# Subsequences and Bolzano-Weierstrass Theorem

## Definition 2.4.1

Let $\{p_n\}$ be a sequence and let $\{n_k\}$ be strictly increasing sequence, i.e. $n_1 < n_2 <n_3 < \cdots$. We call $\{p_{n_k}\}_{k=1}^\infty$ a subsequence of $\{p_n\}$.

## Theorem 2.4.3

Let $\{p_n\}$ be a convergent sequence with $\displaystyle{\lim_{n\to\infty}p_n=p}$. Then for all subsequences of $\{p_n\}$ converges to $p$.

<proof> Let $\epsilon >0$ be given and let $\{p_{n_k}\}$ be a subsequence of $\{p_n\}$. Since $p_n\to p$ as $n\to\infty$, there exists $N\in\mathbb{N}$ such that \begin{align} n\geq N \Rightarrow |p_n-p| <\epsilon \end{align} Now, $n_k \geq k$. So, if $k\geq N$, then $n_k\geq k \geq N$.
$\therefore |p_{n_k}-p|<\epsilon$
$\therefore \displaystyle{\lim_{k\to\infty}}p_{n_k}=p$ $\tag*{\square}$

## Definition 2.4.5

Let $E$ be a subset of $\mathbb{R}$

(a) $p\in\mathbb{R}$ is a limit point of $E$ if $N^\prime_\epsilon (p) \cap E\neq \emptyset$ for all $\epsilon >0$, where $N^\prime_\epsilon (p):= N_\epsilon (p) \setminus \{p \}$ is a deleted $\epsilon$-neighborhood of $p$.

(b) $p\in \mathbb{R}$ is an isolated point of $E$ if $p$ is not a limit point of $E$.

## Examples 2.4.6

(a) $E =\{\frac{1}{n}: n\in\mathbb{N}\}$

Assume $p<0$. Take $\epsilon_0 := \frac{\left|p\right|}{2}$. Then $N^\prime_{\epsilon_0} (p)\cap E=\emptyset$
Assume that $p>1$. Take $\epsilon_0 := |p-1|$. Then $N^\prime_{\epsilon_0} (p)\cap E=\emptyset$.
Assume that $p=\frac{1}{n}$. Take $\epsilon_0 :=\left|\frac{1}{n}-\frac{1}{n+1}\right|$. Then, $N^\prime_{\epsilon_0} (p)\cap E=\emptyset$
Assume that $p\neq \frac{1}{n}, \text{ and } p\in (0,1)$. For this $p$, we can find $n\in\mathbb{N}$ such that $\frac{1}{n+1} < p < \frac{1}{n}$. If we take \begin{align} \epsilon_0 = \min\{\left|p-\frac{1}{n+1}\right|,\left|p-\frac{1}{n}\right| \} \end{align}, then $N^\prime_{\epsilon_0} (p)\cap E=\emptyset$.
Finally assume $p=0$. Let $\epsilon >0$ be given. Then, take $N\in\mathbb{N}$ such that $\frac{1}{N} < \epsilon$ by Archimedean property. In other words, \begin{align} \frac{1}{N} \in (-\epsilon,\epsilon) \setminus \{0\} = N^\prime_\epsilon (0). \end{align} That is $\frac{1}{N}\in N^\prime_\epsilon (0) \cap E$.
$\therefore N^\prime_\epsilon (0)\cap E\neq \emptyset$.
$\therefore 0$ is a limit point of $E$. $\tag*{\square}$

(b) $E:=\{a_n:n\in\mathbb{N}\}, \lim_{n\to\infty}a_n =L.$ $L$ is not a limit point of $E$.

<proof> Since $a_n\to L$ as $n\to\infty$, for any $\epsilon >0$, there exists $N\in\mathbb{N}$ s.t. $n\geq N \Rightarrow \left|a_n-L\right|<\epsilon$. In other words, $a_n \in (L-\epsilon, L+\epsilon)$ for all $n\geq N$, but we cannot guarantee that $a_n \neq L$.
For example, if $a_n=L$ for all $n\in\mathbb{N}$, then $a_n\not\in N^\prime_\epsilon (L)$.

$\therefore N^\prime_\epsilon (L) \cap E =\emptyset$
$\therefore L$ is not a limit point $\tag*{\square}$

(c) $E:=\{a_n:n\in\mathbb{N}\}, \lim_{n\to\infty}a_n =\infty.$ Then there is no limit point of $E$. Let $p\in\mathbb{R}$ be given. Since $a_n\to\infty$ as $n\to\infty$, for all $M>0, N\in\mathbb{N}$ such that $a_n > M$. Take $M_0 >0$ with $M_0 > p$ and define \begin{align} \epsilon_0 := \min\{|p-M_0|, |a_i-p|: a_i\neq p, a_i <M_0 \} \end{align} Then $\epsilon_0 >0$ and $N^\prime_{\epsilon_0} (p) \cap E =\emptyset$.

(d) $E:=[0,1]\cap \mathbb{Q}$. Find all the limit points of $E$.

<proof> Let $q\in [0,1]$ and $\epsilon >0$ be given. Since the rational number is dense in real number, there exists $r\in\mathbb{Q}$ such that $q-\epsilon <r <q+\epsilon$. Definitely $r\in E$.

$\therefore N^\prime_\epsilon (q) \cap E \neq \emptyset$.

$\therefore q \in [0,1]$ is a limit point of $E$.

## Theorem 2.4.7

Let $E$ be a subset of $\mathbb{R}$ and let $p\in \mathbb{R}$ be a limit point of $E$.

(a) Every deleted $\epsilon$-neighborhood of $p$ has infinitely many points of $E$

(b) $\exists \{p_n\} \subset E$ with $p_n\neq p$ for all $n\in\mathbb{N}$ such that $\displaystyle{\lim_{n\to\infty}}p_n=p$.

<proof> (a) Suppose that there is a deleted neighborhood $N$ of $p$ with finitely many points of $E$. Say $N=\{p_1, \ldots, p_n\}$. Take \begin{align} \epsilon :=\min\{\frac{|p_1-p|}{2}, \ldots, \frac{|p_n-p|}{2}\} \end{align} Then, $N^\prime_\epsilon (p) \cap E = \emptyset$, which contradicts to the assumption that $p$ is a limit point.

$\therefore$ There are infinitely many points in the deleted neighborhoods of $p$. $\tag*{\square}$

(b) For $n\in\mathbb{N}$, we can pick a sequence $\{p_n\}\subset E$ such that $|p_n-p|<\frac{1}{n}$ with $p_n\neq p$ because $p$ is a limit point of $E$. Let $\epsilon>0$ be given. Then there is a $N\in\mathbb{N}$ such that $\frac{1}{N}<\epsilon$ by Archimedean property. Thus, for $n\geq N$, \begin{align} |p_n-p| <\frac{1}{n}\leq\frac{1}{N}<\epsilon \end{align} $\therefore \displaystyle{\lim_{n\to\infty}}p_n = p$ $\tag*{\square}$

## Theorem 2.4.10 (Bolzano-Weierstrass Theorem)

Every bounded infinite subset of $\mathbb{R}$ has a limit point.

<proof> Let $S$ be a bounded infinite subset of $\mathbb{R}$. Since it is bounded, $S\subset [a,b]$ for some $a,b\in\mathbb{R}$. Define $I_1:=[a,b]$.

Consider two intervals $[a, (a+b)/2], [(a+b)/2, b]$. Since $S$ is infinite, at least one of them has finitely many points of $S$. We call it $I_2$.

Repeating this process, we get a sequence of intervals $\{I_n\}$ such that
(a) $I_1 \supset I_2 \supset I_3 \supset \cdots$
(b) $($length of $I_n)=(b-a)/2^{n-1}$
(c) $I_n\cap S$ is infinite.

Let $x\in \bigcap_{i=1}^\infty I_n$. We want to show that $x$ is a limit point of $S$. Let $\epsilon >0$ be given. Then there is a $N\in\mathbb{N}$ such that \begin{align} \frac{b-a}{2^{N-1}} < \epsilon \end{align} For all $y\in I_N, |y-x| \leq (b-a)/2^{N-1} < \epsilon.$ Thus, $I_N \subset N_\epsilon (x)$, i.e. $I_N\cap S \subset N_\epsilon (x) \cap S$.

Since $I_N$ has infinitely many points of $S$, there is a $y\in I_N \cap S$ such that $y\neq x$.
$\therefore N^\prime_\epsilon (x) \cap S \neq \emptyset$.

## Corollary 2.4.11

Every bounded sequence has a convergent subsequence.

<proof> Let $\{p_n\}_{n=1}^\infty$ be a bounded sequence and define $S:=\{p_n: n=1,2,\ldots\}$. If $S$ is finite, there is a $n_1<n_2<\cdots$ such that $p_{n_1}=p_{n_2}=\cdots = p$.

$\therefore \displaystyle{\lim_{k\to\infty}}p_{n_k}=p$

If $S$ is infinite, by Bolzano-Weierstrass theorem, there is a limit point $p$ of $S$. By Theorem 2.4.7, there exists $\{p_n\}_{n=1}^\infty \subset S$ with $p_n\neq p$ for all $n\in\mathbb{N}$ such that $\displaystyle{\lim_{n\to\infty}p_n = p}.$

$\tag*{\square}$

## Theorem

Let $(p_n)_{n=1}^\infty$ be a sequence in metric space $X$. If $X$ is compact, then there is a convergence subsequence of $(p_n)_{n=1}^\infty$ in $X$.

<Proof>

Consider the range of $(p_n)_{n=1}^\infty$. Suppose that the range is finite, $\{r_1, \ldots, r_k\}.$

Then one of these values must be obtained infinitely many times. Let the value be $r_1$ and let $\{n_j \}$ be indices in ascending order such that $p_{n_j}=r_1$. Then $p_{n_j}\to r_1$ as $j\to\infty$.

Now, suppose that the range is infinite. By Heine Borel Theorem, there is a limit point of $(p_n)_{n=1}^\infty$ in $X$, denoted as $p$. Since $p$ is a limit point, for every $\epsilon>0, N^\prime_\epsilon(p)$ has infinitely many points of $(p_n)_{n=1}^\infty$.

For each $k\in\mathbb{N}$, take $\epsilon=1/k$. Then there is $p_{n_k}$ such that $0<d( p_{n_k} - p)< \epsilon$ such that $n_{k-1} < n_k$ for $k>1$. Since there are infinitely many points of $(p_n)_{n=1}^\infty$ in $N^\prime_\epsilon (p)$, we can choose $n_k$ such that $n_k > n_{k-1}$.

Let $\epsilon>0$ be given. There is $N\in\mathbb{N}$ such that $\frac{1}{N}<\epsilon$ by Archimedean property. For $k\geq N$,

\begin{align*} d( p_{n_k}-p ) <\frac{1}{k}\leq \frac{1}{N}<\epsilon \end{align*}

$\therefore \displaystyle{\lim_{k\to\infty}p_{n_k}=p}$

$\tag*{\square}$

## Corollary

In $\mathbb{R}$, any bounded sequence has a convergence subsequence.

## Reference

• Manfred Stoll, Introduction to Real Analysis, Pearson

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