Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 3

Proof
Let $\epsilon \in \R_{>0}$ be given.

Since $\left\langle {a_n} \right\rangle$ is Cauchy, a natural number $N$ exists such that:


 * $\left\vert{a_n - a_m}\right\vert < \epsilon$ for every $m, n \ge N$

We aim to show that $\left\langle {a_n} \right\rangle$ converges.

That is, that numbers $a$ in $\R$ and $N'$ in $\N$ exist such that:


 * $\left\vert{a_n - a}\right\vert < \epsilon$ for every $n \ge N'$

Let $\left\langle {\epsilon_i} \right\rangle_{i \in \N}$ be a sequence of strictly positive real numbers that satisfies:


 * $\epsilon_0 = \epsilon$


 * $\epsilon_{i+1} < \epsilon_i$


 * $\displaystyle \lim_{i \to \infty} \epsilon_i = 0$

Since $\left\langle {a_n} \right\rangle$ is Cauchy, for each $\epsilon_i$ a natural number $N_i$ exists such that:


 * $\left\vert{a_n - a_m}\right\vert < \epsilon_i$ for every $m, n \ge N_i$

Let us study the sequence $\left\langle {N_i} \right\rangle_{i \in \N}$.

First, we consider $N_0$.

We can choose $N_0 = N$ because (1) $\left\vert{a_n - a_m}\right\vert < \epsilon$ for every $m, n \ge N$ and (2) $\epsilon = \epsilon_0$.

Next, we consider the relation between $N_{i+1}$ and $N_i$.

We have, for $i \in \N$:


 * $\left\vert{a_n - a_m}\right\vert < \epsilon_i$ for every $m, n \ge N_i$

and:


 * $\left\vert{a_n - a_m}\right\vert < \epsilon_{i+1}$ for every $m, n \ge N_{i+1}$

There are two cases for $\left\vert{a_n - a_m}\right\vert$ when $m, n \ge N_i$.

Either:


 * $\left\vert{a_n - a_m}\right\vert < \epsilon_{i+1}$ for every $m, n \ge N_i$

or:


 * $\left\vert{a_{n'} - a_{m'}}\right\vert \ge \epsilon_{i+1}$ for some $m', n' \ge N_i$

In the first case, we can choose $N_{i+1} = N_i$ since $\left\vert{a_n - a_m}\right\vert < \epsilon_{i+1}$ for every $m, n \ge N_i$.

In the second case, suppose that $m', n' \ge N_{i+1}$.

This cannot be true since we would then have $\left\vert{a_{n'} - a_{m'}}\right\vert < \epsilon_{i+1}$ since $\left\vert{a_n - a_m}\right\vert < \epsilon_{i+1}$ for every $m, n \ge N_{i+1}$.

Therefore, at least one of $m', n'$, call it $k'$, must be less than $N_{i+1}$.

This means that:

Thus, we have established that there exists a sequence $\left\langle {N_i} \right\rangle_{i \in \N}$ that satisfies:


 * $N_0 = N$


 * $N_{i+1} \ge N_i$

Since $\left\langle {a_n} \right\rangle$ is Cauchy, for each $\epsilon_i$ a constant $N_i$ exists such that:


 * $\vert a_n - a_{N_i} \vert < \epsilon_i$ whenever $n \ge N_i$

Negative of Absolute Value: Corollary 1 applied to $\vert a_n - a_{N_i} \vert < \epsilon_i$ yields:


 * $a_{N_i} - \epsilon_i < a_n < a_{N_i} + \epsilon_i$ whenever $n \ge N_i$

Define the sequence $\left\langle {U_i} \right\rangle_{i \in \N}$ by:


 * $U_0 = a_{N_0} + \epsilon_0$


 * $U_{i+1} =\min \left( {U_i, a_{N_{i+1}} + \epsilon_{i+1}} \right)$

We observe that $U_{i+1}$ is the minimum of two numbers, one of which is $U_i$.

Therefore $\left\langle {U_i} \right\rangle$ is decreasing.

$U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$
We prove this by using the Principle of Mathematical Induction.

$a_{N_0} + \epsilon_0$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_0}$ because $a_n < a_{N_0} + \epsilon_0$ whenever $n \ge N_0$.

Therefore, $U_0$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_0}$ since $U_0 = a_{N_0} + \epsilon_0$.

This concludes the first induction step.

We need to prove that $U_{i+1}$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_{i+1}}$ if $U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$.

$U_{i+1}$ equals either $U_i$ or $a_{N_{i+1}} + \epsilon_{i+1}$.

Assume that $U_{i+1} = U_i$.

$U_{i+1}$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ since $U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ by presupposition.

We have that $\left\langle {a_n} \right\rangle_{n \ge N_{i+1}}$ is a subset of $\left\langle {a_n} \right\rangle_{n \ge N_i}$ because $N_{i+1} \ge N_i$.

Therefore, $U_{i+1}$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_{i+1}}$ because $U_{i+1}$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$.

Assume that $U_{i+1} = a_{N_{i+1}} + \epsilon_{i+1}$.

$U_{i+1}$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_{i+1}}$ because $a_n < a_{N_{i+1}} + \epsilon_{i+1}$ whenever $n \ge N_{i+1}$.

This concludes the proof that $U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ for every $i \in \N$.

$\left\langle {U_i} \right\rangle$ is bounded
By Cauchy Sequence is Bounded, $\left\langle {a_n} \right\rangle$ is bounded, and therefore bounded below.

A lower bound for $\left\langle {a_n} \right\rangle$ is also a lower bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ for every $i$ because $\left\langle {a_n} \right\rangle_{n \ge N_i}$ is a subsequence of $\left\langle {a_n} \right\rangle$.

This lower bound is less than or equal to $U_i$ for every $i$ because $U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$.

Therefore, this lower bound is also a lower bound for $\left\langle {U_i} \right\rangle$.

Since $\left\langle {U_i} \right\rangle$ is decreasing, its first element is an upper bound for $\left\langle {U_i} \right\rangle$.

Since $\left\langle {U_i} \right\rangle$ is bounded below and above, it is bounded.

By Monotone Convergence Theorem, a bounded, monotonic sequence converges, and therefore $\left\langle {U_i} \right\rangle$ converges.

Now, define the sequence $\left\langle {L_i} \right\rangle_{i \in \N}$ by:


 * $L_0 = a_{N_0} - \epsilon_0$


 * $L_{i+1} =\max \left( {L_i, a_{N_{i+1}} - \epsilon_{i+1}} \right)$

An analysis of $\left\langle {L_i} \right\rangle$, not given here because it is similar to the one above of $\left\langle {U_i} \right\rangle$, produces the following results:


 * $\left\langle {L_i} \right\rangle$ is increasing


 * $\left\langle {L_i} \right\rangle$ converges


 * $L_i$ is a lower bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ for every $i \in \N$

The limits of $\left\langle {U_i} \right\rangle$ and $\left\langle {L_i} \right\rangle$ as $i \to \infty$ are equal
Since $U_i$ and $L_i$ are, respectively, upper and lower bounds for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ for every $i \in \N$, we have for all $i \in \N$:

which shows that $U_i - L_i \to 0$ as $i \to \infty$ since $\displaystyle \lim_{i \to \infty} \epsilon_i = 0$.

We have:

Let $\displaystyle a := \lim_{i \to \infty} U_i = \displaystyle \lim_{i \to \infty} L_i$.

Because $\left\langle {U_i} \right\rangle$ is decreasing, we have $a \le U_i$ for all $i$.

Because $\left\langle {L_i} \right\rangle$ is increasing, we have $a \ge L_i$ for all $i$.

Putting this together:
 * $\forall i \in \N: L_i \le a \le U_i \iff 0 \le a-L_i \le U_i - L_i$

Also, since $U_i$ and $L_i$ are, respectively, upper and lower bounds for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ for every $i \in \N$, we have for every $n \ge N_i$ for every $i \in \N$:


 * $L_i \le a_n \le U_i \iff 0 \le a_n - L_i \le U_i - L_i$

Now, pick a natural number $j$ such that:
 * $\epsilon_j < \dfrac 1 2 \epsilon$

Since $\displaystyle \lim_{i \to \infty} (U_i - L_i) = 0$ a natural number $k$ exists such that:
 * $\left\vert{U_i - L_i}\right\vert < \dfrac \epsilon 2$

whenever $i \ge k$.

Set $l = \max \left({j, k}\right)$.

We have:

and:

Putting all this together, we find for $n \ge N_l$:

which finishes the proof that $\left\langle {a_n} \right\rangle$ is convergent.