Recent content by 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.)

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.)...

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.

Proofs regarding Cauchy sequences rely on the Completeness Axiom.

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.

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.

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...

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

(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...

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