16 - Sequences in Metric Spaces

Bounded Sequence

Let $(x_n{)}_{n=1}^{\infty }$ be a sequence in some metric space $X$

Sequence is bounded if

$$\text{set }\{x_n:n\ge 1\}\text{ is bounded}$$

In other words there exists $R>0$ and $a$ such that

$$x_n\in B(a,R)\text{ for all }n\ge 1$$

Cauchy Sequence

Let $(x_n{)}_{n=1}^{\infty }$ be a sequence in some metric space $X$

Sequence is Cauchy if for all $ϵ>0$ there exists $N$ such that for all $n,m\ge N$ then

$$d(x_n,x_m)<ϵ$$

Convergent Sequence

Let $(x_n{)}_{n=1}^{\infty }$ be a sequence in some metric space $X$

Sequence is convergent if there exists $a\in X$ such that

$$\underset{n\to \infty }{\lim }{x}_n=a$$

Properties of Subsequence for Bounded, Cauchy, Convergent Sequences

If sequence $\sigma$ have any one of Bounded, Cauchy or Convergent then

$$\text{subsequence }\sigma \text{ also has the property}$$

Relation between Types of Sequences

$$\begin{array}{rlrl}\text{Convergent Sequence}& ⟹\text{Cauchy Sequence}& & (1)\\ \text{Cauchy Sequence}& ⟹\text{Bounded Sequence}& & (2)\end{array}$$

Note that reverse implications don’t hold generally

Proof

Showing reverse implications do not hold through examples

1. Take $X=(0,1]$ and $x_n=\frac{1}{n}$
   Hence $(x_n)$ is Cauchy but not convergent
2. Take $X=(0,1]$ and $x_n=\{\begin{array}{ll}1& \text{ if }n\text{ odd}\\ \frac{1}{2}& \text{ if }n\text{ even}\end{array}$
   Hence $(x_n)$ is bounded but not Cauchy as no $N$ such that

$$d(x_n,x_{n+1})<\frac{1}{2}\text{ for all }n\ge N$$

Showing main implications

1. Suppose $(x_n{)}_{n=1}^{\infty }$ is convergent and ${\lim }_{n\to \infty }{x}_n=a$
   Let $ϵ>0$ so by definition there exists $N$ such that if $n\ge N$ then $d(x_n,a)<\frac{ϵ}{2}$ so
   For $m,n\ge N$ then

$$d(x_m,x_n)\le d(x_n,a)+d(x_m,a)<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$$

Hence $(x_n{)}_{n=1}^{\infty }$ is Cauchy
2) Suppose $(x_n{)}_{n=1}^{\infty }$ is Cauchy so let $ϵ=1$ so there exists $N$ such that for all $m,n\ge N$

$$d(x_m,x_n)<1$$

So then for all $n\ge N$ then $x_n\in B(x_N,1)$
Hence by defining $R:=\max \{d(x_N,x_1)+1,\cdots ,d(x_N,x_{N-1})+1\}$ then

$$x_n\in B(x_N,R)\text{ for all }n\ge 1$$

Thus $(x_n{)}_{n=1}^{\infty }$ is bounded

Convergence in Product Space with Convergence in Metric Space lemma

Let $X,Y$ be metric spaces then

Sequence $((x_n,y_n){)}_{n=1}^{\infty }$ in $X\times Y$ converges if and only if

$$(x_n{)}_{n=1}^{\infty }\text{ converges in }X\text{ and }(y_n{)}_{n=1}^{\infty }\text{ converges in }Y$$

Proof

Projection Maps $p_X:X\times Y\to X$ and $p_Y:X\times Y\to Y$ are continuous
$p_X$ and $p_Y$ are also Lipschitz continuous with constant $1$

If ${\lim }_{n\to \infty }(x_n,y_n)=(a,b)$ then

$$\begin{array}{rl}\underset{n\to \infty }{\lim }{x}_n& =\underset{n\to \infty }{\lim }{p}_X(x_n,y_n)=p_X(a,b)=a\\ \underset{n\to \infty }{\lim }{y}_n& =\underset{n\to \infty }{\lim }{p}_Y(x_n,y_n)=p_Y(a,b)=b\end{array}$$

Conversely if $x_n\to a$ and $y_n\to b$ then

$$d_{X\times Y}((x_n,y_n),(a,b))=\sqrt{d_X(x_n,a{)}^2+d_Y(y_n,b{)}^2}\to 0$$

as $n\to \infty$ hence $(x_n,y_n)\to (a,b)$ as $n\to \infty$