5.8. The Bolzano-Weierstrass Theorem

If g is a limit we know that there are infinitely many sequence members in every

ε

-neigbhourhood of g. Placed however in special way, namely all of them in a row from a certain

n_{0}

upwards! Omitting this special condition weakens the convergence idea to a related concept.

Definition: Let $(a_{n})$ be any sequence and $x \in ℝ$ arbitrary.

x is called an accumulation point of

(a_{n})

[5.8.1]

if there are infinitely many sequence members in every $ε$ -neigbourhood $] x - ε, x + ε [$ of x.

As a notedly difference to a limit the uniqueness property is no longer valid. The sequence $({(- 1)}^{n})$ e.g. has two accumulation points: There are infinitely many members in every $ε$ -neigbourhood of 1 (namely all who's index is even) and similarly we find infinitely many in each $ε$ -neigbourhood of $- 1$ .

Of cause there are sequences, take e.g. $(n)$ , with no accumulation points at all.

Using the following idea an accumulation point may appear as a limit of a suitable sequence.

Definition: Let $(a_{n})$ be any sequence (not necessary in $ℝ$ ). If $(k_{n})$ is strictly increasing in $ℕ^{*}$ the sequence

(a_{k_{n}}) = (a_{n}) \circ (k_{n})

[5.8.2]

is called a subsequence of $(a_{n})$ .

Consider:

The fact that $(k_{n})$ is strictly increasing guarantees, that a subsequence selects infinitely many members in strict succession.

Furthermore an easy proof on induction shows that:

k_{n} \geq n for all n \in ℕ^{*}

[5.8.3]

$(1) = ({(- 1)}^{2 n}) = ({(- 1)}^{n}) \circ (2 n)$ is a subsequence of $({(- 1)}^{n})$ . This proves that divergent sequences are well allowed to have convergent subsequences. Due to the following proposition however this cannot apply to all subsequences.

Proposition:

a_{n} \to g \Leftrightarrow a_{k_{n}} \to g

for each subsequence

(a_{k_{n}})

[5.8.4]

Proof:

" $\Rightarrow$ ": Let $(a_{k_{n}})$ be an arbitrary subsequence. At first we get an $n_{0} \in ℕ^{*}$ for each $ε > 0$ such that

| a_{n} - g | < ε for all n \geq n_{0}

Acoording to [5.8.3] this holds more than ever for all $n \geq k_{n_{0}}$ and especially for all $k_{n} \geq k_{n_{0}}$ . Due to the monotony of $(k_{n})$ we thus conclude:

| a_{k_{n}} - g | < ε for all n \geq n_{0}

" $\Leftarrow$ ": $(a_{n}) = (a_{n}) \circ (n)$ is a subsequence of itself, thus converging to g from the premise.

Accumulation points are nothing else than special limits, more precisely:

Proposition:

x is an accumulation point of

(a_{n}) \Leftrightarrow

there is a subsequence

(a_{k_{n}})

such
that

a_{k_{n}} \to x

[5.8.5]

Proof:

" $\Rightarrow$ ": We construct a strictly increasing sequence ${(k_{n})}_{n \geq 0}$ by recursion with

| a_{k_{n}} - x | < \frac{1}{n} for all n \in ℕ^{*}

From that the nesting theorem [5.5.8] leads to our assertion: $a_{k_{n}} \to x$ . Now to the construction details:

We set $k_{0} ≔ 0$
and for $n \geq 1$ we take the set $M_{n} ≔ {k \in ℕ^{*} | k > k_{n - 1} \land a_{k} \in] x - \frac{1}{n}, x + \frac{1}{n} [}$ . $M_{n} \neq \emptyset$ as there are infinitely many sequence members in every neigbourhood of x. With

$k_{n} ≔ min M_{n}$

we have: $k_{n} \in M_{n},$ i.e. $k_{n} > k_{n - 1}$ and $a_{k_{n}} \in] x - \frac{1}{n}, x + \frac{1}{n} [$ , thus $(k_{n})$ is strictly increasing and satisfies the estimate $| a_{k_{n}} - x | < \frac{1}{n} .$

" $\Leftarrow$ ": If $a_{k_{n}} \to x$ holds for a certain subsequence we have infinitely many sequence members of $(a_{k_{n}})$ in every $ε$ -neigbourhood of x. These members are of course sequence members of $(a_{n})$ as well.

The following theorem assures the existence of accumulation points for bounded real sequences.

Theorem (Bolzano-Weierstrass theorem): The following implication holds for every sequence $(a_{n})$ in $ℝ$ .

(a_{n})

is bounded then

(a_{n})

possesses at least one accumulation point.

[5.8.6]

Proof: We employ the completeness axiom
i

Every non-empty, bounded subset of $ℝ$ has a greatest lower bound, its infimum, and a least upper bound, its supremum.

twice. As $(a_{n})$ is bounded we see that for each $n \in ℕ^{*}$ the set ${a_{k} | k \geq n}$ is a non-empty, bounded subset of $ℝ$ . Thus

x_{n} ≔ \inf {a_{k} | k \geq n}

is well defined. Obviously we have ${a_{k} | k \geq n} \supset {a_{k} | k \geq n + 1}$ so that every lower bound of ${a_{k} | k \geq n}$ is a lower bound of ${a_{k} | k \geq n + 1}$ as well. Especially the infimum $x_{n}$ is to be found among all the lower bounds of ${a_{k} | k \geq n + 1}$ . This implies $x_{n} \leq x_{n + 1}$ , i.e. $(x_{n})$ is increasing. Furthermore $(x_{n})$ is bounded, as $x_{1} \leq x_{n} \leq a_{n} \leq c$
i

$(a_{n})$ is bounded!

, thus the set ${x_{n} | n \in ℕ^{*}}$ is a non-empty, bounded subset of $ℝ$ . Using the completeness theorem a second time we set

x ≔ \sup {x_{n} | n \in ℕ^{*}}

We will now prove that x is an accumulation point of $(a_{n})$ . According to [5.8.5] it is sufficient to construct a stictly increasing sequence ${(k_{n})}_{n \geq 0}$ such that

x - \frac{1}{n} < a_{k_{n}} < x + \frac{1}{n} for all n \in ℕ^{*}

[0]

We do this recursively:

$k_{0} ≔ 0$

For $n \in ℕ^{*}$ the number $x - \frac{1}{n}$ is certainly no upper bound of ${x_{n} | n \in ℕ^{*}}$ (x is the least of these bounds!) thus will be exceeded by at least one $x_{m}$ :

x - \frac{1}{n} < x_{m}

[1]

As $(x_{n})$ is increasing we may assume $m > k_{n - 1}$ without any restriction.

Similarly $x_{m} + \frac{1}{n}$ is no lower bound for ${a_{k} | k \geq m}$ , so there is a $k_{n} \geq m$ such that

a_{k_{n}} < x_{m} + \frac{1}{n} .

[2]

From [1] and [2] (and keeping in mind that $x_{m}$ is a lower bound of ${a_{k} | k \geq m} ∋ a_{k_{n}}$ ) we see that our sequence ${(k_{n})}_{n \geq 0}$ satisfies:

k_{n} \geq m > k_{n - 1}

x - \frac{1}{n} < x_{m} \leq a_{k_{n}} < x_{m} + \frac{1}{n} \leq x + \frac{1}{n}

Thus we have achieved our aim [0].

There are some implications to be drawn from the Bolzano-Weistrass theorem. At first we characterize uniquely the convergent sequences among the bounded ones.

Proposition: For any real sequence $(a_{n})$ we have:

(a_{n})

is convergent

\Leftrightarrow

(a_{n})

is bounded with exactly one
accumulation point

[5.8.7]

Proof:

" $\Rightarrow$ ": Let $a_{n} \to g$ . Then we know that $(a_{n})$ is bounded (cf. [5.5.1]) and that every $ε$ -neigbourhood of g contains infinitely many sequence members, thus g is an accumulation point. It is the only one, because if $x \neq g$ would be a further accumulation point we would have the following contradictory statements for $ε ≔ \frac{| x - g |}{2} > 0$ :

There are inifinitely many sequence members in $] x - ε, x + ε [$ .
There are at most finitely many sequence members lacking in $] g - ε, g + ε [$ .
$] x - ε, x + ε [\cap] g - ε, g + ε [= \emptyset$ as $ε$ is half the distance between x and g.

" $\Leftarrow$ ": Now let g be the only accumulation point of $(a_{n})$ . g proves to be its limit: If $] g - ε, g + ε [$ is an arbitrary $ε$ -neighbourhood of g we need to find an $n_{0} \in ℕ^{*}$ such that

a_{n} \in] g - ε, g + ε [for all n \geq n_{0} .

Suppose there is no such $n_{0}$ . Starting with $k_{0} ≔ 0$ we find (recursively) for each n a $k_{n} > k_{n - 1}$ such that $a_{k_{n}}$ is lacking in $] g - ε, g + ε [$ . Thus we get a subsequence $(a_{k_{n}})$ , bounded as well as $(a_{n})$ is bounded, with an accumulation point x according to the Bolzano-Weierstrass theorem. As there are no sequence members of $(a_{k_{n}})$ at all in $] g - ε, g + ε [$ we conclude that $x \neq g$ . On the other hand x is an accumulation point of $(a_{n})$ as well, so due to uniqueness we have: $x = g$ . Contradiction!

Consider:

[5.8.7] won't work without boundedness. The divergent sequence

$(n (1 + {(- 1)}^{n})) = (0,4,0,8,0,12,0, \dots)$

e.g. has only one accumulation point.

A second application of the Bolzano-Weierstrass theorem results in a new convergence criterion. We know from [5.5.7] that the members of convergent sequences cluster more and more if the index number increases. Surprisingly the reverse is also true provided the sequence is a real sequence.

Theorem (Cauchy test): Let $(a_{n})$ be any sequence in $ℝ$ . If there is an $n_{0} \in ℕ^{*}$ for each $ε > 0$ such that

| a_{n} - a_{m} | < ε for all n, m \geq n_{0}

[5.8.8]

then $(a_{n})$ is convergent.

Proof: At first we see that $(a_{n})$ is bounded: From the premise we find an $n^{*}$ for 1 such that

| a_{n} | - | a_{n^{*}} | \leq | a_{n} - a_{n^{*}} | < 1 for all n \geq n^{*}

Consequently we have: $| a_{n} | \leq \max {1 + | a_{n^{*}} |, | a_{1} |, \dots, | a_{n^{*} - 1} |}$ for all n.

According Bolzano-Weierstrass $(a_{n})$ has an accumulation point g. We show $a_{n} \to g$ and take an arbitrary $ε > 0$ to that end. From the Cauchy condition [5.8.8] we find an $n_{0} \in ℕ^{*}$ such that

| a_{n} - a_{m} | < \frac{ε}{2} for all n, m \geq n_{0} .

As there are inifinitely many sequence members in every neigbourhood of g there will be an $n_{1} \geq n_{0}$ with

| a_{n_{1}} - g | < \frac{ε}{2} .

Thus we have for all $n \geq n_{0} :$

| a_{n} - g | = | a_{n} - a_{n_{1}} + a_{n_{1}} - g | \leq | a_{n} - a_{n_{1}} | + | a_{n_{1}} - g | < \frac{ε}{2} + \frac{ε}{2} = ε

Consider:

The Cauchy test [5.8.8] is often abbreviated to:

Every Cauchy sequence in $ℝ$ is convergent.
The Cauchy test guarantees the existence of a limit but does not provide any information on its value.

The Bolzano-Weierstrass theorem and thus all the conclusions is bound to the reals, more precisely to the completeness axiom. In $ℚ$ e.g. the theorem would fail. When introducing the Babylonian root algorithm (cf. [5.7.11]) we used a bounded rational sequence satisfying the Cauchy test. This sequence has no accumulation point, not to mention a limit in $ℚ$ .

5.7. 5.9.