In this article I will provide some alternative ways to formulate the completeness of real or complex numbers which I have found in the last weeks.
Formulation with epsilon-almost analysis
In my bachelor thesis about \( \epsilon \)-almost analysis I give an alternative completeness axiom for real numbers. It states, that:
Every \( 2\epsilon \)-almost Cauchy sequence converges \( \epsilon \)-almost.
Okay, what the hell are \( 2\epsilon \)-almost Cauchy sequences and \( \epsilon \)-almost convergent sequences? Here are the definitions:
An \( 2\epsilon \)-almost Cauchy sequence is a sequence \( (x_n)_{n\in\N} \) with \( |x_n - x_m| \le 2\epsilon \) for almost all \( n,m \in \N \) which means for all natural numbers \( n \) and \( m \) bigger than an existing \( N \in \N \).
A sequence \( (x_n)_{n\in\N} \) converges \( \epsilon \)-almost, iff there is a \( c \in \R \) and an \( N \in \N \) so that \( |x_n-c|\le\epsilon \) for all natural numbers \( n \ge N \) so iff \( |x_n-c| \le \epsilon \) for almost all \( n\in\N \).
You will find the proof and more informations in my bachelor thesis. Unfortunately it is just available in German yet. Please contact me, if you are interested in an introduction to \( \epsilon \)-almost analysis in German.
Formulation with the Cauchy property of functions
In the article How to proof the convergence of functions without knowing the limit I introduce the concept of functions with the Cauchy property at a special point \( x_0 \). This concept is an extension of the definition for Cauchy sequences to functions. I say that a real or complex valued function \( f : D \rightarrow \K \) with \( D \subset \K \) has the Cauchy property at the point \( x_0 \), iff:
Thereby \( \K \) shall be now and in the following \( \R \) or \( \C \) dependent whether we consider the completeness axiom for real or complex numbers. As I showed in the article a limit for a function exists for \( x \rightarrow x_0 \), if and only if \( f \) has the Cauchy property at \( x_0 \). So it is:
This really looks like the completeness axiom and indeed, it is equivalent to it or better to say the following implication is equivalent to it:
For every function \( f : D \rightarrow \K \) with \( D \subseteq \K \) and every limit point \( x_0 \) of \( \K \) the following implication is true: If \( f \) has the Cauchy property at \( x_0 \), then the limit \( \lim_{x\rightarrow x_0} f(x) \) exists.
Proof: In the article How to proof the convergence of functions without knowing the limit I already gave a proof that the above implication is true in the field of real or complex numbers. So I just have to show, that we can conclude the completeness axiom from the above statement.
Let \( (x_n)_{n\in\N} \) be an arbitrary Cauchy sequence of real or complex numbers. I define \( f : \{ \tfrac 1n | n\in\N \} \rightarrow \K \) with \( f\left(\tfrac 1n\right) := x_n \). Because \( (x_n)_{n\in\N} \) is a Cauchy sequence \( f \) has the Cauchy property at \( x_0 \) and therefore \( \lim_{x\rightarrow 0} f(x) = \lim_{n\rightarrow\infty} f\left(\tfrac 1n\right) = \lim_{n\rightarrow\infty} x_n \) exists.