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?
Best Answer
One thing is to say:
A much stronger statement is
They are far from meaning the same thing. Here's a similar situation: it is true that
But it is false that: