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Cauchy condition: $$\forall\epsilon\gt 0\quad\exists N\in\Bbb{N}:n,m\ge N\implies|a_n-a_m|\lt\epsilon.$$

Theorem: $\Bbb{R}$ is complete with respect to Cauchy in the sense that if $(a_n)$ is a sequence of real numbers obeying a Cauchy condition then it converges to a limit in $\Bbb{R}$.

Proof: Let $A=\{x\in\Bbb{R}:\exists n\in\Bbb{N}\land a_n=x\}.$ We first observe that $A$ is a bounded set in $\Bbb{R}$. Taking $\epsilon=1$ in Cauchy condition, there is an integer $N_1$, then for each $n\ge N_1,|a_n-a_{N_1}|\lt1$.

Clearly the finite set $a_1,a_2,...,a_{N_1},a_{N_1}-1,a_{N_1}+1$ is bounded; say all its elements belong to the interval $[-M,M]$. So $A$ is bounded. Next, consider the set $$S=\{s\in[-M,M]:\exists\text{infinitely many }n\in\Bbb{N},\text{for which }a_n\ge s\}.$$

Clearly $-M\in S$ and $S$ is bounded above by $M$. According to the least upper bound property of $\Bbb{R}$ there exists $b\in\Bbb{R},b=\text{lub }S.$ We claim that the sequence $(a_n)$ converges to $b$.

Given $\epsilon\gt 0$ we must show that there exists an $N$ such that $\forall n\ge N, |a_n-b|\lt\epsilon$. The Cauchy condition provides an $N_2:m,n\ge N_2\implies|a_m-a_n|\lt\frac{\epsilon}{2}.$ Only finitely often does $a_n$ exceed $b+\epsilon/2$. That is, for some $N_3\ge N_2$, $$n\ge N_3\implies a_n\stackrel{?_1}{\le} b+\frac{\epsilon}{2}.$$

Since $b$ is a least upper bound for $S$, some $s\in S$ is $\ge b-\epsilon/2$, which implies that $a_n\ge s\gt b-\epsilon/2$ infinitely often. In particular, there exists $N\ge N_3$ such that $a_N\gt b-\epsilon/2$. Since $N\ge N_3$, $a_N\stackrel{?_2}{\lt} b+\epsilon/2$ and $$a_N\in(b-\epsilon/2,b+\epsilon/2).$$

Since $N\ge N_2$, implies $$|a_n-b|\le|a_n-a_N|+|a_N-b|\lt\epsilon,$$ which verifies convergence. $\quad\square$

$?_1:$ Why it is $\le$, not $\lt$, according to the definition of $S$?

$?_2:$ Then why the author uses $\lt$, not $\le$ here?

Book: Real Mathematical Analysi - Pugh page $18$.

  • 3
    The answer to both your questions is that the choice of a large or strict inequality sign is irrelevant to the proof. – 2017-02-27
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    This kind of choice is always annoying me when reading proofs. It seems like they're equivalent. Can I pick the one I need in the proof? Any advice is appreciated @Did – 2017-02-27
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    Sure, since $\epsilon$ is any positive number and $x\leqslant\epsilon\implies x<\frac12\epsilon$ and $x<\epsilon\implies x\leqslant\epsilon$. – 2017-02-27

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Whatsoever this is my try.

(1) Since $b+\epsilon/2$ is not in the set $S$, only finitely many $a_n\ge b+\epsilon/2$, so there are infinitely many $a_n\lt b+\epsilon/2$, given $n\ge N_3\ge N_2$. (By definition of S, and b is the l.u.b..)

(2) Since $b-\epsilon/2$ is not an upper bound for the set $S$, $\exists s\in S:b-\epsilon/2\lt s\le a_n$. (By the definition of upper bound, $\forall s\in S, s\le U$, so the negation gives $\exists s\in S, s\gt U$.)

We still have infinitely many $a_n$ such that $b-\epsilon/2\lt s\le a_n\lt b+\epsilon/2$, given $n\ge N_3$. So $a_n\in(b-\epsilon/2,b+\epsilon/2)$, given $n\ge N_3$.