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

## Recommended Posts

In reply to both of u please state specifically about what the point is in your questions. I think I may need to include a paragraph or two before my statement or axiom or whatever so u will know exactly my train of thought. It would actually be pretty hard to try to figure out what a definition or theorem means in a textbook without a background.

I'll give an example of a property P: Let x,y ∈ (0,1) with x=0.xxxxx..., y=0.yyy.... and if no digit of y is 5 and no digit of x is 3 or otherwise if x<y. Then we would say x<*y iff it has property P. Note, there is no claim here at all that <*  is or is not a WO.- I believe I've covered all cases here.

Written logically property P could be that 'A and B and C are true.' If any of these are untrue then we would have ~P.

22 minutes ago, uncool said:

I realized that you were trying to refute wtf's implicit claim; it didn't help me figure out how it was supposed to address it.

I guess my argument I kept repeating-to some people-doesn't mean the reals can't be WO. I concluded to back off this claim and simply say my perfectly derived results are inconsistent with the assertion there is a WO for the reals..... Now,I constructed an entirely new argument with the property P thing. So, I have actually 3 arguments for my claim. I think I have enough now to write a paper to submit to a journal. I could also include some of you people's responses.

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

I guess my argument I kept repeating-to some people-doesn't mean the reals can't be WO. I concluded to back off this claim and simply say my perfectly derived results are inconsistent with the assertion there is a WO for the reals

If by "perfectly derived results", you mean something you believe to be proven, then there is no difference between these two statements.

2 hours ago, discountbrains said:

Now,I constructed an entirely new argument with the property P thing.

As both wtf and I have said, what you have written does not reach the level of "argument", because it is incoherent.

3 hours ago, discountbrains said:

The "point" in all of my questions is to get you to fully explain an argument that you insist on handwaving.

It sounds from your post like "Property P" is merely another name for the relation x <* y, and correspondingly, the axiom you seem to claim is that there must be some set S such that for any x and y in S, "x <* y" is not true. If I have that wrong, then please try to write your axiom out explicitly

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

If by "perfectly derived results", you mean something you believe to be proven, then there is no difference between these two statements.

As both wtf and I have said, what you have written does not reach the level of "argument", because it is incoherent.

The "point" in all of my questions is to get you to fully explain an argument that you insist on handwaving.

It sounds from your post like "Property P" is merely another name for the relation x <* y, and correspondingly, the axiom you seem to claim is that there must be some set S such that for any x and y in S, "x <* y" is not true. If I have that wrong, then please try to write your axiom out explicitly

Allow me to examine the concern in your last paragraph more carefully. But, yes I do believe that's what I mean. I believe the example I wrote illustrates this.

23 hours ago, wtf said:

Your earlier arguments that there's no well-order of the reals are wrong. This paragraph is "not even wrong." It's incoherent. It says nothing and means nothing. Is Property P like Preparation H?

Maybe you've never seen this 'property P' idea. I had an instructor for 2 classes who I think used this at least twice. I don't recall seeing it in textbooks. I think I saw it in my Axiomatic Set Theory . I essentially said if we can identify a z in S, and then consider S\{z}. This set has no min for any order. That is with S containing ALL elements of the interval (0,1). You can't refute this! You know what I'm saying; both of u quit acting like u don't understand. This is not hard. My stuff is simple yet profound-very profound.

I don't know what else I can do for you 2 to make this more straightforward. You 2 keep saying its incoherent and badly stated to keep from admitting I'm right.

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

I essentially said if we can identify a z in S, and then consider S\{z}. This set has no min for any order. That is with S containing ALL elements of the interval (0,1). You can't refute this!

It's refuted by the order "1/2 <* 1/4 <* everything else." We've been over this many times already.

If you say, "Ok just apply the idea twice," I'll give you "1/2 <* 1/4 <* 1/8 <* everything else."

If you say, "Apply it countably many times" I'll give you $\omega + 1$. And I can keep on going like this right up to the first uncountable ordinal and beyond. You have NEVER responded satisfactorily to this refutation of your argument.

It's clear that whatever context your prof used this P idea, it wasn't this one.

Edited by wtf

##### Share on other sites

The use of 'property P' is to assign a property to or define the relation, <*, by giving it some arbitrary property, With x<*y both x and y have this property; this may also mean x has one characteristic any y does not. And, then we clearly see we can exhibit a set whose elements have none of these properties.

5 minutes ago, wtf said:

It's refuted by the order "1/2 <* 1/4 <* everything else." We've been over this many times already.

If you say, "Ok just apply the idea twice," I'll give you "1/2 <* 1/4 <* 1/8 <* everything else."

If you say, "Apply it countably many times" I'll give you $\omega + 1$. And I can keep on going like this right up to the first uncountable ordinal and beyond. You have NEVER responded satisfactorily to this refutation of your argument.

It's clear that whatever context your prof used this P idea, it wasn't this one.

AS I TOLD U MY LAST TIME, its reasonable to say that, but I said its not consistent with my derivation. You refute my derivation? Is it wrong? What does it imply?

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

I essentially said if we can identify a z in S, and then consider S\{z}. This set has no min for any order. That is with S containing ALL elements of the interval (0,1). You can't refute this!

We can and have, as wtf keeps pointing out. The order "1/2 <* 1/4 <* 1/8 <* ... <* everything else in the usual order" has 1/2 as a minimum; then S \ {1/2} has minimum 1/4; then S \ {1/2, 1/4} has minimum 1/8, etc.

20 minutes ago, discountbrains said:

AS I TOLD U MY LAST TIME, its reasonable to say that, but I said its not consistent with my derivation. You refute my derivation? Is it wrong? What does it imply?

Edited by uncool

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

seem to claim is that there must be some set S such that for any x and y in S, "x <* y" is not true. If I have that wrong, then please try to write your axiom out explicitly

Yes, I guess that should work. I'm trying my best as quickly as I can to thoroughly, thoroughly examine what I'm saying. If u can give an example where I'm saying nonsense please tell me.

2 minutes ago, uncool said:

The one u just quoted above you're question.

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

The one u just quoted above you're question.

Then the "derivation" is a bare assertion easily shown to be false by the example wtf has provided to you over and over and over again.

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

Then the "derivation" is a bare assertion easily shown to be false by the example wtf has provided to you over and over and over again.

The "bare assertion" must be wrong. Really? Wow! Whatever. No sense in repeating myself from above.

##### Share on other sites

There is no sense in repeating the bare assertion, no. If you want to convince us that your assertion is correct, you can write the proof to your assertion - without handwaving. If you want to show that it might be correct, you could demonstrate that wtf's order isn't a counterexample without repeating the assertion as if it were proven fact. But simply repeating it is a waste of your time and ours, yes.

Edited by uncool

##### Share on other sites
!

Moderator Note

I am closing this pending further review.

Edit: staff have agreed that this thread has well and truly run its course. Discountbrains, this topic of conversation is now closed and you are not permitted to reintroduce it.

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

×