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

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 $i \in \N$:

 $\ds \size {a_n - a_m}$ $<$ $\ds \epsilon_i$ 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

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 $i \in \N$

$\Box$

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.

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 $i \in \N$.

$\Box$

$\sequence {u_i}$ 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.

$\Box$

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

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 $i \in \N$:

 $\ds l_i$ $\le$ $\ds a_n \le u_i$ $\ds \leadsto \ \$ $\ds l_i$ $\le$ $\ds u_i$ $\ds \leadsto \ \$ $\ds 0$ $\le$ $\ds u_i - l_i$

Therefore, 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 $\max \set {l_i, a_{N_{i + 1} } - \epsilon_{i + 1} } \ge a_{N_{i + 1} } - \epsilon_{i + 1}$ $\ds$ $=$ $\ds 2 \epsilon_{i + 1}$

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 \ \$ $\ds 0$ $\le$ $\ds a - l_i \le u_i - l_i$ $\ds \leadsto \ \$ $\ds \size {a - l_i}$ $\le$ $\ds \size {u_i - l_i}$

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 \ \$ $\ds 0$ $\le$ $\ds a_n - l_i \le u_i - l_i$ $\ds \leadsto \ \$ $\ds \size {a_n - l_i}$ $\le$ $\ds \size {u_i - l_i}$

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

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 $\size {a_n - l_j} \le \size {u_j - l_j}$ $\ds$ $\le$ $\ds \size {u_j - l_j} + \size {u_j - l_j}$ as $\size {a - l_j} \le \size {u_j - l_j}$ $\ds$ $=$ $\ds 2 \size {u_j - l_j}$ $\ds$ $<$ $\ds \epsilon$ 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.