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discountbrains

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

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31 minutes ago, discountbrains said:

 

I would like someone to focus more directly on my original claim at the start and show its wrong.

I've already done that several times over. If you have a new idea, you haven't expressed it yet. 

Also, nobody is "accusing" anyone of anything. I've studied some math and you're interested in the well-ordering theorem. I'm happy to share what I know with you, and also use my mathematical training to point out logical errors in your reasoning. It's been a long time in this thread since you've posted any mathematical content. Perhaps you could just restate your argument clearly and we can go over it. 

Edited by wtf

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1 hour ago, uncool said:

I assume this is the post?

You incorrectly attribute it to wtf. But wtf took it as a quote from a quora post by the OP. 

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NO, you are WRONG. I specifically chose sets that are dense. Of course, the natural numbers can be WO. You are making the same mistake as wtf did. You don't understand the problem. The WO theorem states there is at least one ordering that well orders the reals. That is there is some <* that can identify a l.e. for ANY subset of R. Of course, this is not the same number for all subsets. I don't think I saw any explanation from wtf showing me my errors. He at first kept referring to the AC. If u leave the AC out you have nothing.

I question the logic of my own assertion: That is, if u assume u have a method of choosing a l.e. for any subset of R and exhibit a set with its l.e. another subset of this set can naturally be generated for which a l.e. can't be found using your well ordering <* method. This doesn't say the same specific l.e. works for both. My logic is like Wack-a-Mole. I'll call this my "Wack-a-Mole" proof. 

This is an answer to someone new.  I'm not seeing their post right now.

 

Edited by discountbrains

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2 hours ago, discountbrains said:

NO, you are WRONG. I specifically chose sets that are dense. Of course, the natural numbers can be WO. You are making the same mistake as wtf did. You don't understand the problem. The WO theorem states there is at least one ordering that well orders the reals. That is there is some <* that can identify a l.e. for ANY subset of R.

Any nonempty subset.

2 hours ago, discountbrains said:

Of course, this is not the same number for all subsets. I don't think I saw any explanation from wtf showing me my errors. He at first kept referring to the AC. If u leave the AC out you have nothing.

If you leave out AC it is undecidable if a WO of the reals exists.

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2 hours ago, discountbrains said:

NO, you are WRONG. I specifically chose sets that are dense. Of course, the natural numbers can be WO.

Yes, that's my point. What step in your proof fails if you try to apply it to the natural numbers instead of the reals? Where do you use "denseness" in your proof?

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2 hours ago, discountbrains said:

I question the logic of my own assertion: That is, if u assume u have a method of choosing a l.e. for any subset of R and exhibit a set with its l.e. another subset of this set can naturally be generated for which a l.e. can't be found using your well ordering <* method.

Why would you need a "method" of choosing a least element? There is a least element or there is not. If there is one, there is no "choice", since it is already there and it is unique.

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I keep saying over and over again all I need to do to show there is no WO is to exhibit a set for any possible WO you choose  is not well ordered. I need only choose the easist example. I decided that if there exists a well ordering for the reals its really a philosophical question like God exists. To construct a WO is impossible. This is why I said 'a method'. OK, I'll go over my thinking again since I'm getting so much push back.

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3 hours ago, discountbrains said:

 

I keep saying over and over again all I need to do to show there is no WO is to exhibit a set for any possible WO you choose  is not well ordered.

 

Yes; more precisely, once you are given the well-ordering, all you need is a way to produce a set with no minimum. I think your error is "proving" that you have found such a set; I think your proof is fallacious there.

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I  made a mistake. I found my error yesterday. I simply asserted or assumed a set with the form [a, b) with respect to the well ordering <* exists. I believe I need to prove it exists. Otherwise, all subsets with the WO would look like {a1, a2, a3,...} and if an<*an+1 there is no x such that  an<*x<*an+1 for any subset of R. I will try to prove there are.

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I will now prove there is a dense nonempty subset, (S, <*), of that can't be well ordered by <* no matter what <* is. 

Proof: If S is well ordered every subset of it has a minimum element. Suppose for S itself its 'a' and we can form {a}\S which also has to have a minimum element let it be 'b'. We could go on deleting number from each new subset and of course this would result in the discrete string of numbers mentioned several times above. thus a <* b and there cannot be any x with a <* x <*b. We can write a +a <* a + b and get 2a <* a + b or a <* (a + b)/2. It must be that b <* (a+b)/2 because (a + b)/2 can't be between a and b. Thus, 2b <* a+b which implies b <* a and this is a contradiction (the rules for addition and whatnot should be valid for <* because this is the supposed well ordering that theorems based on well ordering also need). Sets containing such exhibited numbers certainly do exist and are subsets of R and their elements can't be arranged as a WO would. So,   can't be well ordered.

I believe I can also write out a proof following very similarly the Cantor method for his uncountable sets.

 

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1 hour ago, discountbrains said:

thus a <* b and there cannot be any x with a <* x <*b

You need to be careful here; there is no such x in S, but there could be in the real numbers. Unless you are choosing S to be the set of all real numbers?

1 hour ago, discountbrains said:

We can write a +a <* a + b

Why? Who said that this ordering respects addition?

 

1 hour ago, discountbrains said:

(the rules for addition and whatnot should be valid for <* because this is the supposed well ordering that theorems based on well ordering also need)

What? Please explain this more fully. 

Edited by uncool

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As for the 2nd question I pondered this for a while myself. I think you can perform these +, -, etc operations independently of the order relation just like you do in balancing both sides of an equation to maintain equality. I can't give a specific theorem that's dependent on WO, but I suspect some <* is assumed and then further on something is manipulated with the ordinary + and - as if it doesn't matter.

I believe I meant x is in S. I'll have to be more clear about that for myself. Some ideas I've played with end up with an element that's not even in the set in question.

Edited by discountbrains
deleted extra word

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11 hours ago, discountbrains said:

As for the 2nd question I pondered this for a while myself. I think you can perform these +, -, etc operations independently of the order relation just like you do in balancing both sides of an equation to maintain equality.

The statement won't be true for all well-orderings, which is what you'd have to prove. There is no connection between the additive structure and the well-ordering on which to base that statement.

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18 hours ago, discountbrains said:

As for the 2nd question I pondered this for a while myself. I think you can perform these +, -, etc operations independently of the order relation just like you do in balancing both sides of an equation to maintain equality.

Let \(\mathbb N \) be ordered as 0, 1, 2, 5, 4, 3, 6, 7, 8, 9, ... Call that order \(<^* \).

Now \(2 <^* 3 \) is true, but \(2 + 2 <^* 2 + 3\) is false where \(+\) is the usual addition. In order for addition to be compatible with \(<^*\) you need to redefine it as \(+^*\) with new rules to account for the reordering.

 I don't think I saw any explanation from wtf showing me my errors.

Perhaps you should read my posts. I've explained your errors multiple times. 

Edited by wtf

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I believe I want to state my claim in a different way:  Let S be the subset above containing a and b with a<* b. Consider (a + b)/2 which is in R. If  a <* (a + b)/2  and (a + b)/2 is in S then b <* (a + b)/2 which leads to b <* a and a <* b as above. But,  as you said this arithmetic may not work with <*. I'm thinking now its really independent of any ordering.  Anyway, I can prove my original claim using Cantor's method. We note he only really generated one new number, but that's not the point. The assumption is this was a list of ALL numbers.  In my case the new generated number can't be compared to any number in the list because its not in the list.

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I'm afraid you're not being clear here. Do you still think this is accurate, or do you accept that arithmetic may not respect an arbitrary inequality?

17 minutes ago, discountbrains said:

If  a <* (a + b)/2  and (a + b)/2 is in S then b <* (a + b)/2 which leads to b <* a and a <* b as above. 

And what is your full argument in the rest of your post?

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And what is your full argument in the rest of your post?

I agree with your latest post which I can't find. For example, 8+5 might = 2 or 15 in an ordering other than <. But, then I started thinking in this case how could 2(a + b)/2 not = a + b and 2b not = b + b etc. So, for the rest of my argument assuming any x in S with a <* x,  we have a <*b <* x which implies  a<* b and b <* a which is a contradiction. Hence, there could be x's between a and b with a WO which contradicts what a WO should look like. I may not be completely through though. I need to prove there exists an x between a and b. I started thinking I could prove my original assertion another way. I'll work on this some more.

What I just said above is anytime x=(a + b)/2 its between a and b; otherwise, of course, if x is in S x could be greater than b.

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1 hour ago, discountbrains said:

 

I agree with your latest post which I can't find. For example, 8+5 might = 2 or 15 in an ordering other than <.

No, that's false. 8 + 5 is always 13 in any order, as long as the symbol '+' has not been redefined. 

You could have an alternate definition of +, which you might call \(+^*\), but that would not be the same as the usual +. 

I hope you can see that just as you are always careful to distinguish \(<\) from  \(<^*\), you have to be just as careful to notationally distinguish alternative definitions of addition from the standard one. 

how could 2(a + b)/2 not = a + b 

Of course 2(a + b)/2 = a + b always. But \((a + b) / 2\) may not be between a and b if + retains its usual meaning and the ordering is nonstandard.

Edited by wtf

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I was doubting myself thinking (a + b)/2 might not be in the same set that has a as its minimum. But, we must consider all subsets of and certainly such sets exist and the WO  <* must work for all sets. "But (a+b)/2 may not be between a and b if + retains its usual meaning and the ordering is nonstandard."  I'll have to think about that. If 'its always greater than b' creates a contradiction and its not between a and b where is it? It has to be one or the other since <* is a linear order relation on R. Otherwise could it be its always less than a and never in S.

Edited by discountbrains

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17 minutes ago, discountbrains said:

I was doubting myself thinking (a + b)/2 might not be in the same set that has a as its minimum. But, we must consider all subsets of and certainly such sets exist and the WO  <* must work for all sets. "But (a+b)/2 may not be between a and b if + retains its usual meaning and the ordering is nonstandard." - For <* to WO  the reals (a + b)/2 must not be between and b which leads  to a contradiction.

I don't understand what you're trying to say. Consider the following well-order of the positive integers:

1, 3, 2, 4, 5, 6, 7, 8, ... 

Now 2 <* 4 but (2 + 4)/2 is not between 2 and 4. What contradiction is there? All we've shown is that the usual addition is not compatible with the nonstandard order. And why should it be? After all, addition is based on the standard order.

Edited by wtf

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1, 3, 2, 4, 5, 6, 7, 8, ... 

Now 2 <* 4 but (2 + 4)/2 is not between 2 and 4.

Any set of positive integers can be ordered any way you choose to get the results you're looking for. This in no way represents all sets that have to be considered. All one needs to do is exhibit just one and subsets of with the right properties certainly do exist.

 

 

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10 minutes ago, discountbrains said:

1, 3, 2, 4, 5, 6, 7, 8, ... 

Now 2 <* 4 but (2 + 4)/2 is not between 2 and 4.

Any set of positive integers can be ordered any way you choose to get the results you're looking for. This in no way represents all sets that have to be considered. All one needs to do is exhibit just one and subsets of with the right properties certainly do exist.

 

 

> Any set of positive integers can be ordered any way you choose to get the results you're looking for.

Exactly. Which is why I'm puzzled by your arguing in terms of the compatibility between standard addition and a nonstandard order. They are not compatible and we can always find such counterexamples. So why are you trying to reason this way? It's not productive and I think you can see that.

> This in no way represents all sets that have to be considered. All one needs to do is exhibit just one and subsets of with the right properties certainly do exist.

Can you exhibit such a subset? You haven't so far. 

Edited by wtf

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I was going to say "thank you" to your post about 3 back because u said the arithmetic worked the same no matter what order  one uses. Then I saw the rest of your post. I'm working on another proof of my claim. I think there may be several possible proofs. I'll continue also trying to present my proof more formally.

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1, 3, 2, 4, 5, 6, 7, 8, ...    Good example, (2 +4)/2 <* 2. In this set 2 is the min, therefore (2 + 4)/2 =3 and is not in the set. (1 + 3)/2 = 2 so its in the set with min=1. 

All I really need to do is consider the set (a,b) and let x=(a+b)/2 with the usual order <. Therefore x is definitely in (a,b) and hence x is in ((a,b), <*). So, I don't have to say If x is in a set with some minimum with respect to <*  which is also in (a,b)-ie, min(a,b) and x are both in (a,b). 

You gave an example of where my claim doesn't work, but I presented a specific kind of subset. 

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On 11/7/2018 at 8:36 AM, discountbrains said:

1, 3, 2, 4, 5, 6, 7, 8, ...    Good example, (2 +4)/2 <* 2. In this set 2 is the min, therefore (2 + 4)/2 =3 and is not in the set. (1 + 3)/2 = 2 so its in the set with min=1. 

All I really need to do is consider the set (a,b) and let x=(a+b)/2 with the usual order <. Therefore x is definitely in (a,b) and hence x is in ((a,b), <*). So, I don't have to say If x is in a set with some minimum with respect to <*  which is also in (a,b)-ie, min(a,b) and x are both in (a,b). 

You gave an example of where my claim doesn't work, but I presented a specific kind of subset. 

1, 3, 2, 4, 5, 6, 7, 8, ...    Good example, (2 +4)/2 <* 2. In this set is the min

Could you please explain what "this set" refers to? I don't know what you mean. And 2 is not the min of the ordered set <3, 2, 4>, 3 is.

I'm going to notate ordered sets differently than sets. The set {1,2,3} has no order on it and none should ever be inferred by how you list them. The curly braces say, "Unordered set." That's the convention.

You have a couple of uses for parens and that's a little confusing. So I'm going to denote an ORDERED SET as <2, 1, 3> for example, as a shorthand for the set {1,2,3} with the order 2 <* 1 <* 3.

> therefore (2 + 4)/2 =3 and is not in the set.

What? 3 is not in WHAT SET?

> (1 + 3)/2 = 2 so its in the set with min=1. 

WHAT SET? 

 

> Therefore x is definitely in (a,b) and hence x is in ((a,b), <*).

But NO. This is as false as false can be. I assume by (a,b) you mean the INTERVAL of elements strictly greater than a and strictly less than b. That's what the notation means. If you want to include a and b, write [a,b].

Now in the usual order <1,2,3>, 2 ∈ (1,3).

But in the order <1, 3, 2>, 2 ∉ (1,3). 

That is the counterexample to what you wrote.

Edited by wtf

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