Sequences

Sequence

A sequence is a function $p:\mathbb{N}\to X$.

Convergence

A sequence $\{p_n\}$ in a Metric Space $X$ is said to converge if there is a point $p\in X$ with the following property:

• For every $\epsilon >0$ there is an integer $N$ such that $n\ge N$ implies that $d(p_n,p)<\epsilon$. (Here $d$ denotes the distance in $X$.)
• $\forall \epsilon >0,\exists N\in \mathbb{N},n\ge N,d(p_n,p)<\epsilon$

In this case, $\{p_n\}$ converges to $p$, or that $p$ is the limit of $\{p_n\}$, and we write $p_n\to p$ or

$$\underset{n\to \infty }{\lim }{p}_n=p$$

If $\{p_n\}$ does not converge, then it must diverge. This sequence depends on the ambient space $X$.

Range

The set of all the points $p_n$ for $(n=1,2,3,\cdots )$ is the range of $\{p_n\}$. The range of a sequence may be a finite set, or it may be infinite.

$$\{p_n\mid n\in \mathbb{N}\}\subset X$$

The sequence $\{p_n{\}}_{n\in \mathbb{N}}$ is bounded if its range is bounded.

Theorem

Let $\{p_n\}$ be a sequence in a Metric Space $X$.

1. $\{p_n\}$ converges to $p\in X$ if and only if every neighborhood of $p$ contains $p_n$ for all but finitely many $n$.
2. If $p\in X,p^{\prime }\in X$ and if $\{p_n\}$ converges to $p$ and to $p^{\prime }$, then $p=p^{\prime }$.
3. If $\{p_n\}$ converges, then $\{p_n\}$ is bounded
4. If $E\subset X$ and if $p$ is a limit point of $E$, then there is a sequence $\{p_n\}$ in $E$ such that

$$p=\underset{n\to \infty }{\lim }{p}_n$$

Proof: $(1)$ Suppose $p_n\to p$. Then,

$$\begin{array}{rl}{p}_n\to p& ⟺\forall \epsilon >0,\exists n\in \mathbb{N},\forall n\ge \mathbb{N},d(p_n,p)<\epsilon \\ & ⟺\forall \epsilon >0,\exists n\in \mathbb{N},\forall n\ge \mathbb{N},p_n\in N_{\epsilon }(p)\\ & ⟺\text{all but finitely many of the n’s satisfy pn∈Nr(ε)}\end{array}$$

$(2)$ Suppose $p_n\to p$ and $p_n\to q$. Suppose $p\ne q$. Then $d(p,q)>0$. Let

$$\epsilon =\frac{d(p,q)}{2}$$

• Then since $p_n\to p$, then $\exists N_1\in \mathbb{N},\forall n\ge N_1$ such that $d(p_n,p)<\epsilon$
• Then since $p_n\to q$, then $\exists N_2\in \mathbb{N},\forall n\ge N_2$ such that $d(p_n,q)<\epsilon$ Let $N=\max \{N_1,N_2\}$. Then $\forall n\ge N$,

$$d(p,q)\le d(p,p_n)+d(p_n,q)<\epsilon +\epsilon =2\epsilon =d(p,q)$$

which is a contradiction.

$(3)$ Convergence $⟹$ bounded Assume $p_n\to p$. Let $\epsilon ={\pi }^{\sqrt{2}}>0$. Then $\exists N\in \mathbb{N}$ such that $\forall n\ge N,d(p_n,p)<{\pi }^{\sqrt{2}}$. Let

$$R_1=\max \{d(p_m,p)\mid m=1,\cdots ,N-1\}$$

Now let $R=\max \{R_1,{\pi }^{\sqrt{2}}\}+1>0$ and $\forall n\in \mathbb{N}$,

$$\begin{array}{rl}d(p_n,p)<R& ⟺p_n\in N_R(p),\forall n\in \mathbb{N}\\ & ⟺\{p_n\}\text{ boudned}\end{array}$$

$(4)$ Let $p\in A^{\prime }$. Then $\forall n\in \mathbb{N},\exists p_n\in N_{1/n}(p)\cap A$. I claim $p_n\to p$. Let $\epsilon >0$ and let $N\in \mathbb{N}$ be such that $\frac{1}{n}<\epsilon$ by Archimedean Property. Then $\forall n\ge N$,

$$p_n\in N_{1/n}(p)⟺d(p_n,p)<\frac{1}{n}\le \frac{1}{N}<\epsilon$$

Use the fact that $N_{1/n}(p)$ is nonempty, and so pick a point for all $n\in \mathbb{N}$ to build a sequence.

Lemma (Algebra of Sequences)

Let $\{a_n\},\{b_n\}$ be sequences in $\mathbb{C}$. Assume $a_n\to a,b_n\to b$. Then,

1. $\{a_n+b_n\}\to a+b$
2. $\{a_n\cdot b_n\}\to ab$
3. Suppose $b\ne 0$. $\frac{a_n}{b_n}\to \frac{a}{b}$

Lemma (Convergence of Vectors)

Let $\{{\vec{x}}_n\}$ be a sequence in ${\mathbb{R}}^d$. Then $\{{\vec{x}}_n\}$ converges to $\vec{x}$ if and only if $x_{j,n}\to x_j$ for all $j=1,2,3,\cdots ,d$.

Component wise convergence.

Subsequences

Let $\{p_n\}$ be a sequence. Suppose $n_1<n_2<n_3<\cdots$ be a set of natural numbers. Then we call $\{p_{n_i}{\}}_{i\in \mathbb{N}}$ a subsequence of $p_n$.

Note that

$$p_1,{\overline{p}}_2,{\overline{p}}_3,p_4,p_5\cdots ⟹p_1=p_{n_1},p_{n_2}=p_4,p_{n_3}=p_5$$

Subsequence Limits

Let $\{p_n\}$ be a sequence. A point $p\in X$ is called a subsequential limit of $\{p_n\}$ if $\exists$ a subsequence $\{p_{n_i}\}$ such that

$$\underset{i\to \infty }{\lim }{p}_{n_i}\to p$$

Lemma (Structure of Subsequences)

Let $\{p_n\}$ be a sequence. Then

$$p_n\to p⟺\text{every subsequence converges to p}$$

$(⟸)$ The full sequence is a subset of itself. Therefore $p_n\to p$.

$(⟹)$ Suppose $p_n\to p$. WTS every subsequence converges to $p$. Let $\{p_{n_k}{\}}_{k\in \mathbb{N}}$ be a subsequence of $\{p_n\}$. The goal is to show $p_{n_k}{\to }_{k\to \infty }p$.

Let $\epsilon >0$. We want to find $K\in \mathbb{N}$ such that $\forall k\ge K$, we have $d(p_{n_k},p)<\epsilon$. Since $p_n\to p,\exists N\in \mathbb{N}$ such that $\forall n\ge \mathbb{N}$ we have $d(p_n,p)<\epsilon$.

Lemma (Subsequences Stabilize)

For any subsequence $\{p_{n_i}\}$, there exists an index $K$ such that for all $k\ge K$ we have $n_k\ge N$.

Proof:

• The subsequence indices are in increasing sequence $n_1<n_2<n_3<\cdots$
• Since $N$ exists for $d(p_n,p)<\epsilon$ for $n\ge N$ we can choose the first $k$ where $n_k\ge N$ and for all $k\ge K$, we have $n_k\ge N$.
• This is by induction.

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Thus, for sufficiently large $K$, we have $n_K\ge N$ where

$$n_1<n_2<\cdots <N<n_K<\cdots$$

now if $k\ge K$, then by stabilizing lemma,

$$n_k\ge n_K\ge N$$

so then $d(p_{n_k},p)<\epsilon$.

Lemma (Bolzano-Weierstrass Theorem)

1. Let $\{p_n\}$ be a sequence in a compact set $K\subset (X,d)$. Then $\{p_n\}$ has a convergent subsequence.
2. Every bounded sequence in ${\mathbb{R}}^d$ has a convergent subsequence.

Proof:

Part $(2)$ follows from $(1)$ and Heine-Borel Theorem.

Part $(1)$: Case 1: If the range of a sequence is finite, then $\exists n_1<n_2<n_3<\cdots$ and some $p\in K$ such that $p_{n_i}=p,\forall i\in \mathbb{N}$. Indeed, every point after $p_{n_i}$ is the same. Hence $p_{n_i}=p\to p$.

Case 2: Let the range be infinite. Let $A=\{p_n\mid n\in \mathbb{N}\}$. Then by corollary and $A$ is infinite, (and compact) then $\exists$ a limit point $p\in A$.

Hence you can construct $\{q_i\}\subset A$ from the range such that $q_i\to p$. But then $\{q_i\}\subset A$ is just a subsequence of $\{p_n\}$ and it converges.

Theorem (Sequential Compactness Criterion)

A Metric Space $X$ is compact $⟺$ every sequence in $X$ has a convergent subsequence.

Proof:

$(⟹)$ Let $X$ be compact. Consider any sequence $\{p_n\}$ in $X$. It must be in a bounded set since it is compact then it is bounded by Lemma (Compact Implies Bounded). By Bolzano-Weierstrass Theorem, $\{p_n\}$ has a convergent subsequence.

$(⟸)$ Let every sequence in $X$ have a convergent subsequence. We can consider infinite set $E\subset X$. Since $E$ is infinite, then we can extract a sequence $\{x_n\}$ form this set. Since this has a limit point $x$ and $x\in E$, every infinite subset has a limit point in it.

Then by Homework 5, Problem 26, it is compact.

Lemma (Closure of Subsequent Limits)

Let $\{p_n\}$ be a sequence in $(X,d)$. Let $Q$ denote the set of subsequence limits of $\{p_n\}$. Then $Q$ is closed.

Special Sequences

1. Let $p>0$. Then ${\lim }_{n\to \infty }\frac{1}{n^p}=0$
2. If $|a|<1$, then ${\lim }_{n\to \infty }{a}^n=0$
3. ${\lim }_{n\to \infty }\sqrt[n]{n}=1$
4. Let $a>0$. Then ${\lim }_{n\to \infty }\sqrt[n]{a}=1$
5. Let $p>0,\alpha \in \mathbb{R}$. Then

$$\underset{n\to \infty }{\lim }\frac{n^{\alpha }}{(1+p{)}^{\alpha }}$$