wtf

12 minutes ago, discountbrains said:

Are u _____ or what? READ MY POST!! I DID read yours. I was trying to convey I was not including your set. You keep repeatedly bringing up an example I'm not even talking about. I wrote I'm talking about subsets (or sets if u prefer) we don't already know a WO for which I was hoping was precise enough. Maybe this is the source of confusion for both of u. 

How does ignoring the counterexamples to your argument help you convince anyone? The examples I'm giving you falsify your argument.

14 minutes ago, discountbrains said:

Think about it this way: You have your set, S, above with all its original numbers in it. You have your <*. You declare z  the min of S. You delete z and are left with all numbers x from S,  z <*x. For your order how do you find the min of these numbers?

I already explained this. It would go better if you'd read my posts. 

Run your argument on the natural numbers in their usual order 0, 1, 2, 3, ...

Take the tail 14, 15, 16, 17, ...

Remove 14 to get 15, 16, 17, ...

What's the smallest element? 15.

Remove 15 to get 16, 17, 18, 19, ...

What's the smallest element? 16

This is exactly what would happen if you have a well-order for the reals. You haven't shown this can't happen. You're still confusing the usual order on the reals with an arbitrary linear order.

It doesn't even have to be a well-order. Say your order is 14, 15, everything else. Then you pick the tail S = 14, 15, everything else. You delete 14. Now you have 15, everything else. The smallest element is 15.

But what if it was 14, 15, 16, everything else. Same thing. No matter how many times you remove an element, there's a linear order that breaks your proof.

On 8/22/2019 at 6:58 AM, discountbrains said:

I'm going to make a new attempt:

> I'm going to make a new attempt: Like I said before there always exists a subset of R like {x: a <*x} for some a for any total, linear order, <*, which has no min for <*.

That's a restatement of the claim that there's no well-order on the reals. A set is well-ordered if every nonempty subset has a smallest element. So if you say that for any linear order there's some nonempty set that has no smalleset element, then there could be no well-order. This is a simple restatement of your claim.

> I think you and uncool want absolute evidence there is such a set.

Yes, because it's your original claim! You have made a claim -- which is false, by the way -- and you must supply a proof. 

> My proof is as follows: So, now lets assume for any <* there are at least a few subsets of the reals  that contain a minimum element. Otherwise, I am done: there are NO subsets with a min. Suppose we have a subset that has a min, z, and its S={x: z ≤*x}. Lets consider S\{z}. This set is {x: z <*x} and is the same form as the set on the first line and has no min with respect to <*.

WHY NOT? Consider the usual order on the natural numbers, which is a well-order. I have formed the set [math]S = \{x : 14 \leq x\}[/math]. No problem there. Now suppose I delete 14 from that set? What is the least element of the new set? It's 15. And if you delete 15, the new smallest element is 16. And this goes on forever, because the usual order on the naturals is a well-order.

Now you claim this can't happen with the reals, but why not? Suppose [math]<^*[/math] happens to be a well-order. Then the exact same thing will happen as it does with the naturals. You take some element, form the set of everything greater than or equal to that element, delete that element, and the new set still has a smallest element; namely, the next element in the well-order.

But it doesn't even have to be a well-order. You might have picked some element, like 14, that JUST HAPPENS to have an immediate successor. Then your proof fails.

A well-order of the reals looks JUST LIKE THE NATURAL NUMBERS. It's one element followed directly by another then by another then by another. Once in a while you have to take a "limit ordinal," as in the process that produces [math]\omega[/math] following all the natural numbers. You just keep on going taking immediate successors and occasionally limits, and you can work your way up to an uncountable ordinal. 

Once you have this picture in your mind your proof fails. You can take a tail of the order, ie the set of elements greater than or equal to some particular element; then you can delete the element; and the NEXT element in the order is now the smallest element of the deleted set. You haven't proved this can't happen and in fact you don't seem to understand this point at all. 

49 minutes ago, discountbrains said:

I hope you understand what I mean by this remark. I believe I was saying we should be able to apply the same general concept we use for '<" to the use of '<*'. On 2nd thought I might have to prove denseness  in the reals for <*.

Your remark just now that <* is dense shows that you don't understand this point. When you well-order the reals, you lose the denseness property. You have to forget about it. A well-order on the reals is one element followed by another. If you delete the smallest element of a tail, the new smallest element is now the next element in the order. 

But you don't even need for <* to be a well-order. If you pick some tail S = {x : z <= x} for some real number z, and you delete z, the new set DOES HAVE A SMALLEST ELEMENT just in case z happens to have a direct successor. You haven't proved that can't happen. And in fact if <* does happen to be a well order, it MUST happen every time. Every element has a direct successor and every "deleted tail" still has a smallest element.

2 hours ago, discountbrains said:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1 hour ago, discountbrains said:

In my last message for

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

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

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

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

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

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

13 minutes ago, discountbrains said:

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

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

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

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

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

Now you have a countably infinite well-ordered subset. 

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

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

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

Please:

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

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

50 minutes ago, discountbrains said:

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

 

...

 

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

I have no comments on matters of physics.

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

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

13 minutes ago, discountbrains said:

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

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

24 minutes ago, discountbrains said:

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

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

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

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

3 hours ago, discountbrains said:

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

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

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

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

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

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

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

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

The idea of the proof seems to be that given an arbitrary linear order, you define [math]T[/math]. 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]<^*[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[/math]. 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]<^*[/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 [math]<^*[/math] 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.

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. 

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?

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.

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) [math]<^*[/math]; 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  [math]a[/math] and  [math]b[/math] he picks, I choose the linear order:

[math]a <^* b <^* \text{everything else in its usual order}[/math]

In this order it's clear that  [math]T = \{x \in \mathbb R : a <^* x <^* b\}[/math] is empty. Therefore there is no  [math]z_0[/math] 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  [math]c[/math] and  [math]d[/math], 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,  [math]T[/math] is empty, so we don't care about it. We'll take some other interval. But I can give you the order

[math]a <^* b <^* c <^* d <^* \text{everything} [/math]

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 [math]\omega + \omega[/math], 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  [math]<^*[/math] happens to actually be a well-order of the reals? It's some uncountable ordinal. You might find some nonempty  [math]T[/math]'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.

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. 

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.

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.

20 minutes ago, discountbrains said:

Yes, I know that's exactly what I'm saying: the reals can't be well ordered. And, you say there is a proof without the AC that says they can. Yet, I'm getting the result I'm getting. I think I can come up with many, many examples of your logic. I might say all my numbers in my set S add up to a whole number like n+m=p (p a whole number). For example 1/3+2/3 =1, 2+1=3... and you say what about 1/2+1/3 etc? And then say my statement is false. This may not be an exact analogy though. Let me reiterate, I believe my last version of <* must include numbers from T. Let me see if this is what u are saying: I say all the car prices on a Chevy dealer's lot are ordered a certain way and u say my statement is invalid because the Ford dealers are not ordered that way.

> And, you say there is a proof without the AC that says they can.

No it's more subtle than that. There is a model of set theory in which the reals can be w.o. but Choice fails. But this is not relevant to our discussion, except that you are looking for a proof that the reals can't be w.o. and there can be no such proof.

The rest of your post I did not find helpful. I don't know why you're adding numbers. Your car analogy doesn't apply. You are trying to prove that no <* can be a w.o. but your proof is faulty, as you already agreed. 

>  Let me reiterate, I believe my last version of <* must include numbers from T.

Which version of <*? It will be very helpful for you to write out a specific proof. 

But it doesn't matter that SOME <* has a nonempty T, that's perfectly fine. Your proof depends on NO <* having a nonempty T and that is false.

7 hours ago, discountbrains said:

Yes, Neither 0 or 1 are in T. 1<*everything else makes any other x and y such that (x,y)∈<* not be in T. So T is empty with your order relation. You are pointing out an imprecision in my argument. That's good. I should restrict any order relation to be if x and y are in T, (x,y)∈<*. Or this could be written in other ways......

No, there are lots of well orderings for a lot of sets. I'm trying to show there are some sets that can't be well ordered.

> Yes, Neither 0 or 1 are in T. 1<*everything else makes any other x and y such that (x,y)∈<* not be in T. So T is empty with your order relation. You are pointing out an imprecision in my argument. That's good

Ok!! Good. We agree that if I am allowed to cherry-pick <*, then I can make T be empty. 

Now YOU are claiming that the reals can't be well-ordered. So your argument is: Given ANY <* and you'll show it's not a well-order.

But I showed that for SOME <* orders, T is empty and your proof is blocked, because you can no longer pick the min of T. 

So now, you need a new argument. Yes?

> I'm trying to show there are some sets that can't be well ordered.

Aren't you trying to show explicitly that the reals can't be? It's impossible to do that because even in the absence of the axiom of choice, it's possible that the reals are well-ordered. But nobody can produce an explicit well order.

Since it is logically consistent for the reals to be well-ordered even in the absence of Choice, you can not possibly prove that the reals aren't well-ordered. We covered this point months ago if I recall.

5 hours ago, discountbrains said:

I admit I should take more time for serious thought on your example, but will need to see if your example is meant to be totally unworkable with my definition of T. Yes, again I can come up with definitions of <* that will make T seem empty right off the bat. So what. If I say T contains all elements of S except a and b  while a and b are also in S then that's what T contains. Again defining <* to make T contain nothing is beside the point because u must have at least one min in T to make T well ordered. You're trying to make T empty so u don't have to answer the question. You could offer to help me define T and <* like suggest I start by saying "if T is a nonempty set".

> I admit I should take more time for serious thought on your example,

That would be a good idea. I'd like you to just say YES or NO to my question: If <* is defined as "0 <* 1 <* everything else", is T = {x : 0 <* x <* 1} empty?

If you agree, then there's something we agree on. If you disagree, we can focus on this example till we agree.

Now having done that, you might argue that I'm wrong to MENTION it. We can argue about its RELEVANCE. But at least we would agree on whether this particular T is empty or not. That is important. 

> but will need to see if your example is meant to be totally unworkable with my definition of T.

We can argue about that LATER. First just say YES or NO as to whether my T is empty or not. Once we do that, you can argue that I had no right to mention it in the first place. But it would be something we agree on. And I can then explain why it's relevant. So please don't let trying to figure out the relevance, get in the way of you just saying YES or NO to this basic question.

> Yes, again I can come up with definitions of <* that will make T seem empty right off the bat. So what. If I say T contains all elements of S except a and b  while a and b are also in S then that's what T contains.

Fine. Let's talk about that LATER. First please answer the question. I for one can not go forward unless we have agreement on that one technical point. I'll justify its relevance later, but first it's important for us to at least agree on the basic fact.

> Again defining <* to make T contain nothing is beside the point because u must have at least one min in T to make T well ordered.

Ah. No I am not trying to show anything is well ordered. I am pointing out that YOUR PROOF FAILS that <* is NOT well-ordered. Do you see that?

> You're trying to make T empty so u don't have to answer the question.

I'm giving an example where T does happen to be empty, because your proof depends on T not being empty. If T is empty even for a SINGLE <*, then your proof fails; because you have to make your proof go through for EVERY linear order <*. Once there exists even a SINGLE <* that makes T empty, your proof fails because the empty set does not have a min. Remember well-order definition is that every NONEMPTY set has a smallest element. This is vital.

> You could offer to help me define T and <* like suggest I start by saying "if T is a nonempty set".

Ah you think I am being deliberately unhelpful to not do so. On the contrary. If I could say T is nonempty I would. But here is the problem:

We are playing the following game. You claim that NO linear order <* can be a well-order. So I am entitled to cherry-pick a <* that makes T empty, which blocks your proof. That's why I'm using that example. Not to be unhelpful. But to be HELPFUL. To show you the error in your proof.

2 hours ago, discountbrains said:

I see in my latest definition of T I made a couple minor errors. Since T doesn't contain a and b T can't = S near the bottom of my proof. Since my initial conditions state all other elements in S are in T clearly T is not empty. However, are you really a math prof? You know and you stated your example is not a well ordering. I don't care about this at all. This is entirely not the point. I care only about what is proposed to be a WO and about 6 lines down I conclude such an ordering implies, in fact, various sets I present must be empty. You do realize if u can produce an ordering that well orders my set thus negating my argument u actually have done something pretty profound.  You will have actually produced a WO for the reals. You seem to be stuck the idea of finding anything to show my T is empty.  Yet I get -32 votes here-amazing.

You seem to have changed the subject and decided not to answer the question I posed to you. Not much for me to say here. Let me know if you want to discuss some math at some point.

>  Since my initial conditions state all other elements in S are in T clearly T is not empty.

You may "state" that T is nonempty but my example shows that it might be. And you're making a point of not responding to that issue. So for the record, I ask again: If we define <* as "0 <* 1 <* everything else in its usual order", is T = {x : 0 <* x <* 1} empty or not?

5 hours ago, discountbrains said:

That's not what I'm trying to say. Clearly, to me I don't intend to cherry pick any order relation. I believe I clearly stated this applies to any order you choose. I should revisit my proof, but I should stipulate a,b are in S, not in T, and T contains all other elements of S. This leaves no doubt T is nonempty.

Interesting some of my replies are missing. Could it be I used a popular slang word that's not demeaning to anyone?

No, no, no, NO! You don't redefine my T!!! What is the issue here is that <* is arbitrary and you can define it any way like. 

 

 

> That's not what I'm trying to say. Clearly, to me I don't intend to cherry pick any order relation. I believe I clearly stated this applies to any order you choose.

That's right. You claim that for ANY linear order, you can prove it's not a well order. Ok, then I get to pick the order. And if you tell me which [math]a[/math] and [math]b[/math] you want to use, I can still modify the order and dare you to make your proof work. That is CORRECT, you do understand that. 

> Interesting some of my replies are missing. Could it be I used a popular slang word that's not demeaning to anyone?

I represent nobody but myself here and have my own issues with the moderators from time to time. I'm the last person who can do anything about your posting issues. 

> No, no, no, NO! You don't redefine my T!!! What is the issue here is that <* is arbitrary and you can define it any way like. 

I'm not redefining your T at all. As you agree, I am allowed to pick the total order <* and then challenge you to prove it's not a well-order. And once you tell me your [math]a[/math] and [math]b[/math], which is an unnecessary complication we could do without, I can just adjust my <* accordingly. 

For example you want [math]0 < a < b < 1[/math] where < is the usual order. Then my <* is just "a < b < everything else" and your T is empty. 

> Three times u redefined my set, T

I never redefined it. I used it exactly as you specified it. You're right that I used 0 and 1 instead of a and b but that's a minor detail. I can use a and b if you like. 

> ....... OK, got it. I will need to take a look at your example of <*.

I would truly ask you to do that. 

> Maybe I can find a way around it.

Give it a try. When you understand my example you'll see why your proof doesn't work.

> Wait a minute, who said the "everything else" is in the usual order?

I said that though I never said it explicitly. My order <* is "0 < 1 < everything else in its usual order". 

So for example 0 <* 1, 1 <* 47, 102 <* 104 and so forth. 

But the numbers in (0,1) are of special interest. We have 1 <* 1/2, 1 <* pi - 3, 1 < sqrt(2) -1, and so forth. 

0 <* 1 <* everything else in its usual order. So this order has a discrete initial segment and a dense tail. 

If you want to call it a and b that's fine but I'll just use a <* b <* everything else in its usual order. 

So in the order 0 < 1 < everything else [can I stop writing "in its usual order" and let that be understood?], can you see that T = {x : 0 <* x <* 1} must be empty? It MUST be since we defined the order exactly so it would be! I am allowed to do that since I get to pick the order and you must supply a proof it's not a well order. 

So to narrow this down to one simply YES or NO question: 

If I define <* as 0 <* 1 <* everything else, would you agree that T = {x : 0 <* x <* 1} must be empty? Just answer this, and then we can move on. We must get clarity on this single point before discussing anything else. 

> I really think I need not even require x<*b in T. All I'm concerned about is the lower bound or the min .

An empty set does not have a lower bound OR a min. And since I can always cook up a <* that makes T empty, your proof fails.

5 hours ago, discountbrains said:

 I know what you're trying to say. Since you are starting with its given every set can be well ordered it must follow my sets are empty. It that not correct? Somebody here doesn't understand what's going on. First of all, I defined S=(0,1) with usual order not your set. Do u not understand I am presenting a set, T, which I showed can't be well ordered. You keep saying my sets are empty. Let me put it this way: we can think of any uncountable subset of the reals as an infinite deck of cards. We can shuffle this deck any way u like and get a new order. If u pick any card and claim there's no other card below it I can always find one that is. Or, lets replace any two digits of some decimal number with two others and then replace the two others with another two. We have a reordering. Or we might change all the digits of every number. Whatever number u say is the first I still can find another number before it. This doesn't matter how u order things.

>  I know what you're trying to say. Since you are starting with its given every set can be well ordered it must follow my sets are empty. It that not correct?

No it is not correct. I am NOT starting with a well-order. And your T sets may be empty even if <* is not a well order. 

> Somebody here doesn't understand what's going on.

That's right. 

> Do u not understand I am presenting a set, T, which I showed can't be well ordered.

For suitable choice of <*, which need not be a well-order, T is empty and your argument fails.

> You keep saying my sets are empty.

T MAY be empty for suitable choice of <* and a and b. But to make your argument work, you have to show that T is nonempty for any possible <*, a, and b. But you can't do that because there is always a counterexample of the form I gave.

> What u are trying to say is my set T does not exist by saying its empty.

It does exist. It just happens to be empty. 

Suppose we take the order "0 < 1 < everything else". Now this is NOT a well-order because the "everything else" is in its usual dense order. But T = {x : 0 <* x <* 1} is empty. DO YOU OR DO YOU NOT AGREE WITH THIS? Let's not discuss anything else till we get clarity since this example falsifies your argument.

2 hours ago, discountbrains said:

Here's another way to look at it: We don't hesitate writing a set like {x: a<x<b} with natural ordering. Isn't it natural to write {x: a<*x<*b} for a different ordering, <*? ....Lets suppose there is some dude in an alternative universe for which <* is the natural order of things and < is an alternative order? So, any of these sets will have plenty of elements in them.

Sure, it's perfectly natural to write that. But let's get clarity on this one thing. 

Suppose <* is defined as: "0 <* 1 <* everything else". Now this is NOT a well order since the "everything else" is all the real numbers in their usual order, with the exception of 0 and 1, which we pulled out and put in front of everything else.

Then T = {x : 0 <* x < 1} is empty. Let's stop right here and drill this down till this is clear to you. Do you see that with this particular <* it must be the case that T is empty?

1 hour ago, discountbrains said:

I did no such thing!

It's really tedious to work this way, I'm not going to mark up your image post like this again. Please write out your argument line by line in a way that can be quoted in the forum. 

It says, plain as day what T is, and then you say a, x, b ∈ T.

Of course x is a variable used to define a set, it's not an element of anything in this context. 

The larger point here is that you are NOT FREE to choose a and b. If you claim <* is not a well-order, you have to make your argument for arbitrary a and b. But you can't possibly do that since "a <* b <* everything else" always results in a T that is empty.

BTW you can use a site like https://math.typeit.org/ for math markup.

We're now at the point where you have made the same faulty argument repeatedly for weeks, and I keep giving the same counterexample. But now you're posting images that are labor-intensive to quote, and you don't seem to be proofreading your own work. We're going backward.