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

Proof
The aim is to define two sequences whose elements are respectively upper and lower bounds to subsequences of the sequence $\sequence {a_n}$.

It is then shown that these two sequences converge to the same limit.

This is used to prove that $\sequence {a_n}$ converges.

Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.

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 $\sequence {\epsilon_i}_{i \mathop \in \N}$ be a sequence of strictly positive real numbers that satisfies:
 * $\epsilon_0 = \epsilon$
 * $\epsilon_{i + 1} < \epsilon_i$ for every $i \in \N$
 * $\ds \lim_{i \mathop \to \infty} \epsilon_i = 0$

$\sequence {a_n}$ is between two other sequences
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 $m, n \ge N_i$

Therefore, we have for every $i \in \N$:

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

First, we consider $N_0$.

We have $\epsilon = \epsilon_0$.

Also, $\sequence {a_n}$ is Cauchy:


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

Also:
 * $\size {a_n - a_m} < \epsilon$ for every $m, n \ge N_0$

Therefore, we can choose:
 * $N_0 = N$

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

We have, for every $i \in \N$:
 * $\size {a_n - a_m} < \epsilon_i$ for every $m, n \ge N_i$

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

We have that $\epsilon_{i + 1} < \epsilon_i$ for every $i \in \N$.

Accordingly, the condition $\size {a_n - a_m} < \epsilon_{i + 1}$ is stricter than $\size {a_n - a_m} < \epsilon_i$.

Therefore, we an choose the value of $N_{i + 1}$ so that:
 * $N_{i + 1} \ge N_i$ for every $i \in \N$

Define a real sequence $\sequence {u_i}_{i \mathop \in \N}$ by:
 * $u_0 = a_{N_0} + \epsilon_0$
 * $u_{i + 1} = \min \set {u_i, a_{N_{i + 1} } + \epsilon_{i + 1} }$ for every $i \in \N$

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 $\sequence {a_n}_{n \mathop \ge N_i}$
We prove this by using the Principle of Mathematical Induction.

We have that $a_n < a_{N_0} + \epsilon_0$ whenever $n \ge N_0$.

Therefore, $a_{N_0} + \epsilon_0$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_0}$.

Furthermore, $u_0$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_0}$ 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 $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$ if $u_i$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$.

Assume that $u_i$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$.

$u_{i + 1}$ equals either $u_i$ or $a_{N_{i + 1} } + \epsilon_{i + 1}$.

First, assume that $u_{i + 1} = u_i$.

We have that $u_i$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$ by presupposition.

Therefore, $u_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$ as $u_{i + 1} = u_i$.

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

Therefore, an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$ is also an upper bound for $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$.

Therefore, $u_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$ because $u_{i +

1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$.

Now, assume that $u_{i + 1} = a_{N_{i + 1} } + \epsilon_{i + 1}$.

We have that $a_n < a_{N_{i + 1} } + \epsilon_{i + 1}$ whenever $n \ge N_{i + 1}$.

Therefore, $u_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$ as $u_{i + 1} = a_{N_{i + 1} } + \epsilon_{i + 1}$.

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

$\sequence {u_i}$ converges
$\sequence {a_n}$ is bounded by Real Cauchy Sequence is Bounded.

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

$\sequence {a_n}_{n \mathop \ge N_i}$ is a subsequence of $\sequence {a_n}$ for every $i \in \N$.

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

We have that $u_i$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$.

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.

We have that$ \sequence {u_i}$ is bounded and decreasing.

Therefore $\sequence {u_i}$ converges by the Monotone Convergence Theorem.

Now, define a real sequence $\sequence {l_i}_{i \mathop \in \N}$ by:
 * $l_0 = a_{N_0} - \epsilon_0$
 * $l_{i + 1} = \max \set {l_i, a_{N_{i + 1} } - \epsilon_{i + 1} }$ for every $i \in \N$

$l_i$ is a lower bound for $\sequence {a_n}_{n \mathop \ge N_i}$, and $\sequence {l_i}$ converges
An analysis of $\sequence {l_i}$ is similar to the one above of $\sequence {u_i}$.

Therefore, it is not given here.

It produces the following results:


 * $\sequence {l_i}$ is increasing


 * $l_i$ is a lower bound for $\sequence {a_n}_{n \mathop \ge N_i}$ 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
We have that $u_i$ and $l_i$ are, respectively, upper and lower bounds for $\sequence {a_n}_{n \mathop \ge N_i}$ for every $i \in \N$.

Therefore, we have for every $i \in \N$:

This shows that $u_i - l_i \to 0$ as $i \to \infty$ since $\ds \lim_{i \mathop \to \infty} \epsilon_i = 0$.

We have:

$\sequence {a_n}$ converges
Let $\ds a = \lim_{i \mathop \to \infty} u_i = \lim_{i \mathop \to \infty} l_i$.

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

Also, we have for every $n \ge N_i$ 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 $n \ge N_j$:

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