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

## Recommended Posts

Whatever it is you're trying to claim, you have to prove it. You can't handwave it with "this should be allowed in our meta math system."

##### Share on other sites

f by "might", you mean it happens in some orders, then yes, I agree. But you seem to be claiming this must be true for all orders, and that's something you'll have to prove (in fact, it's false), so no, I do not agree.

Don't think I have a problem with this.

##### Share on other sites

So what is your current proof, then?

##### Share on other sites
On 8/22/2019 at 4:01 PM, uncool said:
On 8/22/2019 at 4:01 PM, uncool said:

"As we might always have an x such that x < z for any z we should have an x <* z for any z" That is if this z is supposed to be a min for a set.

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 <*.

And, what do u mean by z1 and z2? Is z1 <* z2? Once you identify these numbers  how do find the min of the set with these deleted?

You keep saying I have proved nothing. Just what is wrong with my latest attempts? You never have said exactly what is wrong. It sounds like you already have a notion of what a proof-if possible-should look like.

##### Share on other sites

Please try to correct your quoting; you have put your response as if I said it. I know, the quoting software for this forum is awful, but it's hard to tell what you are responding to.

19 minutes ago, discountbrains said:

On 2nd thought I might have to prove denseness  in the reals for <*.

If you want to claim it, then yes, you will have to prove it. And if by "denseness in the reals", you mean "between any two reals (as defined by any order <*), there is another real", then you will be trying to prove a false statement.

19 minutes ago, discountbrains said:

Just what is wrong with my latest attempts?

I have explained what I thought is wrong rather explicitly each time. If you want me to do so again, then post a proof - in full, no handwaving - and I will see.

Edited by uncool

##### Share on other sites
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 $S = \{x : 14 \leq x\}$. 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 $<^*$ 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 $\omega$ 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.

Edited by wtf

##### Share on other sites

I'm sorry I'm wasting your time by my dumbness. I have always been thinking of a subset of for example S=[0,1] (with the usual order, <) which would be an interval or segment from [0,1]. Surely, if one considers {1/2, 1/3, 1/4,...} and they delete 1/2 there is a next number etc so it too a well ordered subset of S. This type of set I'm not thinking of Should I say "start with a set for which there is not already a known well order for?" If we use (0,1) then clearly < doesn't well order it. What order relation will? Here's the way I visualize it: What if you have a set you have no WO for and you pick some number, z, out of it and claim z is a <*min for it then you delete z. You then are left with all numbers from S greater than z with respect to <*. What is your <* min for this set.

I am well aware of what wtf says. This is what motivated my whole notion, there is no WO for all sets.

Here's my order which should better be put in some sort of an array: Lets use numbers in [0,1] we have 0, 1, 2, 3,..., 1/2, 1/3, ..., 2/3, 3/4,...,whatever,.... irrational numbers,..., then you get to 0.0023...,...0.000042... You just get to numbers you can't find the least of or they become incomparable. You finally end up with sets of numbers that you either can't find the smallest or you don't know which is bigger. There are several unusual ways of ordering numbers, but they always end up like this. As the sets of these groups of numbers go along it gets so you can't find a min.

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?

Correction last line first paragraph:  Let this 'set' be S than the statement is correct.

##### Share on other sites
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.

Edited by wtf

##### Share on other sites

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.

Edited by discountbrains
midding text

##### Share on other sites
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.

Edited by wtf

##### Share on other sites

Those are not counterexamples to my sets. I'll reread what u said; maybe I missed something. It looks like the same obvious old thing over and over again.

##### Share on other sites
1 minute ago, discountbrains said:

Those are not counterexamples to my sets. I'll reread what u said; maybe I missed something. It looks like the same obvious old thing over and over again.

I agree with that. You keep making the same logic error, I keep correcting it.

But what if we run the argument on the natural numbers in their usual order? Doesn't every deleted tail still have a smallest element? And where's your proof that this can't happen with an arbitrary linear order on the reals?

##### Share on other sites

Is this your counterexample? Let me say the set for consideration is the usual (0,1). My <* applies to this set and for more numbers outside this set if someone want's to use them in an argument. You can't be bringing up your special set of 1, 2, 3,... I really, really, really don't think u understand this subject. My <* applies to each and every number in (0,1). If we delete a number from (0,1) we still have a very large  uncountable number of elements. Maybe uncool can explain your error to you.... I keep telling u  the only thing of concern is finding ONE set u can't find a min of, geez! ....No one answered my opening question. I'll look up various real math societies or maybe see if I can get a real prof to answer me. These groups have got to be some sort of scam.

##### Share on other sites

LOLOL All the best.

But since you asked I'll explain. The 1, 2, 3, ... example is ... an example. An illustration of how to conceptualize well-ordered sets. You remove the first element and there's another next element. You remove that one and there's another.

It's perfectly true that we can't visualize an uncountable well-ordered set. But its existence can be proved even without the axiom of choice. You have an intuition that there are "too many" reals to well-order, but you haven't got a proof.

Edited by wtf

##### Share on other sites

The wiki site is saying the ordinals are well ordered in respect to set inclusion. "ω1 is a well-ordered set, with set membership ("∈") serving as the order relation. ω1 is a limit ordinal, i.e. there is no ordinal α with α + 1 = ω1." Don't know quite what this means. I guess α + 1 is still a countable number or set and thus can't equal ω1. That these topologies are not metrizable I'm a bit curious about.

If I say A and B and C ¬D and you say D ⇒D you are inserting information which is not there.

You're saying a WO relation is a relation so I must automatically include it in my set of relations on my set S even though I don't know it exists. I can make all kinds of manipulations with order relations without even exhibiting them and the WO fails.

##### Share on other sites
38 minutes ago, discountbrains said:

You're saying ...

You're addressing me but you've indicated a disinclination to hear any more from me. I prefer not to play that game. For the record I'm done here unless something new and/or interesting gets said.

I'll leave it with this. If -- I'm not saying there is, but if -- there were a well-order on the reals, it would look just like the well-order on the natural numbers: one element after another. The only difference is that you'd have to periodically take limits.

A limit ordinal is an ordinal that doesn't have an immediate predecessor. $\omega$ is a limit ordinal. It's the upward limit of 0, 1, 2, 3, ... Technically it's implemented as the set-theoretic union of all the preceding ordinals. Alternatively, it's the set of all the preceding ordinals. Those two are the same since ordinals are transitive sets.

So if you take all the countable ordinals, and take their upward limit (as their union, or by taking the set of all the countable ordinals) you have an ordinal that can't be one of the countable ordinals, so it's an uncountable ordinal.

I completely agree with you that the idea of an uncountable well-ordered set is mind boggling. But that's not a proof against it. Rather, it's another one of those counterintuitive things in math that we have to just "get used to," as John von Neumann said.

You have an intuition that there's no uncountable well-ordered set, and you think your intuition is a proof. But you haven't got a proof. Only an intuition which turns out to be false.

Edited by wtf

##### Share on other sites

You are saying I am making statements about ordinals. They never crossed my mind. After some thought I could be wrong trying to prove a negative. Empirically it often is impossible  at least: Many public figures concluded there were no wmd's in Iraq because they never found any. Logically its wrong to say this-not taking sides though. Was trying to find an analogy of  my type of statement that no WO existed for R. I know of a theorem due to Von Neumann who you mentioned that states "No two variables can be maximized at the same time". So, your equivalent response to him would be "that YOU know of a function that does this". So, let me end by changing my statement to 'the rules on order relations on the reals lead to inconsistent  results with the notion there exists a well ordering' for the reals. This would make my statement similar to some others in set theory, symbolic logic etc.

My source of much of my knowledge of set theory is a book-long out of print-called Axiomatic Set Theory by Rubin and Rubin. In the book they discussed what you were saying about ZFC, ZF, and whatever models of set theory. They used a lot of symbolic logic notation and got a little into meta mathematics. I think my statements are a little on the meta math side.

##### Share on other sites
43 minutes ago, discountbrains said:

So, let me end by changing my statement to 'the rules on order relations on the reals lead to inconsistent  results with the notion there exists a well ordering' for the reals.

What, precisely, do you mean by that? Do you mean that there are statements about the usual order relation on the reals that cannot apply to a well-ordering?

##### Share on other sites
4 hours ago, discountbrains said:

You are saying I am making statements about ordinals. They never crossed my mind.

That's like me saying I'm interested in a scoop of tuna salad between two slices of bread, but that I am NOT interested in a tuna salad sandwich! The two things are synonymous. If you have a well-ordered set, its order type is an ordinal. If you have an ordinal, it's a well-ordered set. It's not possible to be interested in well-orders and not be interested in ordinals. I can't fathom where you're coming from with a statement like this, after I've been explaining it to you for a year

The rest of your post is wildly off the mark. You keep posting an erroneous proof and I keep refuting it with an example. You say that if you delete the first element from the tail of a linear order (of the reals), the rest of the set has no smallest element. But that's false. "0, 1, everything" refutes it. You say ok take the next deleted tail. "0, 1, 2, everything" refutes it. You say ok do it countably many times. Then "1, 2, 3, 4, ... 0, everything" refutes it. That's the ordinal $\omega + 1$. You can keep going to "odds, evens, everything." That's $\omega + \omega$. I can just keep walking through the countable ordinals. You claim a deleted tail has no first element but clearly these examples refute your claim. That's all that's going on here. You have a faulty proof and I keep showing it's faulty.

Edited by wtf

##### Share on other sites
16 hours ago, uncool said:

What, precisely, do you mean by that? Do you mean that there are statements about the usual order relation on the reals that cannot apply to a well-ordering?

No

##### Share on other sites

Then I ask again: what, precisely, do you mean?

##### Share on other sites
12 hours ago, wtf said:

That's ω+ω . I can just keep walking through the countable ordinals. You claim a deleted tail has no first element but clearly these examples refute your claim. That's all that's going on here. You have a faulty proof and I keep showing it's faulty.

Don't know how to answer u. Your position is there is an established well accepted theory of ordinals that contradicts my claims. My argument stands on itself; everything follows logically from each proceeding statement in the argument. It is what it is without your fancy argument. If u didn't already had this bag of tricks back in your 1st few responses to me why didn't u use them then? You keep saying "everything else". How are u going to find a min let alone a least element of that? If this is a well order for the reals than u have ACTUALLY exhibited one.

I will now give yet another argument that's probably non-rigorous nor intuitive:

If you reorder a set S=(0,1) you realize all numbers in S are infinite strings of digits (countable number of digits). Some end in an infinite string of 0s of course. A string might look like 0.045xxxx0....xx. To produce ANY reordering in the most general way so no one can say you skipped any all digits need to be replaced by another not = to it or of course some or all digits could stay the same. Now consider any possible reprdering of these numbers. You would have to rearrange the numbers one by one. Of course we know even if there were a countable number of arrangements we couldn't do this, but theoretically its conceivable. We know, however, there is an uncountable number of ways to do this. We need the Axiom of Choice to do this. We can't even produce all reorderings without the AC. We can make a few reorders like reversing the order of all numbers etc.

10 minutes ago, uncool said:

Then I ask again: what, precisely, do you mean?

I was trying to model my logic after this Goerdel, Continuum Hypothesis stuff. I forget how all that stuff comes out. I mean that if u follow the basic rules of order relations and basic definitions of sets without the concept of WO you must conclude there is a conflict here if you start with the premise every set can be well ordered.

##### Share on other sites

The conclusion with the continuum hypothesis is that it was independent of the other axioms. That is, that ZFC + CH is consistent, and ZFC + ~CH is consistent. The closest analog here would be that ZFC + AoC is consistent, and ZFC + ~AoC is consistent. Which is true, but doesn't match what you've claimed - your claims seem to say that there is an inconsistency.

So yet again, I have to ask: what, precisely, do you mean?

Edited by uncool

##### Share on other sites

"So yet again, I have to ask: what, precisely, do you mean?"

An order relation must satisfy 3 properties. Sets and manipulations of sets are understood to take a certain form and meaning. I only used these basic ideas and ended up with my result. That's all I'm reduced by critics to saying at this point. Your question to me makes me ask just precisely is your issue with my train of thought? Point to exactly what I said that u can say is not true and correct me.

Wtf's examples seem vague to me; might actually be more imprecise than he accuses me of. My sense now of ordinals is that its almost obvious they are well ordered.

How did u like my ordering of numbers in (0,1)?

##### Share on other sites
1 hour ago, discountbrains said:

I only used these basic ideas and ended up with my result.

And what, precisely, is the claimed result?

1 hour ago, discountbrains said:

Your question to me makes me ask just precisely is your issue with my train of thought?

You have had several trains of thought, and I have pointed out specific issues in each of them. My main issues are meta-issues: you are often unwilling to deal with questions with the necessary precision, and you are often too willing to handwave the key parts in your claimed proofs. Both of which are why wtf and I are constantly asking you to write out your precise proof.

1 hour ago, discountbrains said:

My sense now of ordinals is that its almost obvious they are well ordered.

That sense is correct; they are defined to be well-ordered.

1 hour ago, discountbrains said:

How did u like my ordering of numbers in (0,1)?

I don't know which ordering you are referring to; however, if you are still claiming that there cannot be a well-order on the real numbers, then telling us that a specific ordering isn't a well-ordering is not informative.

A request for you: what is the precise statement you think you have proven, and what is the precise proof of that statement?

##### Share on other sites
This topic is now closed to further replies.

×