5.8. The BolzanoWeierstrass 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 $({a}_{n})$ be any sequence and $x\in \mathbb{R}$ arbitrary.
x is called an accumulation point of $({a}_{n})$,

[5.8.1] 
if there are infinitely many sequence members in every $\epsilon $neigbourhood $]x\epsilon ,x+\epsilon [$ 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As a notedly difference to a limit the uniqueness property is no longer valid.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
$\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. $(n)$, 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 $\mathbb{R}$). If $({k}_{n})$ is strictly increasing in ${\mathbb{N}}^{\ast}$ 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:
${k}_{n}\ge n\text{for all}n\in {\mathbb{N}}^{\ast}$

[5.8.3] 

$(1)=({(1)}^{2n})=({(1)}^{n})\circ (2n)$ 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:
${a}_{n}\to g\iff {a}_{{k}_{n}}\to g$ for each subsequence $({a}_{{k}_{n}})$

[5.8.4] 
Proof:
"$\Rightarrow $": Let $({a}_{{k}_{n}})$ be an arbitrary subsequence. At first we get an ${n}_{0}\in {\mathbb{N}}^{\ast}$ for each $\epsilon >0$ such that
${a}_{n}g<\epsilon \text{for all}n\ge {n}_{0}$.
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 $({k}_{n})$ we thus conclude:
${a}_{{k}_{n}}g<\epsilon \text{for all}n\ge {n}_{0}$.
"$\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})\iff $ 
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 ${({k}_{n})}_{n\ge 0}$ by recursion with
${a}_{{k}_{n}}x<\frac{1}{n}\text{for all}n\in {\mathbb{N}}^{\ast}$.
From that the nesting theorem [5.5.8] leads to our assertion: ${a}_{{k}_{n}}\to x$. Now to the construction details:
"$\Leftarrow $": If ${a}_{{k}_{n}}\to x$ holds for a certain subsequence we have infinitely many sequence members of $({a}_{{k}_{n}})$ in every $\epsilon $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 (BolzanoWeierstrass theorem): The following implication holds for every sequence
$({a}_{n})$ in $\mathbb{R}$.
If $({a}_{n})$ is bounded then $({a}_{n})$ possesses at least one accumulation point.

[5.8.6] 
Proof: We employ the
completeness axiom
i 
Every nonempty, bounded subset of $\mathbb{R}$ 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 $n\in {\mathbb{N}}^{\ast}$ the set $\{{a}_{k}k\ge n\}$ is a nonempty, bounded subset of $\mathbb{R}$. Thus
${x}_{n}\u2254\mathrm{inf}\{{a}_{k}k\ge n\}$.
is well defined. Obviously we have $\{{a}_{k}k\ge n\}\supset \{{a}_{k}k\ge n+1\}$ so that every lower bound of $\{{a}_{k}k\ge n\}$ is a lower bound of $\{{a}_{k}k\ge n+1\}$ as well. Especially the infimum ${x}_{n}$ is to be found among all the lower bounds of $\{{a}_{k}k\ge n+1\}$. This implies ${x}_{n}\le {x}_{n+1}$, i.e. $({x}_{n})$ is increasing. Furthermore $({x}_{n})$ is bounded, as
${{x}_{1}\le {x}_{n}\le {a}_{n}}{\le c}$
i 
$({a}_{n})$ is bounded!

, thus the set $\{{x}_{n}n\in {\mathbb{N}}^{\ast}\}$ is a nonempty, bounded subset of $\mathbb{R}$. Using the completeness theorem a second time we set
$x\u2254\mathrm{sup}\{{x}_{n}n\in {\mathbb{N}}^{\ast}\}$.
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
${({k}_{n})}_{n\ge 0}$ such that
$x\frac{1}{n}<{a}_{{k}_{n}}<x+\frac{1}{n}\text{for all}n\in {\mathbb{N}}^{\ast}$

[0]

We do this recursively:

${k}_{0}\u22540$

For $n\in {\mathbb{N}}^{\ast}$ the number $x\frac{1}{n}$ is certainly no upper bound of $\{{x}_{n}n\in {\mathbb{N}}^{\ast}\}$ (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 $m>{k}_{n1}$ without any restriction.
Similarly $x}_{m}+\frac{1}{n$ is no lower bound for $\{{a}_{k}k\ge m\}$, 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 $\{{a}_{k}k\ge m\}\ni {a}_{{k}_{n}}$ ) we see that our sequence
${({k}_{n})}_{n\ge 0}$ satisfies:
${k}_{n}\ge m>{k}_{n1}$
$x\frac{1}{n}<{x}_{m}\le {a}_{{k}_{n}}<{x}_{m}+\frac{1}{n}\le x+\frac{1}{n}$
Thus we have achieved our aim [0].

There are some implications to be drawn from the BolzanoWeistrass 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$\iff $

$({a}_{n})$ 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 $\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 $\epsilon \u2254\frac{xg}{2}>0$:

There are inifinitely many sequence members in $]x\epsilon ,x+\epsilon [$.

There are at most finitely many sequence members lacking in $]g\epsilon ,g+\epsilon [$.

$]x\epsilon ,x+\epsilon [\cap ]g\epsilon ,g+\epsilon [=\varnothing $ as $\epsilon $ 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 $]g\epsilon ,g+\epsilon [$ is an arbitrary $\epsilon $neighbourhood of g we need to find an ${n}_{0}\in {\mathbb{N}}^{\ast}$ such that
${a}_{n}\in ]g\epsilon ,g+\epsilon [\text{for all}n\ge {n}_{0}\text{.}$
Suppose there is no such ${n}_{0}$. Starting with ${k}_{0}\u22540$ we find (recursively) for each n a
${k}_{n}>{k}_{n1}$ such that ${a}_{{k}_{n}}$ is lacking in $]g\epsilon ,g+\epsilon [$.
Thus we get a subsequence $({a}_{{k}_{n}})$, bounded as well as $({a}_{n})$ is bounded, with an accumulation point x according to the BolzanoWeierstrass theorem.
As there are no sequence members of $({a}_{{k}_{n}})$ at all in $]g\epsilon ,g+\epsilon [$ we conclude that $x\ne g$. 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
$(n(1+{(1)}^{n}))=(\mathrm{0,4,0,8,0,12,0,}\dots )$
e.g. has only one accumulation point.
A second application of the BolzanoWeierstrass 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 $\mathbb{R}$. If there is an ${n}_{0}\in {\mathbb{N}}^{\ast}$ for each $\epsilon >0$ such that
${a}_{n}{a}_{m}<\epsilon \text{for all}n,m\ge {n}_{0}$

[5.8.8] 
then $({a}_{n})$ is convergent.
Proof: At first we see that $({a}_{n})$ is bounded: From the premise we find an ${n}^{\ast}$ for 1 such that
${a}_{n}{a}_{{n}^{\ast}}\le {a}_{n}{a}_{{n}^{\ast}}<1\text{for all}n\ge {n}^{\ast}$
Consequently we have: ${a}_{n}\le \mathrm{max}\{1+{a}_{{n}^{\ast}},{a}_{1},\dots ,{a}_{{n}^{\ast}1}\}$ for all n.
According BolzanoWeierstrass $({a}_{n})$ 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 ${n}_{0}\in {\mathbb{N}}^{\ast}$ such that
${a}_{n}{a}_{m}<\frac{\epsilon}{2}\text{for all}n,m\ge {n}_{0}\text{.}$
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 $\mathbb{R}$ is convergent.
The Cauchy test guarantees the existence of a limit but does not provide any information on its value.
The BolzanoWeierstrass theorem and thus all the conclusions is bound to the reals, more precisely to the completeness axiom. In
$\mathbb{Q}$ 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 $\mathbb{Q}$.
