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

Theorem
Let $\left\langle {a_n} \right\rangle$ be a Cauchy sequence in $\R$.

Then $\left\langle {a_n} \right\rangle$ is convergent.

Proof
Let $\epsilon > 0$ be given.

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


 * $\vert a_n - a_m \vert < \epsilon$ whenever $m$, $n \ge N$

We aim to show that $\left\langle {a_n} \right\rangle$ converges which means that numbers $a$ in $\R$ and $N'$ in $\N$ exist such that:


 * $\vert a_n - a \vert < \epsilon$ whenever $n \ge N'$

Consider a sequence of epsilons, $\left\langle {\epsilon_i} \right\rangle_{i \in \N}$, satisfying:


 * $\epsilon_0 = \epsilon$


 * $\epsilon_i < \epsilon_{i-1}$ for $i \ge 1$


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

We note that $\left\langle {\epsilon_i} \right\rangle$ is decreasing and convergent.

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 $n \ge N_i$

We choose $N_i$ such that it satisfies:


 * $N_0 = N$


 * $N_i \ge N_{i-1}$ for $i \ge 1$

Here, the choice $N_i \ge N_{i-1}$ is allowed because if $N_{i-1}$ were to be greater than $N_i$, $N_{i-1}$ could be set equal to $N_i$ since the value of $N_i$ works for $N_{i-1}$ as well.

This is because $\vert a_n - a_{N_i} \vert < \epsilon_i$ whenever $n \ge N_i$ implies that $\vert a_n - a_{N_i} \vert < \epsilon_{i-1}$ whenever $n \ge N_i$ since $\epsilon_{i-1} > \epsilon_i$.

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


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


 * $U_i = min(U_{i-1}, a_{N_i} + \epsilon_i)$ for $i \ge 1$

Since $U_i$ for $i \ge 1$ is the minimum of two numbers, one of which is $U_{i-1}$, $\left\langle {U_i} \right\rangle$ is decreasing.

$U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$
We prove this by using proof by 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 $n \ge N_0$.

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

This concludes the first induction step.

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

$U_i$ equals either $U_{i-1}$ or $a_{N_i} + \epsilon_i$.

Suppose $U_i = U_{i-1}$.

Since by presupposition $U_{i-1}$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_{i-1}}$, so is $U_i$.

Because $N_i \ge N_{i-1}$, it follows that $\left\langle {a_n} \right\rangle_{n \ge N_i}$ is a subset of $\left\langle {a_n} \right\rangle_{n \ge N_{i-1}}$

Because $U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_{i-1}}$, $U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$ as well.

Suppose $U_i = a_{N_i} + \epsilon_i$.

Because $a_n < a_{N_i} + \epsilon_i$ whenever $n \ge N_i$, $U_i$ is an upper bound for $\left\langle {a_n} \right\rangle_{n \ge N_i}$.

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


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


 * $L_i = max(L_{i-1}, a_{N_i} - \epsilon_i)$ for $i \ge 1$

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 $i$:

which shows that since $\epsilon_i \to 0$ as $i \to \infty$ so does $U_i - L_i$.

Using this we find:

from which we deduce

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

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 $b \ge L_i$ for all $i$.

Moreover, since $a = b$ we have for all $i$:

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 $i \in \N$:

Now, pick a natural number $j$ such that

Since $\displaystyle \lim_{i \to \infty} (U_i - L_i) = 0$ a natural number $k$ exists such that

Set $l = max(j,k)$.

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.