# Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition

## 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 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, 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:

\(\ds \forall n > N: \, \) | \(\ds \size {a_n - l}\) | \(=\) | \(\ds \size {a_n - a_{n_r} + a_{n_r} - l}\) | |||||||||||

\(\ds \) | \(\le\) | \(\ds \size {a_n - a_{n_r} } + \size{a_{n_r} - l}\) | Triangle Inequality for Real Numbers | |||||||||||

\(\ds \) | \(<\) | \(\ds \dfrac \epsilon 2 + \dfrac \epsilon 2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \epsilon\) |

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 $\sequence {a_n}$ 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.

$\blacksquare$

## Proof 4

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

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

Thus $\sequence {a_n}$ is both bounded above and bounded below.

Let us create a monotone subsequence $\sequence {b_n}$ of $\sequence {a_n}$ using the following construction:

For each $m \in \N$, let $S_m$ denote the set of elements of $\sequence {a_n}$ from $m$ onwards:

- $S_m = \set {a_n: n \ge m}$

From Real Cauchy Sequence is Bounded, we have that $S_m$ is also bounded.

Hence, by the Continuum Property, $\sup S_m$ exists.

Let $b_m := \sup S_m$.

Because $S_{m + 1} \subseteq S_m$, it follows from Supremum of Set of Real Numbers is at least Supremum of Subset that:

- $\sup S_{m + 1} \le \sup S_m$

Thus $\sequence {b_m}$ is decreasing.

By definition of $b_m$, we also have that:

- $b_m \ge a_m$

So, as $\sequence {a_n}$ is bounded below, so is $\sequence {b_m}$.

So, by the Monotone Convergence Theorem (Real Analysis), $\sequence {b_m}$ is convergent.

Let $\sequence {b_m}$ converge to $l$.

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

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

We also have from Subsequence of Real Cauchy Sequence is Cauchy that $\sequence {b_n}$ is also a Cauchy sequence in $\R$.

Then there exists:

- $N_1 \in \N$ such that $\size {a_m - a_n} < \epsilon$ for $m, n \ge N_1$
- $N_2 \in \N$ such that $\size {l - b_n} < \epsilon$ for $m \ge N_2$

Let $N = \max \set {N_1, N_2}$.

By definition of $b_N$, we have that $b_N - \epsilon$ is not an upper bound of $S_N = \set {a_n: n \ge N}$.

Thus there exists $M \ge N$ such that $a_M > b_N - \epsilon$.

We note also that $a_M \le b_N$ because $b_N$ is an upper bound for $S_N$.

Now let $n \ge N$.

We have that:

\(\ds \size {a_n - l}\) | \(=\) | \(\ds \size {a_n - a_M + a_M - b_N + b_N - l}\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \size {a_n - a_M} + \size {a_M - b_N} + \size {b_N - l}\) | Triangle Inequality | |||||||||||

\(\ds \) | \(<\) | \(\ds 3 \epsilon\) |

Hence by Convergence by Multiple of Error Term we have that $\sequence {a_n}$ converges to $l$.

$\blacksquare$

## Also known as

**Cauchy's Convergence Criterion** is also known as the **Cauchy convergence condition**.