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

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

Since $\sequence {a_n}$ is Cauchy, a natural number $N$ exists such that:


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

We aim to show that $\sequence {a_n}$ converges.

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


 * $\size {a_n - a} < \epsilon$ for every $n > N'$

Let $\sequence {\epsilon_i}_{i \in \N}$ 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 $\sequence {a_n}$ is Cauchy, for each $\epsilon_i$ a natural number $N_i$ exists such that:


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

Let us study the sequence $\sequence {N_i}_{i \in \N}$.

First, we consider $N_0$.

We can choose $N_0 = N$ because $(1)$ $\V{a_n - a_m} < \epsilon$ 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 $i \in \N$:


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

and:


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

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

Either:


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

or:


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

In the first case, we can choose $N_{i + 1} = N_i$ since $\size {a_n - a_m} < \epsilon_{i+1}$ 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 $\size {a_{n'} - a_{m'} } < \epsilon_{i + 1}$ since $\size {a_n - a_m} < \epsilon_{i + 1}$ 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 $\sequence {N_i}_{i \in \N}$ that satisfies:


 * $N_0 = N$


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

Since $\sequence {a_n}$ is Cauchy, we have for every $i \in \N$:

Define a real sequence $\sequence {U_i}_{i \in \N}$ by:


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


 * $U_{i + 1} = \min \set {U_i, a_{N_{i + 1} } + \epsilon_{i + 1} }$

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

Therefore $\sequence {U_i}$ is decreasing.

$U_i$ is an upper bound for $\sequence {a_n}_{n > N_i}$
We prove this by using the Principle of Mathematical Induction.

$a_{N_0} + \epsilon_0$ is an upper bound for $\sequence {a_n}_{n > N_0}$ because $a_n < a_{N_0} + \epsilon_0$ whenever $n > N_0$.

Therefore, $U_0$ is an upper bound for $\sequence {a_n}_{n > N_0}$ 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 $\sequence {a_n}_{n > N_{i + 1} }$ if $U_i$ is an upper bound for $\sequence {a_n}_{n > N_i}$.

$U_{i + 1}$ 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 $\sequence {a_n}_{n > N_i}$ since $U_i$ is an upper bound for $\sequence {a_n}_{n > N_i}$ by presupposition.

We have that $\sequence {a_n}_{n > N_{i + 1} }$ is a subset of $\sequence {a_n}_{n > N_i}$ because $N_{i + 1} \ge N_i$.

Therefore, $U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n > N_{i + 1} }$ because $U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n > N_i}$.

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

$U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n > N_{i + 1} }$ because $a_n < a_{N_{i + 1} } + \epsilon_{i + 1}$ whenever $n > N_{i + 1}$.

This concludes the proof that $U_i$ is an upper bound for $\sequence {a_n}_{n > N_i}$ for every $i \in \N$.

$\sequence {U_i}$ converges
By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded.

Therefore, $\sequence {a_n}$ has a lower bound $b$.

For every $i \in \N$, $\sequence {a_n}_{n > N_i}$ is a subsequence of $\sequence {a_n}$.

Therefore, $b$ is also a lower bound for $\sequence {a_n}_{n > N_i}$.

$U_i$ is an upper bound for $\sequence {a_n}_{n > N_i}$.

Therefore, $b$ is less than or equal to $U_i$ for every $i$.

So, $\sequence {U_i}$ is bounded below.

Since $\sequence {U_i}$ is decreasing, its first element is an upper bound for $\sequence {U_i}$.

Since $\sequence {U_i}$ is bounded below and above, it is bounded.

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

Now, define a real sequence $\sequence {L_i}_{i \in \N}$ by:


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


 * $L_{i + 1} = \max \set {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} }$

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


 * $\sequence {L_i}$ is increasing


 * $L_i$ is a lower bound for $\sequence {a_n}_{n > N_i}$ for every $i \in \N$


 * $\sequence {L_i}$ converges

The limits of $\sequence {U_i} $ and $\sequence {L_i}$ as $i \to \infty$ are equal
Since $U_i$ and $L_i$ are, respectively, upper and lower bounds for $\sequence {a_n}_{n > N_i}$ 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 $\displaystyle a := \lim_{i \mathop \to \infty} U_i = \lim_{i \mathop \to \infty} L_i$.

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

Also, we have for every $n > N_i$ for every $i \in \N$:

A natural number $j$ exists such that, for every $i \ge j$:

Putting all this together, we find for every $n > N_j$:

which finishes the proof that $\sequence {a_n}$ is convergent.