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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).

Tao, Analysis I