Has the Completeness Axiom Been Disproved?

[mathbuster]

Has the completeness Axiom been disproved? see completenessaxiom.com .

 

[Hurkyl]

Congratulations, you have proven that second-order logic and integer arithmetic are inconsistent with each other! Assuming your proof is valid, of course.

I stopped looking at it once I saw the error in Theorem C Claim 1, where you confuse the notions of "supremum" and "maximum". In particular "Statement C" does not imply "Not Statement A".

Do you see why?

 

[mathbuster]

(Statement C) states: "For all elements b in E, b<M."
(Not Statement A) states: "For all elements b in E, b≤M."


If (Statement C) is true and (Not Statement A) is true, then "For all elements b in E, b<M" is true.


(I believe the above alteration to the proof of claim 1 of Thereom C fixes the problem. -You have sharper eyes than I do!)

 

[mathbuster]

As far as I can tell, the Completeness Axiom contradicts the Ordred Field Axioms.

 

[losiu99]

Theorem A, claim 2: If (Statement B) implies b ≤ r for fixed r and all b, then it's written in a wrong way. "For all b there exist r ..." means that r has such property for a given b, not for all b. It should be "There exist r, such that for all b...", but then (Not Statement B) is no longer negation of (Statement B).
(Statement B) as it is now doesn't imply that M is not a supremum.
M being supremum doesn't imply (Not Statement B) as it is now.

 

[mathbuster]

You have an excellent point!

Hows this?


Let E be a nonempty subset of the Real numbers and let E be bounded above and let M be a Real number.


(Statement B): For all elements b in E, there exists a Real number r such that M>r≥b.

Suppose that none of the real numbers r, which are specified in (Statement B), are upper bounds of E.

Then there exists a b in E such that b>r for all r in (Statement B). If not so then there must exist an r in (Statement B) which is an upper bound of E. However, this is a contradiction since, for all elements b in E, there exists a Real number r such that r≥b.

Then there must exist an r in (Statement B) which is an upper bound of E. This means that M cannot be the supremum of E since r is an upper bound of E which is less then M.

Therefore, if (Statement B) is true, then M is not the supremum of E.

 

[losiu99]

"Then there exists a b in E such that b>r for all r in (Statement B)."
For every r there exists such b, in general there is no universal b for all r.

 

[JThompson]

Which Ordered Fields axioms in particular do you think it contradicts?

 

[adriank]

The completeness axiom is clearly not inconsistent with ordered field axioms, since we have a model of a complete ordered field (such as the field consisting of equivalence classes of Cauchy sequences of rational numbers, which surely exists).

 

[Wizlem]

I believe the error occurs in the line "This would mean that there exists an element b in E such that for each Real number r, either r<b or r>b. In this case, b would not be a real number because for each Real number r, either r<b or r>b." Like losiu99 said, for every r you find can find a b special to that r so for each r you can always find a b that is either less than r or is greater than r. In essence, you can always find a b that is not equal to r. There is no contradiction.

 

[mathbuster]

"Then there exists a b in E such that b>r for all r in (Statement B)."
For every r there exists such b, in general there is no universal b for all r.

Suppose the set E:={1,2,3}. Suppose r=5 and M=6. Then for every specified 'r' there does not exist an element b in E such that b>r.

 

[mathbuster]

Which Ordered Fields axioms in particular do you think it contradicts?

I believe that the Ordered Field Axioms which are used to prove that if a<b, then there exits a real number c=(a+b)/2 such that a<c<b are the Axioms that cause the contradictions. There could be more.

 

[mathbuster]

The completeness axiom is clearly not inconsistent with ordered field axioms, since we have a model of a complete ordered field (such as the field consisting of equivalence classes of Cauchy sequences of rational numbers, which surely exists).

Proofs regarding Cauchy sequences rely on the Completeness Axiom.

 

[mathbuster]

I believe the error occurs in the line "This would mean that there exists an element b in E such that for each Real number r, either r<b or r>b. In this case, b would not be a real number because for each Real number r, either r<b or r>b." Like losiu99 said, for every r you find can find a b special to that r so for each r you can always find a b that is either less than r or is greater than r. In essence, you can always find a b that is not equal to r. There is no contradiction.

If there exists a specific b such that no real number is equal to it, then I cannot say that b is a real number.

 

[losiu99]

Suppose the set E:={1,2,3}. Suppose r=5 and M=6. Then for every specified 'r' there does not exist an element b in E such that b>r.

"Suppose that none of the real numbers r, which are specified in (Statement B), are upper bounds of E. "
Isn't 5 an upper bound of this E?
Let E = (0, 1), M = 1. For given b in E there is such r, there is no such r for all b in E. Or, since you doubt in validity of definition of real numbers, let E = {x in Q: x in (0, 1)}.

Proofs regarding Cauchy sequences rely on the Completeness Axiom.

How about Dedekind cuts?

Edit: Besides, Charles Chapman's "Real Mathematical Analysis" in chap.2 describes Cauchy completion of rationals, proving completeness directly.

 

[JG89]

In fact, Charles Pugh gives two proofs

EDIT: Made a mistake, only one proof

 

[D H]

Cauchy completion is a slightly different subject from the completeness axiom. I don't have Chapman's book; did he prove Cauchy completion by assuming the completeness axiom?

That said, the OP's theorem C is incorrect. As losiu99 already pointed out, any rational greater than or equal to 1 is an upper bound on the set of rationals between 0 and 1 exclusive. 1 is obviously the least upper bound, aka the supremum, and it is equally obviously strictly greater than all rationals between 0 and 1 exclusive.

 

[HallsofIvy]

Yes, the "Cauchy Criterion" (Cauchy completeness) and the "least upper bound property" are equivalent- either can be proved assuming the other.

In fact, there are several other equivalent statements that can be used as the basis for the real numbers:

If a set of real numbers has an upper bound, then it has a least upper bound.

If a set of real numbers has a lower bound, then it has a greatest lower bound.

If an increasing sequence of real numbers has an upper bound, then it converges.

If a decreasing sequence of real numbers has a lower bound, then it converges.

If a sequence of real numbers has both upper and lower bounds, then has a convergent subsequence.

Every Cauchy sequence of real numbers converges.

The set of real numbers, with the usual topology (the metric topology with d(x, y)= |x- y|) is a connected set.

If A is a closed and bounded set of real numbers (with the usual topology) then A is compact.

Any one of those can be used to prove all of the others. They are not true, of course, if "real numbers" is replaced by "rational numbers".

 

[losiu99]

D H, he doesn't assume the completeness axiom; he explicitly constructs the sequence - binary expansion of the supremum, using nothing more than well-ordering of positive integers.

Edit: http://books.google.pl/books?id=R_Z...EwAQ#v=onepage&q=second construction&f=false" (first result)

 

[D H]

Thanks, both.

So, back to the topic at hand: Is there any reason to leave this thread open?

 

[mathbuster]

The Completeness Axiom states that all nonempty subsets of the Real numbers which is bounded above has a supremum. If you do not assume that this is true then you cannot create the real numbers just by using the supremums. (Some nonempty bounded subsets of the reals will not have a supremum.)

The purpose of the Completeness Axiom is to ensure that there are no 'holes' in the number line.

Consider the following sets: A:={c: c<1} , E:={b: 1<b<=3/2 - c/2 for all c in A}

Using the Completeness Axiom, Set E has a Supremum and therefore is empty. However, assuming that some nonempty subsets of the reals do not have a supremum, E is not empty.

Does the Completeness Axiom ensure 'no holes' - or does it create some?

 

[Hurkyl]

The Completeness Axiom states that all nonempty subsets of the Real numbers which is bounded above has a supremum. If you do not assume that this is true

Your next comment suggests you mean that "if you assume this is false" rather than "do not assume this is true".

But anyways, if you assume the completeness axiom is false, then you are working with a mathematical system that is not the real numbers.



You're beginning to lose the coherency you originally had. I let this thread stay open because you were remarkably clear in your website, and it was actually looked like a good logical argument marred by a minor, yet fatal mistake, and you looked like someone who could acknowledge the error and learn.

But now, you look like someone grasping at straws to defend a preconceived notion. What is your intent now?

 

[JThompson]

Consider the following sets: A:={c: c<1} , E:={b: 1<b<=3/2 - c/2 for all c in A}

Using the Completeness Axiom, Set E has a Supremum and therefore is empty. However, assuming that some nonempty subsets of the reals do not have a supremum, E is not empty.

The set E is not bounded above. Do you see why? 3/2-c/2 gets larger when c is negative. The Completeness Axiom is irrelevant here.

 

[mathbuster]

I will have to think about that one for a little while. I do value the feed back that I have received from this forum. Most especially from Hurkyl. While sometimes pointed, it is always informative! (It's like sitting in a room full of math professors.)