# 5.8. The Bolzano-Weierstrass Theorem

If g is a limit we know that there are infinitely many sequence members in every $\epsilon$-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 $\left({a}_{n}\right)$ be any sequence and $x\in ℝ$ arbitrary.

 x is called an accumulation point of  $\left({a}_{n}\right)$, [5.8.1]

if there are infinitely many sequence members in every $\epsilon$-neigbourhood $\right]x-\epsilon ,x+\epsilon \left[$ of x.

As a notedly difference to a limit the uniqueness property is no longer valid.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

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

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

Definition:  Let $\left({a}_{n}\right)$ be any sequence (not necessary in $ℝ$). If $\left({k}_{n}\right)$ is strictly increasing in ${ℕ}^{\ast }$ the sequence

 $\left({a}_{{k}_{n}}\right)=\left({a}_{n}\right)\circ \left({k}_{n}\right)$ [5.8.2]

is called a subsequence of $\left({a}_{n}\right)$.

Consider:

• The fact that $\left({k}_{n}\right)$ is strictly increasing guarantees, that a subsequence selects infinitely many members in strict succession.

Furthermore an easy proof on induction shows that:

 [5.8.3]

• $\left(1\right)=\left({\left(-1\right)}^{2n}\right)=\left({\left(-1\right)}^{n}\right)\circ \left(2n\right)$ is a subsequence of $\left({\left(-1\right)}^{n}\right)$. 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:

 ${a}_{n}\to g⇔{a}_{{k}_{n}}\to g$   for each subsequence  $\left({a}_{{k}_{n}}\right)$ [5.8.4]

Proof:

"$⇒$":   Let  $\left({a}_{{k}_{n}}\right)$ be an arbitrary subsequence. At first we get an ${n}_{0}\in {ℕ}^{\ast }$ for each $\epsilon >0$ such that

.

Acoording to [5.8.3] this holds more than ever for all $n\ge {k}_{{n}_{0}}$ and especially for all ${k}_{n}\ge {k}_{{n}_{0}}$. Due to the monotony of $\left({k}_{n}\right)$ we thus conclude:

.

"$⇐$":   $\left({a}_{n}\right)=\left({a}_{n}\right)\circ \left(n\right)$ 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 $\left({a}_{n}\right)⇔$ there is a subsequence $\left({a}_{{k}_{n}}\right)$ such that  ${a}_{{k}_{n}}\to x$ [5.8.5]

Proof:

"$⇒$":   We construct a strictly increasing sequence ${\left({k}_{n}\right)}_{n\ge 0}$ by recursion with

.

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 $n\ge 1$ we take the set  ${M}_{n}≔\left\{k\in {ℕ}^{\ast }|k>{k}_{n-1}\wedge {a}_{k}\in \right]x-\frac{1}{n},x+\frac{1}{n}\left[\right\}$. ${M}_{n}\ne \varnothing$ as there are infinitely many sequence members in every neigbourhood of x. With

${k}_{n}≔min{M}_{n}$

we have: ${k}_{n}\in {M}_{n}\text{,}$ i.e. ${k}_{n}>{k}_{n-1}$ and ${a}_{{k}_{n}}\in \right]x-\frac{1}{n},x+\frac{1}{n}\left[$, thus  $\left({k}_{n}\right)$ is strictly increasing and satisfies the estimate $|{a}_{{k}_{n}}-x|<\frac{1}{n}\text{.}$

"$⇐$":   If ${a}_{{k}_{n}}\to x$ holds for a certain subsequence we have infinitely many sequence members of $\left({a}_{{k}_{n}}\right)$ in every $\epsilon$-neigbourhood of x. These members are of course sequence members of $\left({a}_{n}\right)$ 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 $\left({a}_{n}\right)$ in $ℝ$.

 If $\left({a}_{n}\right)$ is bounded then $\left({a}_{n}\right)$ 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 $\left({a}_{n}\right)$ is bounded we see that for each $n\in {ℕ}^{\ast }$ the set $\left\{{a}_{k}|k\ge n\right\}$ is a non-empty, bounded subset of $ℝ$. Thus

${x}_{n}≔\mathrm{inf}\left\{{a}_{k}|k\ge n\right\}$.

is well defined. Obviously we have $\left\{{a}_{k}|k\ge n\right\}\supset \left\{{a}_{k}|k\ge n+1\right\}$ so that every lower bound of $\left\{{a}_{k}|k\ge n\right\}$ is a lower bound of  $\left\{{a}_{k}|k\ge n+1\right\}$ as well. Especially the infimum ${x}_{n}$ is to be found among all the lower bounds of $\left\{{a}_{k}|k\ge n+1\right\}$. This implies ${x}_{n}\le {x}_{n+1}$, i.e. $\left({x}_{n}\right)$ is increasing. Furthermore $\left({x}_{n}\right)$ is bounded, as ${{x}_{1}\le {x}_{n}\le {a}_{n}}{\le c}$
 i $\left({a}_{n}\right)$ is bounded!
, thus the set $\left\{{x}_{n}|n\in {ℕ}^{\ast }\right\}$ is a non-empty, bounded subset of $ℝ$. Using the completeness theorem a second time we set

$x≔\mathrm{sup}\left\{{x}_{n}|n\in {ℕ}^{\ast }\right\}$.

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

 [0]

We do this recursively:

• ${k}_{0}≔0$
• For $n\in {ℕ}^{\ast }$ the number $x-\frac{1}{n}$ is certainly no upper bound of  $\left\{{x}_{n}|n\in {ℕ}^{\ast }\right\}$ (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 $\left({x}_{n}\right)$ is increasing we may assume $m>{k}_{n-1}$ without any restriction.

Similarly ${x}_{m}+\frac{1}{n}$ is no lower bound for $\left\{{a}_{k}|k\ge m\right\}$, so there is a  ${k}_{n}\ge m$ such that

 ${a}_{{k}_{n}}<{x}_{m}+\frac{1}{n}\text{.}$ [2]

From [1] and [2] (and keeping in mind that ${x}_{m}$ is a lower bound of  $\left\{{a}_{k}|k\ge m\right\}\ni {a}_{{k}_{n}}$ ) we see that our sequence ${\left({k}_{n}\right)}_{n\ge 0}$ satisfies:

${k}_{n}\ge m>{k}_{n-1}$

$x-\frac{1}{n}<{x}_{m}\le {a}_{{k}_{n}}<{x}_{m}+\frac{1}{n}\le x+\frac{1}{n}$

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 $\left({a}_{n}\right)$ we have:

 $\left({a}_{n}\right)$ is convergent$⇔$ $\left({a}_{n}\right)$ is bounded with exactly one accumulation point [5.8.7]

Proof:

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

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

"$⇐$":   Now let g be the only accumulation point of  $\left({a}_{n}\right)$. g proves to be its limit: If $\right]g-\epsilon ,g+\epsilon \left[$ is an arbitrary $\epsilon$-neighbourhood of g we need to find an ${n}_{0}\in {ℕ}^{\ast }$ such that

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 $\right]g-\epsilon ,g+\epsilon \left[$. Thus we get a subsequence $\left({a}_{{k}_{n}}\right)$, bounded as well as $\left({a}_{n}\right)$ is bounded, with an accumulation point x according to the Bolzano-Weierstrass theorem. As there are no sequence members of  $\left({a}_{{k}_{n}}\right)$ at all in $\right]g-\epsilon ,g+\epsilon \left[$ we conclude that $x\ne g$. On the other hand x is an accumulation point of  $\left({a}_{n}\right)$ as well, so due to uniqueness we have: $x=g$.   Contradiction!

Consider:

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

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

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 $\left({a}_{n}\right)$ be any sequence in $ℝ$. If there is an ${n}_{0}\in {ℕ}^{\ast }$ for each $\epsilon >0$ such that

 [5.8.8]

then $\left({a}_{n}\right)$ is convergent.

Proof:  At first we see that $\left({a}_{n}\right)$ is bounded: From the premise we find an ${n}^{\ast }$ for 1 such that

Consequently we have:  $|{a}_{n}|\le \mathrm{max}\left\{1+|{a}_{{n}^{\ast }}|,|{a}_{1}|,\dots ,|{a}_{{n}^{\ast }-1}|\right\}$ for all n.

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

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

$|{a}_{{n}_{1}}-g|<\frac{\epsilon }{2}\text{.}$

Thus we have for all $n\ge {n}_{0}\text{:}$

$|{a}_{n}-g|=|{a}_{n}-{a}_{{n}_{1}}+{a}_{{n}_{1}}-g|\le |{a}_{n}-{a}_{{n}_{1}}|+|{a}_{{n}_{1}}-g|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$.

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.