# Cauchy Sequence Converges on Real Number Line

## Theorem

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

Then $\sequence {a_n}$ is convergent.

In other words, $\struct {\R, \size {\,\cdot\,}}$ is a Banach space.

## Proof 1

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

We have the result Real Number Line is Metric Space.

Hence by Convergent Subsequence of Cauchy Sequence in Metric Space, it is sufficient to show that $\sequence {a_n}$ has a convergent subsequence.

Since $\sequence {a_n}$ is Cauchy, by Real Cauchy Sequence is Bounded, it is also bounded.

By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence.

Hence the result.

$\blacksquare$

## Proof 2

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

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

By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence $\sequence {a_{n_r} }$.

Let $a_{n_r} \to l$ as $r \to \infty$.

It is to be shown that $a_n \to l$ as $n \to \infty$.

Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.

Then $\dfrac \epsilon 2 > 0$.

Hence:

$(1): \quad \exists R \in \R: \forall r > R: \size {a_{n_r} - l} < \dfrac \epsilon 2$

We have that $\sequence {a_n}$ is a Cauchy sequence.

Hence:

$(2): \quad \exists N \in \R: \forall m > N, n > N: \size {x_m - x_n} \le \dfrac \epsilon 2$

Let $n > N$.

Let $r \in \N$ be sufficiently large that:

$n_r > N$

and:

$r > R$

Then $(1)$ is satisfied, and $(2)$ is satisfied with $m = n_r$.

So:

 $\, \displaystyle \forall n > N: \,$ $\displaystyle \size {a_n - l}$ $=$ $\displaystyle \size {a_n - a_{n_r} + a_{n_r} - l}$ $\displaystyle$ $\le$ $\displaystyle \size {a_n - a_{n_r} } + \size{a_{n_r} - l}$ Triangle Inequality for Real Numbers $\displaystyle$ $<$ $\displaystyle \dfrac \epsilon 2 + \dfrac \epsilon 2$ $\displaystyle$ $=$ $\displaystyle \epsilon$

So, given $\epsilon > 0$, we have found $n \in \R$ such that:

$\forall n > N: \size {a_n - l} < \epsilon$

Thus:

$x_n \to l$ as $n \to \infty$.

$\blacksquare$

## Proof 3

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:

$\epsilon_0 = \epsilon$
$\epsilon_{i + 1} < \epsilon_i$
$\displaystyle \lim_{i \mathop \to \infty} \epsilon_i = 0$

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 $\sequence {N_i}_{i \mathop \in \N}$.

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 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:

 $\displaystyle N_i$ $\le$ $\displaystyle k' < N_{i + 1}$ $\displaystyle \leadsto \ \$ $\displaystyle N_i$ $<$ $\displaystyle N_{i + 1}$

Thus, we have established that there exists a sequence $\sequence {N_i}_{i \mathop \in \N}$ that satisfies:

$N_0 = N$
$N_{i + 1} \ge N_i$

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

 $\displaystyle \size {a_n - a_m}$ $<$ $\displaystyle \epsilon_i$ for every $m, n \ge N_i$ $\displaystyle \leadsto \ \$ $\displaystyle \size {a_n - a_{N_i} }$ $<$ $\displaystyle \epsilon_i$ for every $n \ge N_i$ by choosing $m = N_i$ $\displaystyle \leadsto \ \$ $\displaystyle \size {a_n - a_{N_i} }$ $<$ $\displaystyle \epsilon_i$ for every $n > N_i$ $\displaystyle \leadsto \ \$ $\displaystyle a_{N_i} − \epsilon_i$ $<$ $\displaystyle a_n < a_{N_i} + \epsilon_i$ for every $n > N_i$ by Negative of Absolute Value: Corollary 1

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

$U_0 = a_{N_0} + \epsilon_0$
$U_{i + 1} = \min \set {U_i, a_{N_{i + 1} } + \epsilon_{i + 1} }$

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 $\sequence {a_n}_{n \mathop > N_i}$

We prove this by using the Principle of Mathematical Induction.

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

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

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

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

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

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

$U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop > N_{i + 1} }$ 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 $\sequence {a_n}_{n \mathop > N_i}$ 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 $i \in \N$, $\sequence {a_n}_{n \mathop > N_i}$ is a subsequence of $\sequence {a_n}$.

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

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

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.

$\Box$

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

$L_0 = a_{N_0} - \epsilon_0$
$L_{i + 1} = \max \set {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} }$

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 $\sequence {a_n}_{n \mathop > N_i}$ 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 $\sequence {a_n}_{n \mathop > N_i}$ for every $i \in \N$, we have for all $i \in \N$:

 $\displaystyle 0$ $\le$ $\displaystyle U_{i + 1} - L_{i + 1}$ $\displaystyle$ $=$ $\displaystyle \min \set {U_i, a_{N_{i + 1} } + \epsilon_{i + 1} } - \max \set {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} }$ Definitions of $U_{i + 1}$ and $L_{i + 1}$ $\displaystyle$ $\le$ $\displaystyle a_{N_{i + 1} } + \epsilon_{i + 1} - \max \set {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} }$ because $\min \set {U_i, a_{N_{i + 1} } + \epsilon_{i + 1} } \le a_{N_{i + 1} } + \epsilon_{i + 1}$ $\displaystyle$ $\le$ $\displaystyle a_{N_{i + 1} } + \epsilon_{i + 1} - \paren {a_{N_{i + 1} } - \epsilon_{i + 1} }$ because $\max \size {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} } \ge a_{N_{i + 1} } - \epsilon_{i + 1}$ $\displaystyle$ $=$ $\displaystyle 2 \epsilon_{i + 1}$

which shows that $U_i - L_i \to 0$ as $i \to \infty$ since $\displaystyle \lim_{i \mathop \to \infty} \epsilon_i = 0$.

We have:

 $\displaystyle \lim_{i \mathop \to \infty} U_i - \lim_{i \mathop \to \infty} L_i$ $=$ $\displaystyle \lim_{i \mathop \to \infty} (U_i - L_i)$ Combined Sum Rule for Real Sequences $\displaystyle \leadstoandfrom \ \$ $\displaystyle \lim_{i \mathop \to \infty} U_i - \lim_{i \mathop \to \infty} L_i$ $=$ $\displaystyle 0$ as $\displaystyle \lim_{i \mathop \to \infty} (U_i - L_i) = 0$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \lim_{i \mathop \to \infty} U_i$ $=$ $\displaystyle \lim_{i \mathop \to \infty} L_i$

$\Box$

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

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

 $\displaystyle L_i$ $\le$ $\displaystyle a \le U_i$ as $\sequence {L_i}$ is increasing and $\sequence {U_i}$ is decreasing $\displaystyle \leadstoandfrom \ \$ $\displaystyle 0$ $\le$ $\displaystyle a - L_i \le U_i - L_i$

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

 $\displaystyle L_i$ $\le$ $\displaystyle a_n \le U_i$ as $L_i$ is a lower bound for $\sequence {a_n}_{n > N_i}$ and $U_i$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle 0$ $\le$ $\displaystyle a_n - L_i \le U_i - L_i$

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

 $\displaystyle \size {U_i - L_i}$ $<$ $\displaystyle \dfrac \epsilon 2$ as $\displaystyle \lim_{k \mathop \to \infty} (U_k - L_k) = 0$

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

 $\displaystyle \size {a_n - a}$ $=$ $\displaystyle \size {a_n - L_j + L_j - a}$ $\displaystyle$ $\le$ $\displaystyle \size {a_n - L_j} + \size {L_j - a}$ Triangle Inequality for Real Numbers $\displaystyle$ $\le$ $\displaystyle \size {U_j - L_j} + \size {L_j - a}$ as $0 \le a_n - L_j \le U_j - L_j$ $\displaystyle$ $\le$ $\displaystyle \size {U_j - L_j} + \size {U_j - L_j}$ as $0 \le a - L_j \le U_j - L_j$ $\displaystyle$ $=$ $\displaystyle 2 \size {U_j - L_j}$ $\displaystyle$ $<$ $\displaystyle \epsilon$ as $\size {U_j - L_j} < \frac \epsilon 2$

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

$\blacksquare$
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