Lecture 15 - Cauchy Convergence

I did this solely from the textbook.

Cauchy Sequence

A Sequence $(a_{n})$ is called a Cauchy Sequence if, $\forall ϵ > 0$ there $\exists N \in N$ where any $m, n \geq N$ follows that:
$| a_{n} - a_{m} | < ϵ$

Consider the definition of convergence:

Convergence of a sequence

A Sequence $(a_{n})$ of real numbers converges to some $L \in R$ if:
$\forall ε > 0 \exists N \in N \forall n \geq N (| a_{n} - L | < ε)$

The definition of convergence asserts that, given an arbitrary positive $ϵ$ we can find a point in the sequence $N$ such that all points after $n \geq N$ are closer to the limit $a$ than the given $ϵ$ .

On the other hand, a sequence is Cauchy if, for every $ϵ$ there is a point in the sequence after which the terms are all closer to each other than the given $ϵ$ .

We'll argue that the two definitions are equivalent: convergent sequences are cauchy, and a cauchy sequence is convergent. The significance of this definition is that there is no mention of the value of the limit, similar to the Monotone Convergence Theorem. See Cauchy Sequences Converge.

Theorem

A sequence converges iff it is a Cauchy Sequence.

Proof

( $\Rightarrow$ ): See Cauchy Sequences Converge.

( $\Leftarrow$ ): Start with the Cauchy sequence $(x_{n})$ . Using Cauchy Sequences Are Bounded then $(x_{n})$ is bounded. We can use the Balzano-Weierstrass Theorem to produce a convergent subsequence $(x_{n_{k}})$ . Set:
$x = lim_{n \to \infty} x_{n_{k}}$
The idea is to show that $x_{n} \to x$ . We'll use the triangle inequality for this argument. We know the terms in the subsequence are getting close to the limit $x$ , and the assumption that $(x_{n})$ is Cauchy implies the terms in the "tail" of the sequence are close to each other. Thus, we want to make each of these distances less than half of the prescribed $ϵ$ .

Let $ϵ > 0$ . Because ( $x_{n}$ ) is Cauchy then $\exists N$ where:
$| x_{n} - x_{m} | < \frac{ϵ}{2}$
whenever $m, n \geq N$ . Now we also know $(x_{n_{k}}) \to x$ so choose a term in this subsequence, denoted $x_{n_{K}}$ with $n_{K} \geq N$ and:
$| x_{n_{K}} - x | < \frac{ϵ}{2}$
To see that $N$ has the desired property for the original sequence $(x_{n})$ observe that any $n > N$ has:
$\begin{aligned} | x_{n} - x | & = | x_{n} - x_{n_{K}} + x_{n_{K}} - x | \\ \leq | x_{n} - x_{n_{K}} | + | x_{n_{K}} - x | \\ < \frac{ϵ}{2} + \frac{ϵ}{2} \\ = ϵ \end{aligned}$
☐