# Proof explanation: Every Cauchy sequence has a limit

I have a bit of a problem with understanding the parts of the proof as stated in my book:

Lemma: In a complete ordered field, every Cauchy sequence has a limit.

Proof: Let $$\left(a_{n}\right)$$ be a Cauchy sequence in $$F$$. By the argument of lemma $$9.15$$ of chapter 9 (carried out in $$F$$ ) the sequence is bounded. Hence so is every subset of elements in the sequence. Define $$b_{N}=\text { the least upper bound of }\left\{a_{N}, a_{N+1}, a_{N+2}, \ldots\right\}$$ This exists by completeness. Clearly $$b_{0} \geq b_{1} \geq b_{2} \geq \cdots$$ and the sequence $$\left(b_{n}\right)$$ is bounded below-say, by any lower bound for $$\left(a_{n}\right)$$. Hence we can define $$c=$$ the greatest lower bound of $$\left(b_{n}\right)$$.

We claim that $$c$$ is the limit of the original sequence $$\left(a_{n}\right)$$ To prove this, let $$\varepsilon>0$$. Suppose that there exist only finitely many values of $$n$$ with $$c-\frac{1}{2} \varepsilon Then we may choose $$N$$ such that for all $$n>N$$, $$a_{n} \leq c-\frac{1}{2} \varepsilon \text { or } a_{n} \geq c+\frac{1}{2} \varepsilon$$ But there exists $$N_{1}>N$$ such that if $$m, n>N_{1}$$ then $$\left|a_{m}-a_{n}\right|<\frac{1}{2} \varepsilon$$. Hence $$\text { for all } n>N_{1}, a_{n} \leq c-\frac{1}{2} \varepsilon$$ Or $$\text { for all } n>N_{1}, a_{n} \geq c+\frac{1}{2} \varepsilon$$ The latter condition implies that there exists some $$m$$ with $$a_{n}>b_{m}$$ for all $$n>N_{1}$$, which contradicts the definition of $$b_{m}$$. But the former implies that we may change $$b_{N_{1}}$$ to $$b_{N_{1}}-\frac{1}{2} \varepsilon$$, which again contradicts the definition of $$b_{N_{1}}$$. It follows that for any $$M$$ there exists $$m>M$$ such that $$c-\frac{1}{2} \varepsilon Since $$\left(a_{n}\right)$$ is Cauchy, there exists $$M_{1}>M$$ such that $$\left|a_{n}-a_{m}\right|<\frac{1}{2} \varepsilon$$ for $$m, n>M_{1} .$$ Hence for $$n>M_{1}$$, $$c-\varepsilon But this implies that $$\lim a_{n}=c$$ as claimed.

Specifically I don't understand why the 2 bolded sentences are needed in the proof.

Don't we already establish in the previous assumption (Suppose that there exist only finitely...) that for all $$n>N, a_{n} \leq c-\frac{1}{2} \varepsilon \text { or } a_{n} \geq c+\frac{1}{2} \varepsilon$$ and from that, the paragraph afterwards (The latter condition...) follows? Why does the Cauchy criterion need to be invoked?

• +1 for your effort. – Sebastiano Jun 7 at 21:27

One thing is to say:

There is some $$N\in\Bbb N$$ such that, for each $$n\geqslant N$$, $$\displaystyle a_n\leqslant c-\frac\varepsilon2$$ or $$\displaystyle a_n\geqslant c+\frac\varepsilon2$$.

A much stronger statement is

There is some $$N\in\Bbb N$$ such that (for each $$n\geqslant N$$, $$\displaystyle a_n\leqslant c-\frac\varepsilon2$$) or (for each $$n\geqslant N$$, $$\displaystyle a_n\geqslant c+\frac\varepsilon2$$).

They are far from meaning the same thing. Here's a similar situation: it is true that

Each natural number is even or odd.

But it is false that:

Every natural number is even or every natural number is odd.

• Oops, completely missed that "little" detail. Thank you! – Treex Jun 8 at 6:38