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

== Theorem ==
== 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.
A sequence $\sequence {\epsilon_i}$ is introduced that can be considered a sequence of values or levels that a positive, real variable $\epsilon$ ($ \ge 1$) passes through on its way towards $0$.
At each level an upper bound and a lower bound for a subsequence of $\sequence {a_n}$ are defined.
All such upper bounds form a sequence of upper bounds and the lower bounds a sequence of lower bounds.
These two sequences are the ones that are used to show that $\sequence {a_n}$ converges.
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
Since $\sequence {a_n}$ is Cauchy, for every $\epsilon \in \R_{>0}$ 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 a real number $a$ exists and for every $\epsilon \in \R_{>0}$ a natural number $N'$ exists 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:
:$\epsilon_0 = 1$
:$\epsilon_{i + 1} < \epsilon_i$ for every $i \in \N$
:$\ds \lim_{i \mathop \to \infty} \epsilon_i = 0$
=== $a_n$ is between and arbitrarily close to elements of two 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 $\sequence {N_i}_{i \mathop \in \N}$.
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, $N_{i + 1}$ should be at least as great as $N_i$.
Therefore, we 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 $\sequence {a_n}_{n \mathop \ge N_i}$.
This concludes the first part of the second induction step.
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 $n \ge N_{i + 1}$.
In other words, $a_{N_{i + 1} } + \epsilon_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$.
Therefore, $u_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$ 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 $i \in \N$.
This means that, for every $n \ge N_i$ 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 $i \in \N$:
Let $\epsilon \in \R_{>0}$.
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.
Category:Cauchy's Convergence Criterion