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 $$ c-\frac{1}{2} \varepsilon<a_{n}<c+\frac{1}{2} \varepsilon $$ 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 $$ c-\frac{1}{2} \varepsilon<a_{m}<c+\frac{1}{2} \varepsilon $$ 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 $n>M_{1}$, $$ c-\varepsilon<a_{n}<c+\varepsilon $$ 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?

  • 2
    $\begingroup$ +1 for your effort. $\endgroup$ – 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.

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

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