to show that your argument is false. S\{a} has a least element. That falsifies your claim.

Just revisiting this thread now. I don't think I will go over everything above. I just recently concluded my argument of constructing new set S\{a} by continually deleting the min, a, from each previous set is flawed. I was trying to use an assumption to prove itself. I don't recall seeing u presenting anything showing my error-I have to grade my own work. All u did is claim I'm wrong and then said u showed proof I was wrong. 

I will state again that 'the reals cannot be Well Ordered' by the simple fact that for any order relation, ,<*, I choose, a set {x: a <*x<*b} can be constructed which clearly has no least element for <*. Can we not construct such a set? I believe I stated this at the very beginning.