17 - Completeness

Completeness

Metric Space is complete if

$$\text{every Cauchy Sequence converges }$$

Relation between Completeness and Closure

Let $X$ be a complete metric space and $Y\subseteq X$

$$Y\text{ is Complete}⟺Y\text{ is Closed}$$

Proof

Let $X$ be a complete metric space and let $Y\subseteq X$ then

Suppose $Y$ is closed then
Let $(y_n{)}_{n=1}^{\infty }$ be a Cauchy Sequence in $Y$ then it is a Cauchy Sequence in $X$
Since $X$ is complete, it converges so suppose ${\lim }_{n\to \infty }{y}_n=a$
By Interior points and convergent sequences then $a\in Y$ hence $Y$ is complete

Suppose $Y$ is complete then
Let $(y_n{)}_{n=1}^{\infty }$ be a sequence of elements of $Y$ with ${\lim }_{n\to \infty }{y}_n=a$ for some $a\in X$ then
$(y_n{)}_{n=1}^{\infty }$ is a Cauchy Sequence in $Y$ so by completeness it converges to an element in $Y$
By uniqueness of limits then $a\in Y$ thus $Y$ contains limits of such sequence
So by Interior points and convergent sequences then $Y$ is closed

Diameter

Let $X$ be a metric space and $Y\subseteq X$ a non-empty subset then

Diameter of $Y$ is defined as

$$\mathrm{diam}(T):=\sup \{d(x,y):x,y\in Y\}$$

when the set is bounded otherwise it is infinity

Cantor's Intersection Theorem lemma

Let $X$ be a complete metric space and suppose that $S_1\supseteq S_2\supseteq \cdots$
such that it forms a nested sequence of non-empty closed sets in $X$ with property

$$\mathrm{diam}(S_n)\to 0\text{ as }n\to \infty$$

Then

$$\bigcap _{n=1}^{\infty }{S}_n\text{ contains a unique point a}$$

Proof

For each $n\in N$, pick $x_n\in S_n$
Let $ϵ>0$ so there exists large $N>0$ such that

$$diam(S_N)<ϵ$$

If $n,m\ge N$ then since $S_i$ are nested then

$$x_n,x_m\in S_n$$

By definition of diameter then

$$d(x_n,x_m)\le \mathrm{diam}(S_N)<ϵ$$

Hence $(x_n{)}_{n=1}^{\infty }$ is Cauchy

As $X$ is complete then ${\lim }_{n\to \infty }{x}_n=a$ for some $a\in X$
For each $i$, by the nesting property of sets $S_i$ then

$$x_n\in S_i\text{ for all }n\ge i$$

Since $S_i$ is closed then by Interior points and convergent sequences then

$$a\in S_i$$

As this holds true for all $i$ then $a\in {\bigcap }_{i=1}^{\infty }{S}_i$

Suppose there exists $b\in {\bigcap }_{i=1}^{\infty }{S}_i$ then

$$d(a,b)\le \mathrm{diam}(S_i)\text{ for all }i$$

Since $diam(S_i)\to 0$ then $d(a,b)=0$ so $a=b$

Hence $a$ is unique