Consider the set of all Cauchy sequences of rational numbers: concretely, the set

Define an equivalence relation on this set, by $(a_{n}) \sim (b_{n})$ if and only if, for every rational $ϵ > 0$ , there is a natural number $N$ such that for all $n \in N$ bigger than $N$ , we have $| a_{n} - b_{n} | < ϵ$ . This is an equivalence relation (exercise).

Show solution

It is symmetric, because $| a_{n} - b_{n} | = | b_{n} - a_{n} |$ .

It is reflexive, because $| a_{n} - a_{n} | = 0$ for every $n$ , and this is $< ϵ$ .

It is transitive, because if $| a_{n} - b_{n} | < \frac{ϵ}{2}$ for sufficiently large $n$ , and $| b_{n} - c_{n} | < \frac{ϵ}{2}$ for sufficiently large $n$ , then $| a_{n} - b_{n} | + | b_{n} - c_{n} | < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ$ for sufficiently large $n$ ; so by the triangle_inequality, $| a_{n} - c_{n} | < ϵ$ for sufficiently large $n$ .

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:

$[a_{n}] + [b_{n}] := [a_{n} + b_{n}]$

$[a_{n}] \times [b_{n}] := [a_{n} \times b_{n}]$

$[a_{n}] \leq [b_{n}]$ if and only if $[a_{n}] = [b_{n}]$ or there is some $N$ such that for all $n > N$ , $a_{n} \leq b_{n}$ .

This field structure is well-defined (proof).

Examples

Any rational number $r$ may be viewed as a real number, being the class $[r]$ (formally, the equivalence class of the sequence $(r, r, \dots)$ ).

The real number $π$ is indeed a real number under this definition; it is represented by, for instance, $(3, 3.1, 3.14, 3.141, \dots)$ . It is also represented as $(100, 3, 3.1, 3.14, \dots)$ , along with many other possibilities.

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Real number (as Cauchy sequence)

Examples