Section 11

The book does not explicitly state this, but almost every result in this chapter only works for infinite sequences and infinite sub sequences. Subsequences are always infinite through the function definition.

A subsequence is taking elements of another sequence in order. The sub sequence of $s_n$ is defined to be $s_{n_k}$.

To drive this home, the index of the elements of a subsequence $s_{n_k}$ is $a_k$ for $k\in \mathbb{N}$, and each $n_k$ is associated with some $n$ in the parent sequence. Basically, for some sequence $s_n=(a_1,a_2,a_3,a_4....)$ We have a sub sequence $s_{n_k}=(a_2,a_3....)$ where $a_2$ is at the $n_1$ position of $s_{n_k}$. So $n_1=2$. It is important to note that $n_k\ge n$ as a sub sequence is either more along or at the same position, never behind.

You can also think of a sub sequence $t$ of $s_n$ as $t=s\circ \omega$ where $\omega$ takes the index of the subsequence to its index in the parent sequence, and then $s$ just behaves as normal like we defined before ($s(n)=s_n$). So to recap, $\omega (a_k)=n$ and $s(n)=s_n$, leading $s\circ \omega (k)=s_n$. This can, for the most part, be ignored but is good to understand the definition.

Here is an example of the function being used (they use different greek letters) In this example, $\omega (k)=2k$ and then $s(2k)=(2k{)}^2(-1{)}^{2k}$. We just replaced the $n$ with $2k$ but you understand what is going on.

Sequence theorems:

For sequence $s_n$

1. Say $t$ is in $\mathbb{R}$, then there is a subsequence of $s_n$ converging to $t$ $⟺$ the set $\{n\in \mathbb{N}:|s_n-t|<ϵ\}$ is infinite for all $ϵ>0$
2. If the sequence $s_n$ is unbounded above it has a subsequence going to infinity
3. If the sequence $s_n$ is unbounded bellow it has a subsequence going to -infinity Additionaly, the subsequence is monotonic

The first one is just saying there are infinite points of the sequence close to $t$. I am only going to do one part of this proof as it is insanely complex. Proof:

1. For the $\Rightarrow$ of (i) notice that if we had a subsequence $s_{n_k}\to t$ then for every $ϵ>0$ there are only finite elements (namely the ones before $N$) that are not within $ϵ$. The infinate elements after $N$ are clearly within $ϵ$ of $t$ by definition of convergence.

Examples of this theorem: I want to provide concrete sequences and examples for why each one is important. For (i) notice the sequence $a_n=(-1{)}^n$. This sequence has infinate points around both $1$ and $-1$. Thus we can create sub sequences that converge to both (Indeed (1,1,1,1,…) and (-1,-1,-1,…)).

For (ii) and (iii) notice the sequence $s_n=n(-1{)}^n$. This is just a sequence where all the odd terms are negative and the even terms are positive. By (ii) the sequence is not bounded above, so clearly we can take all the even terms and create a sub sequence going to $\infty$. Similarly, by (iii) the sequence is not bounded bellow so we can take the odd terms and go to $-\infty$.

Importance in $\mathbb{Q}$.

Firstly, not that $\mathbb{Q}$ is countable (there is an example and diagram in the book of how to do it.) Because of this, it can be created into a sequence $r_n$. Now, remember that between any two real numbers there is a rational number. Because of this, for any real number $a$ there are infinite rational points near it $(a-ϵ)$ and $(a+ϵ)$both have infinite points..

Additionally there are inflate rationals between any two real points (rational points are also real points). This is a really easy proof.

Another example

Suppose sequence $s_n$ contains only positive numbers such that $\inf s_n=0$. Then we know that it has a subsequence converging monotonically to 0.

To prove this we have to show that for any $ϵ>0$, the set $\{n\in \mathbb{N}:|s_n-0|<ϵ\}$ is infinite. This just means for any $ϵ$ there are infinite elements near $0$. Keep in mind $\inf$ means least upper bound. For the sake of contradiction, assume this is not true and there is some ${ϵ}_1$ where there are not infinite points near $0$. As the set is finiate in $\mathbb{R}$ there is clearly a minimum element of the set. Then this element would be the inf, of the parent set thus we hit our contradiction. If the set is empty, then ${ϵ}_0$ would be the new lower bound which is greater than 0. This in both examples, there is a subsequence converging monotonically to 0 as there are infinite elements of $s_n$ near 0.

Sequence convergence

If a sequence $s_n$ converges to $n$, then every subsequence converges to the same $n$.

To prove this it is actually quite simple. Assume that $s_n\to L$, then we know that, by definition, for any $ϵ>0\exists N:n>N⟹|s_n-L|<ϵ$. Now, notice if we take the same index in the sequence and subsequence, $k=n$, then we have it so $n_k\ge n$. This is because the subsequence is always the same ammount or moure further along. Thus, $n_k>N$ holds the same property for any epsilon. So we can define a new $N_1=k:n_k>N$ and we have the same property for any $ϵ>0$ $k>N_1⟹|s_{n_k}-L|<ϵ$.

Monotonic subsequence

Every sequence $s_n$ has a monotonic subsequence.

Proof:

1. Lets define a peak to be $a_n$ such that $a_n>a_m$ for all $m>n$
   1. Then, suppose we have infinite peaks. These are all elements that are larger than the elements after them. If we take all infinite peaks, in order, then we have a monotonic decreasing subsequence
   2. Suppose we have finite peaks, and $a_N$ is the final peak. Then starting with $a_{N+1}$ we always have an element which is larger, as if there were no larger elements $a_{N+1}$ would be a peak. Lets create a subsequence, starting with $a_{N+1}$ and then appending larger elements to the sequence. As there are no more peaks, we know there is always a larger element. This will create an increasing monotonic subsequence.
   3. Finally, if there are no peaks we can do the same thing as $2$ but just letting $a_{n_k}=a_1$.

The book follows a similar proof, but utilizes dominant terms instead.

Bolzano-Weierstrass

Every bounded sequence has a convergent subsequence Proof:

1. By the proof above, every sequence $s_n$ has a monotonic subsequence. Furthermore, if it is bounded above or bellow by some bound, then the sub sequence is too, as it can not have any elements not in the original sequence (i.e. past this bound). Then, we have a bounded monotonic sequence which converges.
2. To quickly prove that bounded monotonic subsequences converge, assume $(s_n)$ is increasing monotonic and bounded. Then, as it is bounded there exists a $\sup s_n$ where all elements in $s_n$ are less than $\sup s_n$. Now, let $ϵ>0$. Then, notice for some $\sup s_n-ϵ$ we still have some element (say $s_N$) such that $s_N>\sup s_n-ϵ$. If this were not the case $\sup s_n-ϵ$ would be the supremum. Now, as $s_n$ is increasing, we know that $n>N⟹s_n\ge s_N>\sup s_n-ϵ$. However, they can not pass $\sup s_n$. Thus, for any $ϵ>0$ there exists an $N$ such that $n>N⟹|s_n-\sup s_n|<ϵ$.

Subsequential limit

A subsequential limit if $s_n$ is an element that is the limit of some subsequence of $s_n$. A set of subsequential limits has all possible subsequential limits.

We proved before that if some sequence has a limit, all its subsequences have the same limit. Thus the set of subsequential limits for this sequence $s_n\to s$ would be $\{s\}$

Lets look at some examples of sub sequential limits

$s_n=n(-1{)}^n$. This is one we used before, and note that all subsequences here either diverge, or go to either infinity. Thus, the subsequential limits of this sequence are $\{-\infty ,\infty \}$. It is worth noting, if we create a subsequence such that $n_k=3k$ then we have it so the subsequence also does not have a defined limit.

Take a sequence with repeating values, maybe something like $(1,2,3,1,2,3,....)$. Then, the only possible subsequence limits are $1,2,3$ as they are the only points which can possibally have infiniate elements of the sequence within some $ϵ>0$. Thus this is the set of subsequence limits

Remember how $(r_n)$ has a subsequence converging to every real number $a$? Thus, the set of subsequence limits of $r_n$ is $\mathbb{R}\cup \{\infty ,-\infty \}$. Infinity and -infinity for unbounded sequences.

Connection to $\lim \sup$ and $\lim \inf$

Let $s_n$ be any sequence. Then there exists a monotonic subsequence whos limit is $\lim \sup s_n$ and there exists a monotonic subsequence whos limit is $\lim \inf s_n$.

Note: both exist.

This kinda makes sense without a formal proof. Just note that if there is some $\lim \sup s_n$ then there exists some subsequence with elements only going towards that supremum. Similar for $\lim \inf s_n$. Just pick those said terms, and you can make a monotonic sequence out of any subsequence.

Now, let $s_n$ be a sequence in $\mathbb{R}$ and let $S$ denote the list of subsequential limits in $s_n$. Then

1. $S$ is non-empty
2. $\sup S=\lim \sup S_n$ and $\inf S=\lim \inf S_n$
3. The limit exists $⟺$ $S$ has only one element (limit of $s_n$)

1 holds from just $\lim \sup$ and $\lim \inf$ which always exists. 2 This is important to go backwords and find the $\lim \sup$ and $\lim \inf$ 3 holds from a previous proof regarding convergence of subsequences.

Lets take a look at the example $s_n=n(-1{)}^n$ where we noted $S=\{-\infty ,\infty \}$. Then, we know from 2 in the previous proof that the $\lim \sup s_n=\infty$ and $\lim \inf s_n=-\infty$.

Similarly, for the sequence $(1,2,3,1,2,3,....)$ we noted $S=\{1,2,3\}$, thus $\lim \sup s_n=3$ and $\lim \inf s_n=1$

Lastly, remember in the case of $(r_n)$ we know $S=\mathbb{R}\cup \{-\infty ,\infty \}$. Thus, by the same logic we know that $\lim \sup r_n=\infty$ and $\lim \inf s_n=-\infty$