Sequences

Overview §

A sequence is a function whose domain is the natural numbers. Sequences can diverge (e.g., $x_n=n$), converge (e.g., $x_n=1/n$), or oscillate (e.g., $x_n=(-1{)}^n$).

A convergent sequence has a limit $x$ such that the values $x_n$ can be made arbitrarily close to $x$ for large values of $n\in \mathbb{N}$. Every convergent sequence is also Cauchy.

Related: Convergence of functions, Bounded sequences

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In metric spaces §

Sequences and subsequences §

Definition: Sequence in a metric space

Given a metric space $(X,d)$, a sequence, often denoted $(x_n{)}_{n\in \mathbb{N}}$, is a discrete function $\mathbb{N}\to \mathbb{X}$ which maps each natural number $n$ to some $x_n\in X$.

Definition: Subsequence

If $(x_n{)}_{n\in \mathbb{N}}$ is a sequence and $(n_k)\in \mathbb{N}$ is a sequence of natural numbers with $n_{k+1}>n_k$, then $(x^{\prime }{)}_k=x_{n_k}$ is a subsequence of $(x_n{)}_{n\in \mathbb{N}}$.

Definition: Bounded sequences in metric spaces

Let $(X,d)$ be a metric space. We say a sequence $(x_n{)}_{n\in \mathbb{N}}$ in $X$ is bounded if there exist $x\in X$ and $r>0$ such that for all $n\in \mathbb{N}$, we have $d(x_n,x)<r$.

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Convergent sequences and limits §

Definition: Convergent sequence in a metric space

Given a metric space $(X,d)$, a sequence $(x_n{)}_{n\in \mathbb{N}}$ is called convergent in $X$ if there exists $x\in X$ such that

$${\forall }_{ϵ>0}{\exists }_{N\in \mathbb{N}}{\forall }_{n\in \mathbb{N}}[n\ge N⟹d(x_n,x)<ϵ].$$

In this case, we say $x$ is the limit of $(x_n{)}_{n\in \mathbb{N}}$ and write ${\lim }_{n\to \infty }{x}_n=x$. A sequence is divergent if it does not converge to some $x$.

Proposition: Facts about convergent sequences in metric spaces

• (i) Convergent sequences converge to a unique limit.
• (ii) If a sequence converges, then it is also bounded.
• (iii) Every convergent sequence is Cauchy.

Proposition: The set of sub-sequential limits is closed.

Let $A$ be the set of limits of subsequences of a sequence $(x_n{)}_{n\in \mathbb{N}}$ in $\mathbb{R}$. Then $A$ is closed.

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In topological spaces §

Definition: Convergent sequence in a topological space

A sequence $x_1,x_2,\dots$ of points in an arbitrary topological space $X$ converges to the point $x$ if, for each neighborhood $U$ of $x$, there exists a positive integer $N$ such that $x_n\in U$ for all $n\ge N$.

The key difference is that sequences can converge to more than one point in an arbitrary space!

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Review §

Modern Analysis I

• State whether the following sequences are convergent or divergent: $x_n=1/n$, $(-1{)}^n$, $n$.

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Proof appendix §

Proposition: Facts about convergent sequences in metric spaces

• (i) Convergent sequences converge to a unique limit.
• (ii) If a sequence converges, then it is also bounded.
• (iii) Every convergent sequence is Cauchy.

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Proof of (i).

For all $ϵ>0$, exist $N_x,N_y\in \mathbb{N}$ such that for all $n>N_x$, we have $d(x_n,x)<ϵ/2$, and similarly for all $n>N_y$. Then for all $n>\text{max}(N_x,N_y)$, we have

$$d(x,y)\le d(x_n,x)+d(x_n,y)=ϵ,$$

so $d(x,y)=0$ and $x=y$.

Proof of (iii).

Since $(x_n{)}_{n\in \mathbb{N}}$ is convergent, for all $ϵ>0$, there exists $N\in \mathbb{N}$ such that $d(x_n,x)<ϵ/2$ for all $n>N$ and some $x\in X$. By the triangle inequality, for all $n, m > N,

$$d(x_n,x_m)\le d(x_n,x)+d(x,x_m)<ϵ,$$

which is precisely what it means for $(x_n{)}_{n\in \mathbb{N}}$ to converge.

Proposition: The set of sub-sequential limits is closed.

Let $A$ be the set of limits of subsequences of a sequence $(x_n{)}_{n\in \mathbb{N}}$ in $\mathbb{R}$. Then $A$ is closed.

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Proof.

It suffices to show that every convergent sequence in $A$ has a limit in $A$. Let $(a_i{)}_{i\in \mathbb{N}}$ be a sequence in $A$. Then for each $i\in \mathbb{N}$, there exists a subsequence $(x_{n(m)}{)}_{m\in \mathbb{N}}$ such that $x_{n(m_i)}$ converges to $a_i$ as $m_i\to \infty$.

$$\begin{array}{cccc}i=1& i=2& i=3& \cdots \\ x_{n(1{)}_1}& x_{n(1{)}_2}& x_{n(1{)}_3}& \cdots \\ x_{n(2{)}_1}& x_{n(2{)}_2}& x_{n(2{)}_3}& \cdots \\ x_{n(3{)}_1}& x_{n(3{)}_2}& x_{n(3{)}_3}& \cdots \\ ⋮& ⋮& ⋮& \\ a_1& a_2& a_3& \cdots \end{array}$$

Now let $n\in \mathbb{N}$. Since $a_i\to a$, there exists some $a_I$ such that $d(a_I,a)<1/(2n)$.#wip

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Notes §

Honors Mathematics B §

• (Proposition 3.1) Continuous functions take convergent sequences to convergent sequences. If $f:D\to \mathbb{R}$ is continuous and $x_n\in D$ is a convergent sequence $x_n\to x$ with $x\in D$ as well, then $f$ sends $x_n$ to another convergent sequence $f(x_n)\to f(x)$ .
• (Proposition 3.2) Convergent sequences converge to a unique limit. If $x_n\to x$ and $x_n\to x^{\prime }$, then $x=x^{\prime }$.
• (Lemma 3.3) Bounded monotone sequences converge. Suppose $x_n$ is monotone increasing, so $x_n+1\ge x$ for all $n\in \mathbb{N}$, and bounded above, so there exists an $x$ with $x_n\le x$ for all $n\in \mathbb{N}$. Then $x_n$ converges to some $x^{\prime }$.
• (Theorem) All bounded sequences have a convergent subsequence (Bolzano-Weierstrass).