Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 3
Theorem
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be a Cauchy sequence.
Then $\sequence {x_n}$ is convergent.
Proof
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 $m, n \ge N_i$
Let us study the sequence $\sequence {N_i}_{i \mathop \in \N}$.
First, we consider $N_0$.
We can choose $N_0 = N$ because $(1)$ $\size {a_n - a_m} < \epsilon$ 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 $\size {a_n - a_m} < \epsilon_{i + 1}$ 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:
\(\ds N_i\) | \(\le\) | \(\ds k' < N_{i + 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds N_i\) | \(<\) | \(\ds N_{i + 1}\) |
Thus, we have established that there exists a sequence $\sequence {N_i}_{i \mathop \in \N}$ that satisfies:
- $N_0 = N$
- $N_{i + 1} \ge N_i$ for every $i \in \N$
Since $\sequence {a_n}$ is Cauchy, we have for every $i \in \N$:
\(\ds \size {a_n - a_m}\) | \(<\) | \(\ds \epsilon_i\) | for every $m, n \ge N_i$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {a_n - a_{N_i} }\) | \(<\) | \(\ds \epsilon_i\) | for every $n \ge N_i$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a_{N_i} - \epsilon_i\) | \(<\) | \(\ds a_n < a_{N_i} + \epsilon_i\) | for every $n \ge N_i$ by Negative of Absolute Value: Corollary 1 |
$\Box$
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.
$a_{N_0} + \epsilon_0$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_0}$ because $a_n < a_{N_0} + \epsilon_0$ whenever $n \ge N_0$.
Therefore, $u_0$ 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 $\sequence {a_n}_{n \mathop \ge N_i}$.
$u_{i + 1}$ equals either $u_i$ or $a_{N_{i + 1} } + \epsilon_{i + 1}$.
First, assume that $u_{i + 1} = u_i$.
$u_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$ since $u_i$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$ by presupposition.
We have that $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$ is a subset of $\sequence {a_n}_{n \mathop \ge N_i}$ because $N_{i + 1} \ge N_i$.
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 $u_{i + 1} = a_{N_{i + 1} } + \epsilon_{i + 1}$.
$u_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_{i + 1} }$ 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 $\sequence {a_n}_{n \mathop \ge N_i}$ for every $i \in \N$.
$\Box$
$\sequence {u_i}$ converges
By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded.
Therefore, $\sequence {a_n}$ has a lower bound $b$.
For every $i \in \N$, $\sequence {a_n}_{n \mathop \ge N_i}$ is a subsequence of $\sequence {a_n}$.
Therefore, $b$ is also a lower bound for $\sequence {a_n}_{n \mathop \ge N_i}$.
$u_i$ 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.
$\Box$
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
$\Box$
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 $\sequence {a_n}_{n \mathop \ge N_i}$ for every $i \in \N$, we have for every $i \in \N$:
\(\ds 0\) | \(\le\) | \(\ds u_{i + 1} - l_{i + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {u_i, a_{N_{i + 1} } + \epsilon_{i + 1} } - \max \set {l_i, a_{N_{i + 1} } - \epsilon_{i + 1} }\) | Definitions of $u_{i + 1}$ and $l_{i + 1}$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds a_{N_{i + 1} } + \epsilon_{i + 1} - \max \set {l_i, a_{N_{i + 1} } - \epsilon_{i + 1} }\) | because $\min \set {u_i, a_{N_{i + 1} } + \epsilon_{i + 1} } \le a_{N_{i + 1} } + \epsilon_{i + 1}$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds a_{N_{i + 1} } + \epsilon_{i + 1} - \paren {a_{N_{i + 1} } - \epsilon_{i + 1} }\) | because $\max \size {l_i, a_{N_{i + 1} } - \epsilon_{i + 1} } \ge a_{N_{i + 1} } - \epsilon_{i + 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \epsilon_{i + 1}\) |
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:
\(\ds \lim_{i \mathop \to \infty} u_i - \lim_{i \mathop \to \infty} l_i\) | \(=\) | \(\ds \lim_{i \mathop \to \infty} (u_i - l_i)\) | Combined Sum Rule for Real Sequences as $\sequence {u_i}$ and $\sequence {l_i}$ converge | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{i \mathop \to \infty} u_i - \lim_{i \mathop \to \infty} l_i\) | \(=\) | \(\ds 0\) | as $\ds \lim_{i \mathop \to \infty} (u_i - l_i) = 0$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \lim_{i \mathop \to \infty} u_i\) | \(=\) | \(\ds \lim_{i \mathop \to \infty} l_i\) |
$\Box$
$\sequence {a_n}$ converges
Let $\ds a = \lim_{i \mathop \to \infty} u_i = \lim_{i \mathop \to \infty} l_i$.
We have for every $i \in \N$:
\(\ds l_i\) | \(\le\) | \(\ds a \le u_i\) | as $\sequence {l_i}$ is increasing and $\sequence {u_i}$ is decreasing | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 0\) | \(\le\) | \(\ds a - l_i \le u_i - l_i\) |
Also, we have for every $n \ge N_i$ for every $i \in \N$:
\(\ds l_i\) | \(\le\) | \(\ds a_n \le u_i\) | as $l_i$ is a lower bound for $\sequence {a_n}_{n \ge N_i}$ and $u_i$ is an upper bound for $\sequence {a_n}_{n \mathop \ge N_i}$ | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 0\) | \(\le\) | \(\ds a_n - l_i \le u_i - l_i\) |
A natural number $j$ exists such that, for every $i \ge j$:
\(\ds \size {u_i - l_i}\) | \(<\) | \(\ds \dfrac \epsilon 2\) | as $\ds \lim_{k \mathop \to \infty} (u_k - l_k) = 0$ |
Putting all this together, we find for every $n \ge N_j$:
\(\ds \size {a_n - a}\) | \(=\) | \(\ds \size {a_n - l_j + l_j - a}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {a_n - l_j} + \size {l_j - a}\) | Triangle Inequality for Real Numbers | |||||||||||
\(\ds \) | \(\le\) | \(\ds \size {u_j - l_j} + \size {l_j - a}\) | as $0 \le a_n - l_j \le u_j - l_j$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \size {u_j - l_j} + \size {u_j - l_j}\) | as $0 \le a - l_j \le u_j - l_j$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \size {u_j - l_j}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) | as $\ds \size {u_j - l_j} < \frac \epsilon 2$ |
This finishes the proof that $\sequence {a_n}$ is convergent.
$\blacksquare$