3.4 Subsequences and the Bolzano-Weierstrass Theorem

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Summary

Definition

• Subsequence
• Limit superior and limit inferior

Theorem

• $X$ sequence of real numbers
• $X^{\prime }$ subsequence of $X$

1. $(\lim X=x)$ $\to$ $(\lim X^{\prime }=x)$
2. $X$ divergent if either
   • $X$ unbounded
   • $X$ has 2 subsequences with inequal limits
3. $X$ $\to$ $\exists X^{\prime }$ monotone
4. $X$ bounded $\to$ $\exists X^{\prime }$ convergent
5. $X$ bounded, all convergent subsequence converges to $x$ $\to$ $\lim X=x$

3.4.1 Definition: Subsequence

Let $X=(x_n)$ be a sequence of real numbers and let $(n_k)$ be an increasing sequence of natural numbers.

Then the sequence $X^{\prime }=(x_{n_k})$ given by

$$(x_{n_1},x_{n_2},\dots ,x_{n_k},\dots )$$

is called a subsequence of $X$.

For example, if $X:=\left(\frac{1}{1},\frac{1}{2},\frac{1}{3},\dots ,\frac{1}{n},\dots \right)$, then the subsequence given by $n_k=2k$ is

$$X^{\prime }=\left(\frac{1}{2},\frac{1}{4},\frac{1}{6},\dots ,\frac{1}{2k},\dots \right)$$

A tail of a sequence is a special type of subsequence.

In fact, the $m$-tail corresponds the subsequence with indexes $n_k=m+k$

3.4.2 Theorem: Convergence of a subsequence

If a sequence $(x_n)$ of real numbers converges to a real number $x$,

then any subsequence of $(x_n)$ also converges to $x$.

3.4.4 Theorem: Diverging sequence equivalency

Let $X=(x_n)$ be a sequence of real numbers. Then the following are equivalent:

1. The sequence $X=(x_n)$ does not converge to $x\in \mathbb{R}$.
2. There exists an $\epsilon >0$ such that for any $k\in \mathbb{N}$, there exists $n_k\in \mathbb{N}$ such that $n_k\ge k$ and $|x_{n_k}-x|\ge {\epsilon }_0$.
3. There exists an ${\epsilon }_0>0$ and a subsequence $X^{\prime }=(x_{n_k})$ of $X$ such that $|x_{n_k}-x|\ge {\epsilon }_0$ for all $k\in \mathbb{N}$.

3.4.5: Theorem: Diverging sequence criteria

If a sequence $X=(x_n)$ of real numbers has either of the following properties, then $X$ is divergent:

1. $X$ has two convergent subsequences $X^{\prime }=(x_{n_k})$ and $X^{\prime \prime }=(x_{r_k})$ whose limits are not equal.
2. $X$ is unbounded.

3.4.7 Theorem: Monotone subsequence

If $X=(x_n)$ is a sequence of real numbers,

then there exists a subsequence of $X$ that is monotone.

3.4.8 Theorem: The Bolzano Weierstrass Theorem

A bounded sequence of real numbers has a convergent subsequence.

3.4.9 Theorem

Let $X=(x_n)$ be a bounded sequence of real numbers and let $x\in \mathbb{R}$ have the property that every convergent subsequence of $X$ converges to $x$.

Then the sequence $X$ converges to $x$.

3.4.10 Definition: Limit superior and limit inferior

Let $X=(x_n)$ be a bounded sequence of real numebers.

1. The limit superior of $(x_n)$ is the infimum of the set $V$ of $v\in \mathbb{R}$ such that $v<x_n$ for at most a finite number of $n\in \mathbb{N}$. It is denoted by

$$\mathrm{limsup}(x_n)\text{or}\mathrm{limsup}X\text{or}\mathrm{limsup}(x_n)$$

Basically supremum/infimum for sequences

Largest/smallest point of a sequence.

Proofs

3.4.2

1. Premise

Let $\epsilon >0$ be given and let $K(\epsilon )$ be such that if $n\ge K(\epsilon )$, then $|x_n-x|<\epsilon$.

2. From 1

Since $n_1<n_2<\cdots <n_k<\dots$ is an increasing sequence of natural numbers, it is easily proved (by Induction) that $n_k\ge k$.

3. From 1, 2

Hence, if $k\ge K(\epsilon )$, we also have $n_k\ge k\ge K(\epsilon )$ so that $|x_{n_k}-x|<\epsilon$.

4. Conclusion

Therefore the subsequence $(x_{n_k})$ also converges to $x$.

NOTE

Note that $(n_k)$ from 1 is a sequence of natural numbers, and is strictly increasing.

Note that $k$ is also a natural number.

Therefore, values of the sequence $n_1<n_2<\cdots <n_k<\dots$ is AT “MINIMUM”

$$\underset{k=1}{1}<\underset{k=2}{2}<\cdots <\underset{k=3}{3}<\dots$$

and we can trivially see that $k\le n_k$ is always true.