discountbrains

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.

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.

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 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.

On 8/10/2019 at 4:24 PM, wtf said:

I found a mistake. I found the same mistake multiple times over the past several months that this thread has been going on.

Your claim is that there is NO well order of the reals. Your proof consists of picking SOME order and showing it's not a well-order.

But you have to prove that EVERY order is not a well-order. So you have to consider the case where T might be empty. And since you never consider that case -- indeed, since you give no appearance of even understanding the point -- you have not proven your claim.

BTW pasting an image makes it impossible for me to reply line by line. But your approach is wrong. You have to prove that <* can NEVER be a well-order. You can't cherry-pick some <* where T is nonempty. Because there are perfectly obvious <* orders, and choices of a and b, for which T is empty. Your proof fails in your second sentence.

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?

2 hours ago, wtf 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?

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.

 

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?

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. 

Three times u redefined my set, T....... OK, got it. I will need to take a look at your example of <*. Maybe I can find a way around it. Wait a minute, who said the "everything else" is in the usual order? I really think I need not even require x<*b in T. All I'm concerned about is the lower bound or the min .

19 hours ago, wtf said:

ps -- You define T = {x : a <* x <* b} and then in your next sentence you claim a and b are elements of T. That's ridiculous.

I did no such thing!

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.

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

Here it is. If you find any little mistakes let me know. For reasons I gave above how can we imagine the only sets that satisfy my definition above of T have to be empty.

 z₀  z₁ z₂ 

 OK, here's my proof for your perusal:   z₀  z₁ z₂ 

The proof is by contradiction. Let S =(a,b) where a,b∈ R, 0< a <b. Let T= {x: a<*x<*b} where <* is a total order relation other than, <. I define S this way so all z∈S are in T. S is an denumerable set so T is also. If <* well orders T then ∃z∈T such that z₀ ≤*x ∀x∈T. It must be that a<*z₀  and {x: a<*x <*z₀ } is an empty set

z₀ ≤*x ∀x∈T T..... I'm going to have to write this out on paper and scan it as an image. This is just too hard. I was nearing the end and it erased my work. I am proving by induction we can generate a denumerable string of minimums of every subset of T arising from after the previous minimum was deleted. This is a denumerable set and actually contains all elements of T. If we choose a z in T it is the min of any subset of T of numbers greater or equal to it. Therefore its also a member of a string of minimums of T. So, T is = to this denumerable string of minimums. T thus is denumerable. Contradiction.

Using the induction theorem might be a little tricky for a lot of proofs. One has to be careful. 

Here's what I think should be sufficient to show your example doesn't fit my set definition: Lets pick any arbitrary z in your 'E, everything else'.  So, there could be some x<*z in E and clearly there are x's in E such that z<*x and we generate the set T={x: z<*x}. You must have a min, m, in T such that m <* or = all x in T and it has to be that z<*m and a set {x: z<*x<*m} must be an empty set (you maintain). Keep in mind T is all x in S such that z<*x. S might be R or (a,b)-with usual order-a subset of R, a and b are real numbers....... I think I know what you are saying. That is, how do I know my set (S,<*) can be constructed for any order relation <* or that {x: z<*x<*m} has to contain numbers? I can still build such sets from your 'everything else'. This has always been my premise from the beginning that for any possible order relation sets of the above form can always be constructed.

I need to look up and see if the union of any collection of any countable sets is still countable and maybe use this result to help prove my claim.

I keep telling u over and over that the set I'm presenting is a set that cannot be WO. You keep presenting a different set. This is not my set.

Do u think I can use the induction theorem implies your "everything else" does not include all x in S? I think I can; I'm putting my argument together to post.

Yes, I quoted you, but deleted your quote and then gave my response which made my reply confusing. 

What you are saying then is that you already assume S is WO and hence u can get the result it is WO. I'll have to think about your claim that the set we are arguing about is empty. If u can give an example set where u say its empty then u can deny my claim. you are giving an example of a set you concocted; I'm presenting a different set.

Yes, that's what I have read about ordinals before: that each number is really represents the set of its predecessors. What puzzles me is what is the purpose of all this.

Anyway, I'll think about all this.

On 7/19/2019 at 4:34 PM, wtf said:

I agree with what you say. Defining S=(0,1) is not needed or helpful. Let S be a subset of R defined as S={x: a<*x<*b} . Both of you here agree S exists and is nonempty. Here we have a is the greatest lower bound of S with respect to <*. You say all subsets of R must have a minimum, M. Now M is in S and a<*M. Therefore the set {x: a<*x<*M} is empty. There can be no elements between a and M. Carrying this further this implies there are sequences in S, a<*b<*c<*... which have no numbers x, y, z,...such that a<*x<*b<*y<*c<*... This doesn't really constitute a proof of my assertion at the start above, though. Whether we believe math is discovered or invented do we want to restrict our notion of it like this?

You talked about ω + ω. This is the problem I'm having with this ordinals: How do you determine when one sequence ends and the next begins?

OK, what I said is not acceptable. There other ways to go about this like u said. 

What I said later about the functions is simply how derivatives have been calculated for centuries-nothing new here. It is interesting though to think of a function expanding or contracting instead of a specific value  y=f(x).

Edit: I might say (S,<*) is S with a and b removed and = {x: a<*x<*b}. 

On 7/10/2019 at 1:41 PM, wtf said:

Then {x : 0 <* x <* 1} is no longer of interest in that notation, is that correct? What exactly is (S,<*)?

Yes, I am just getting back to this. What I have done is use 'S" for both sets and defining (S,<*) like I did I actually made a and b the min and max of (S,<*). I don't want to do this of course. There are other straight forward ways of defining it. 

I'm thinking of moving on. Because as I realized long ago if you try to reorder a line segment [0,1] on the x axis you end up with a plane [0,1]x[0,1] of an infinite number of points in XxY or if you draw a squiggly curve you pass through the same numbers repeatedly which is not an order relation. So, what value is this anyway? This gave me another thought which is by drawing a curve f(x) on the y axis of x and if d(x_n,x_m) is a metric you would get d(f(x_n), f(x_m)) being greater or less than  d(x_n,x_m). This might lead to another way to finding the max or min of a curve. That is where d(f(x_n), f(x_m))=0 with  d(x_n,x_m)  not =0 for some m and n. I wonder if this has ever been looked at.

Whoops, I hoped I would correct my mistake before someone else did. L might say S=(0,1) with usual order and (S,<*)={x: a<*x<*b where a,b ∈ (0,1) so I don't restrict myself too much where a counter example can be made. Lets suppose you exhibit an ordering that produces a min for my set I believe I can show it doesn't work for all sets. This may lead to a new theorem that for any set a ordering exists that produces a min for that set, but it won't for all sets. This is leading to a lot to think about.

"

Why do you think 0 is the greatest lower bound? It's clear that 1/2 is a lower bound for S and is greater than 0. So 1/2 is the greatest lower bound of the set S. It's 

a) A lower bound for S; and

b) Greater than every other lower bound for S"

But, I stated at the outset S = (0,1) with the usual order. (S, <*) is strictly made up of numbers from S. 1000, -5/2, etc are not in S.

23 hours ago, uncool said:

Why?

Surely u know why. Maybe you're looking for more precision. Precision is required in math. I believe all I need to do is state <* is just like the usual order, <, only its a different ordering.

OK, how are arbitrary orders defined? I need to know this. Is wtf's ordering above acceptable? That is, 1, 1/3, 1/5,...1/2, 1/4, ... etc I've always wondered how u get from numbers in the first sequence to the next sequence. Might this be in the study of ordinals? In a case like his sequences the order might be as f(n), f(n+1), ...

Here's an interesting order: How do they determine all the barcodes, UPC codes, for products in stores like Walmart? Some items in ones home might have to be partially ordered.

wtf made an error-a fatal error. He supposes 1/2 is the min of S. To be such 1/2 has to be in S ⇒ 0 <*1/2 <*x for all other x in S. Since 0 is the greatest lower bound of S, 1/2 is no lower bound and hence not a min of S.

To uncool I'll have to think about if the way I defined S is sufficient or do I need all this other stuff which is unconventional and needs some cleaning up.

You meant S={x: 0<*x<*1} right? 0 is the greatest lower bound and 0 is not in S. There is no <* lower bound in S.  To answer wtf if 1/2 is the min of S then 1/2 <* x for all x in S not = to 1/2. I never really wanted to do this, but lets go back to the very primitive notion of arithmetic-back to ancient Egypt or even the cave man. He might think the fish he caught is longer than all the others. He has to cut some notches in a stick to measure and compare each fish. We can order things anyway we like, but we are using math when we assign numbers to, lets say, a and b. We say a<*b if there is some way to get from a to be. We don't care how. We say a<*b iff we make a symbol, +*, such that a+*σ=b. We can continue with this idea and say σ=q+*q. It follows that a+*q <*b. So if 1/2 is in S we can write there is a q such that 0<*0+*q<*1/2 and of course 0+*q is in S by the definition of S. Therefore 1/2 can't be the min of S.

σ is a number -σ is not defined for this. Lets say σ >*0 and σ > 0.

47 minutes ago, wtf said:

51 minutes ago, wtf said:

 

"You'd do well to study the ordinals. I suggested that a couple of months ago. I reiterate the suggestion. Your m, n, o argument shows that you don't understand the ordinals. If you spent a day reading up on the ordinals all this would become clear to you."

 

I tried to discard everything I said there.

 

 

14 hours ago, uncool said:

Why not?

Then prove it.

This was my original concept. The idea is given any order relation a set, S, can be described containing a number, a, such that for any number you choose, M, in the set not equal a,  a<*M. The opposite notion to this is given any subset of R you can describe an order relation that produces a minimum in this set. The problem is the order relation might only produce this for this set. I know what you are saying. That I must produce every possible ordering <* and prove it doesn't produce a minimum M for S. I believe all I need to do is state what seems obvious that a set {x: 0<*x<*1} exists for any <*. Set theory allows us to produce any set we can think of.

On 7/5/2019 at 2:31 PM, uncool said:

 

Your attempted proof seems to come down to the claim that any infinite well-ordering is isomorphic to the usual ordering on the positive integers. That is, if a well-ordering is infinite, then it looks like m<n<o<..., and every element must be represented in that sequence. Correct?

I keep getting myself sidetracked. I'll go back to my very first original premise. That is, let (S,<*)={x: 0<*x<*1}. Such a set can be described and by its very nature has no minimum with respect to <*. I know  this is extremely simple and unsophisticated. You admitted such a set existed. This is true for any ordering <* of course.