Math and science::Analysis::Tao::05. The real numbers

# Cauchy sequences

A sequence \( (a_n)_{n=m}^{\infty} \) of rational numbers is a Cauchy Sequence iff the sequence is eventually ε-steady for every rational \( \varepsilon > 0 \).In other words, a sequence \( a_0, a_1, a_2, ... \) of rational numbers is a Cauchy Sequence iff for every rational \( \varepsilon > 0 \) there exists an integer N such that \( d(a_j - a_k) \le \varepsilon \) for every \( j, k \ge N \). Where \( d(a,b) = |a - b| \).

#### From sequences to reals

sequence → ε-steady sequence → eventually ε-steady sequence →*Cauchy sequence*→ ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.

### Example

Proposition

The sequence \( a_1, a_2, a_3, ... \) defined by \( a_n := \frac{1}{n} \) (i.e., the sequence \( 1, \frac{1}{2}, \frac{1}{3}, ... \)) is a Cauchy sequence.

Proof

We have to show that for every \( \varepsilon > 0 \) there exists an integer \( N > 0 \) such that the sequence \( (a)_{N}^{\infty} \) is ε-steady, which would mean that \( d(a_j, a_k) \le \varepsilon \) for all \( j, k \ge N \):

\[ | \frac{1}{j} - \frac{1}{k}| \le \varepsilon \text{ for every } j, k \le N \]

As \( j, k \ge N \), we know that \( 0 \le \frac{1}{j}, \frac{1}{k} \le \frac{1}{N} \), so \( |\frac{1}{j} - \frac{1}{k}| \le \frac{1}{N} \). Thus, to force \( |\frac{1}{j} - \frac{1}{k}| \) to be less than or equal to ε, it is sufficient to choose an N such that \( \frac{1}{N} \) is less than ε, or in other words, an N such that \( N \gt \frac{1}{\varepsilon} \), and this can be done due to Prop. 4.4.1 (interspersing of integers by rationals).