Real number (as Cauchy sequence)

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Edited by Patrick Stevens, et al. last updated 6th Jul 2016

Requires: Math 3

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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 , there is a natural number such that for all bigger than , we have . 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
< \epsilon
.
It is transitive, because if
|a_n - b_n| < \frac{\epsilon}{2}
for sufficiently large
n
, and
|b_n - c_n| < \frac{\epsilon}{2}
for sufficiently large
n
, then
|a_n - b_n| + |b_n - c_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
for sufficiently large
n
; so by the triangle_inequality,
|a_n - c_n| < \epsilon
for sufficiently large
n
.

Write

[a_n]

for the equivalence class of . (This is a slight abuse of notation, omitting the brackets that indicate that is actually a sequence rather than a rational number.)

The set of real numbers is the set of equivalence classes of

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
\pi
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.

Summaries

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.

Parents:

Real number

Children:

The reals (constructed as classes of Cauchy sequences of rationals) form a field