Talk:Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 3

I assume you don't expect me to do anything for the time being. Ivar Sand (talk) 11:01, 28 November 2013 (UTC)


 * That is correct. We'll get to it eventually. &mdash; Lord_Farin (talk) 11:03, 28 November 2013 (UTC)

Background for the proof
Once a friend of mine said that a Cauchy sequence convergences for complete spaces. Moreover, he said that a Cauchy sequence converges for the reals because the space of real numbers is complete. I myself knew nothing about complete spaces since I had only knowledge of elementary real analysis.

However, I felt that his reference to complete spaces should be unnecessary, in other words, that there should be no mandatory logical link between the Cauchy criterion and complete spaces or other advanced concepts. So, I started constructing a proof that the Cauchy criterion and the standard convergence criterion are equivalent using only elementary real analysis. To prove that a convergent sequence is Cauchy is easy, as is well known. The challenging part was to prove that a Cauchy Sequence is convergent. --Ivar Sand (talk) 07:33, 27 November 2018 (EST)

It needs to be pointed out that this proof has a student's perspective. It may not be of interest to the rest of the mathematical community. --Ivar Sand (talk) 10:09, 22 November 2019 (EST)

Suggestions of improvement
I've been looking at this to see why it needs to be so complicated and why you cannot use the Bolzano-Weierstrass Theorem. My question is: is the sticking point the Continuum Property? --prime mover (talk) 02:29, 21 June 2019 (EDT)


 * I wrote the proof many years ago, in my student days I believe. I wanted to produce a direct proof of the theorem without reference to any advanced concept. I saw the challenge of producing a proof using only my own knowledge of mathematics. I thought that was fun, and I felt satisfied when I succeeded.


 * But I am aware of the fact that my competence of mathematics is limited even today. I would not be surprised if others with more knowledge of mathematics and who are smarter than I, would be able to change the proof into one that is shorter and better. Therefore, I am not convinced that the proof needs to be so complicated as the one that I produced.


 * You are mentioning the Bolzano-Weierstrass Theorem. I cannot see exactly how it can be used here, but, as I indicated above, others might understand that better (see proofs 1 and 2).


 * Regarding whether the sticking point is the Continuum Property (or rather the Monotone Convergence Theorem), yes I suppose it could be. But to me, the essential part of the proof is the construction of the sequences ${U_i}$ and ${L_i}$ and the proof that they converge to the same limit. --Ivar Sand (talk) 03:37, 1 July 2019 (EDT)


 * The point is that every single step in this proof is developed painstakingly from first principles (and in a very unwieldy way -- what is the point of setting up a sequence of values of $\epsilon_i$ when just stating one value will do the job well enough?) It is seen to be of the same structure as Proof 2, but the latter merely invokes a bunch of previously-demonstrated results which you have proved step by step. One of those results is effectively the B-WT, but as the latter relies upon the Continuum Property and yours does not specifically invoke that, I wondered (sorry, I don't have the patience to check it myself, life's too short) whether you could not actually use B-WT because you specifically did not want to use Continuum Property? --prime mover (talk) 03:51, 1 July 2019 (EDT)


 * Note that $\epsilon$ is a constant, it could have been replaced by the number 1. It is tempting to let $\epsilon$ approach zero, but the definitions of ${U_i}$ and ${L_i}$ depend upon $\epsilon$, and I would not like to risk jeopardising the convergences of ${U_i}$ and ${L_i}$ by letting $\epsilon$ approach zero.


 * I am sorry, but I believe I misunderstood what you meant by sticking point. I knew the Continuum Property as an axiom of the real number system, so I did not have a problem with that. My problem was connected to the concept of complete spaces since I had not been taught it, so I considered it an advanced concept.


 * I still cannot see how Proof 3 could benefit from the use of the Bolzano-Weierstrass Theorem. Proof 3 could benefit from the use of the Squeeze Theorem though, but this would not have made it much shorter. --Ivar Sand (talk) 03:45, 3 July 2019 (EDT)


 * No, sorry, I give up. --prime mover (talk) 04:25, 3 July 2019 (EDT)