Real number (as Cauchy sequence)

by Patrick Stevens Jul 2 2016 updated Jul 5 2016

There are several ways to construct real numbers; this is the most natural way to use them in computations.

[summary: If we want to construct the real numbers in terms of simpler objects (such as the rationals), one way to do it is to take our putative real number and consider sequences of rational numbers which in some sense "get closer and closer" to the real number.]

[summary(Technical): The real numbers can be constructed as a field consisting of all Cauchy sequences of rationals, quotiented by the equivalence relation given by "two sequences are equivalent if and only if they eventually get arbitrarily close to each other".]

Consider the set of all Cauchy sequences of rational numbers: concretely, the set $$~$X = \{ (a_n)_{n=1}^{\infty} : a_n \in \mathbb{Q}, (\forall \epsilon \in \mathbb{Q}^{>0}) (\exists N \in \mathbb{N})(\forall n, m \in \mathbb{N}^{>N})(|a_n - a_m| < \epsilon) \}$~$$

Define an Equivalence relation on this set, by $~$(a_n) \sim (b_n)$~$ if and only if, for every rational $~$\epsilon > 0$~$, there is a Natural number $~$N$~$ such that for all $~$n \in \mathbb{N}$~$ bigger than $~$N$~$, we have $~$|a_n - b_n| < \epsilon$~$. This is an equivalence relation (exercise). %%hidden(Show solution):

Write $~$[a_n]$~$ for the equivalence class of $~$(a_n)_{n=1}^{\infty}$~$. (This is a slight abuse of notation, omitting the brackets that indicate that $~$a_n$~$ is actually a sequence rather than a rational number.)

The set of real numbers is the set of equivalence classes of $~$X$~$ under this equivalence relation, endowed with the following totally ordered field structure:

This field structure is well-defined (proof).



Kevin Clancy

The title mentions Cauchy sequences, but the body does not. Doesn't this definition consider classes of non-converging sequences as real numbers?