The set $\mathbb{Q}$ of rational numbers is not complete (or not a continuum) since it has gaps or holes. For instance, $\sqrt{2}$ is not in $\mathbb{Q}$. On the other hand, the set $\mathbb{R}$ of real numbers has no gaps or holes, so it is complete (or is a continuum). Let $(x_n)$ be a sequence of real numbers. Suppose that $(x_n)$ converges to a real number $x$. Then by the triangle inequality, for any $m,n\in\mathbb{N}$, we have

$$|x_m-x_n|\leq |x_m-x|+|x-x_n|.$$

Hence, $\displaystyle\lim_{m,n\to\infty}|x_m-x_n|=0$, i.e. $(x_n)$ is a Cauchy sequence. Conversely, Georg Cantor introduced the completeness axiom that every Cauchy sequence of real numbers converges and defined a real number as the limit of a Cauchy sequence of rational numbers. For instance, consider the Cauchy sequence $(x_n)$ defined by

$$x_1=1,\ x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n},\ \forall n\geq 2.$$

If $(x_n)$ converges to a number $x$, it would satisfy $x^2=2$ i.e. $(x_n)$ converges to $\sqrt{2}$. There is another way to obtain the completeness of $\mathbb{R}$ by a Dedekind cut, though we are not going to delve into that here.

More generally, one can also consider a complete metric space and that is what we are going to study in this lecture.

*Definition*. A sequence $(x_n)$ is a metric space $(X,d)$ is said to converge or to be convergent to $x\in X$ if

$$\lim_{n\to\infty}d(x_n,x)=0.$$

$x$ is called the limit if $(x_n)$ and we write

$$\lim_{n\to\infty}x_n=x\ \mbox{or}\ x_n\rightarrow x.$$

If $(x_n)$ is not convergent, it is sad to be divergent. We can generalize the definiton of the convergence of a sequence we learned in calculus in terms of a metric as:

*Definition*. $\displaystyle\lim_{n\to\infty}d(x_n,x)=0$ if and only if given $\epsilon>0$ $\exists$ a positive integer $N$ s.t. $x_n\in B(x,\epsilon)$ $\forall n\geq N$.

A nonempty subset $M\subset X$ is said to be bounded if

$$\delta(M)=\sup_{x,y\in M}d(x,y)<\infty.$$

*Lemma*. Let $(X,d)$ be a metric space.

(a) A convergent sequence in $X$ is bounded and its limit is unique.

(b) If $x_n\rightarrow x$ and $y_n\rightarrow y$, then $d(x_n,y_n)\rightarrow d(x,y)$.

*Proof*. (a) Suppose that $x_n\rightarrow x$. Then one can find a positive integer $N$ such that $d(x_n,x)<1$ $\forall n\geq N$. Let $M=2\max\{d(x_1,x),\cdots,d(x_{N-1},x),1\}$. Then for all $m,n\in\mathbb{N}$,

\begin{align*}

d(x_m,x_n)&\leq d(x_m,x)+d(x,x_n)\ (\mbox{ (M3) triangle inequality)}\\

&\leq M.

\end{align*}

This means that $\delta((x_n))\leq M<\infty$ i.e. $(x_n)$ is bounded.

Suppose that $x_n\rightarrow x$ and $x_n\rightarrow y$. Then

\begin{align*}

0\leq d(x,y)&\leq d(x,x_n)+d(x_ny)\\

&\rightarrow 0

\end{align*}

as $n\to\infty$. So, $d(x,y)=0\Rightarrow x=y$ by (M1).

(b) By (M3),

$$d(x_n,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n)$$

and so we obtain

$$d(x_n,y_n)-d(x,y)\leq d(x_n)+d(y,y_n).$$

Similarly, we also obtain the inequality

$$d(x,y)-d(x_n,y_n)\leq d(x,x_n)+d(y_n,y).$$

Hence,

$$0\leq |d(x_n,y_n)-d(x,y)|\leq d(x_n,x)+d(y_n,y)\rightarrow 0$$

as $n\to\infty$.

*Definition*. A sequence $(x_n)\subset (X,d)$ is said to be Cauchy if given $\epsilon>0$ $\exists$ a positive integer $N$ such that

$$d(x_m,x_n)<\epsilon\ \forall m,n\geq N.$$

The space $X$ is said to be complete if every Cauchy sequence in $X$ converges.

*Examples*. The real line $\mathbb{R}$ and the complex plane $\mathbb{C}$ are complete.

*Theorem*. Every convergent sequence is Cauchy.

*Proof*. Suppose that $x_n\rightarrow x$. Then given $\epsilon>0$ $\exists$ a poksitive integer $N$ s.t. $d(x_n,x)<\frac{\epsilon}{2}$ for all $n\geq N$. Now, $\forall m,n\geq N$

$$d(x_m,x_n)\leq d(x_m,x)+d(x,x_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$

Therefore, $(x_n)$ is Cauchy.

*Theorem*. Let $M$ be a nonempty subset of a metric space $(X,d)$. Then

(a) $x\in\bar M\Longleftrightarrow \exists$ a seqence $(x_n)\subset M$ such that $x_n\rightarrow x$.

(b) $M$ is closed $\Longleftrightarrow$ given a sequence $(x_n)\subset M$, $x_n\rightarrow x$ implies $x\in M$.

*Proof*. (a) ($\Longrightarrow$) Since $x\in\bar M$, $\forall n\in\mathbb{N}$ $\exists x_n\in B\left(x,\frac{1}{n}\right)\cap M\ne\emptyset$. Let $\epsilon>0$ be given. Then by the Archimedean property, $\exists$ a positive integer $N$ s.t. $N\geq\frac{1}{\epsilon}$. Now,

$$n\geq N\Longrightarrow d(x_n,x)<\frac{1}{n}\leq\frac{1}{N}<\epsilon.$$

($\Longleftarrow$) Suppose that $\exists$ a sequence $(x_n)\subset M$ s.t. $x_n\rightarrow x$. Then given $\epsilon>0$ $\exists$ a positive integer $N$ s.t. $x_n\in B(x,\epsilon)$ $\forall n\geq N$. This means that $\forall\epsilon>0$, $B(x,\epsilon)\cap M\ne\emptyset$. So, $x\in\bar M$.

(b) ($\Longrightarrow$) Clear

($\Longleftarrow$) It suffices to show that $\bar M\subset M$. Let $x\in\bar M$. Then $\exists$ a sequence $(x_n)\subset M$ such that $x_n\rightarrow x$. By assumption, $x\in M$.

*Theorem*. A subspace $M$ of a complete metric space $X$ itself is complete if and only if $M$ is closed in $X$.

*Proof*. ($\Longrightarrow$) Let $M\subset X$ be complete. Let $(x_n)$ be a sequence in $M$ such that $x_n\rightarrow x$. Then $(x_n)$ is Cauchy. Since $M$ is complete, every Cauchy sequence must converge and hence $x\in M$. This means that $M$ is closed.

($\Longleftarrow$) Suppose that $M\subset X$ is closed. Let $(x_n)$ be a Cauchy sequence in $M\subset X$. Since $X$ is complete, $\exists x\in X$ such that $x_n\rightarrow x$. Since $M$ is closed, $x\in M$. Therefore, $M$ is complete.

*Example*. In $\mathbb{R}$ with Euclidean metric, the closed intervals $[a,b]$ are complete. $\mathbb{Z}$, the set of integers is also complete by the above theorem since it is closed in $\mathbb{R}$. One can directly see why $\mathbb{Z}$ is complete without quoting the theorem though. Let $(x_n)$ be a Cauchy sequence in $\mathbb{Z}$. Then we see that there exists a positive integer $N$ such that $x_N=x_{N+1}=x_{N+2}=\cdots$. Hence any Cauchy sequence in $\mathbb{Z}$ is a convergent sequence in $\mathbb{Z}$. Therefore, $\mathbb{Z}$ is complete.

*Theorem*. A mapping $T: X\longrightarrow Y$ is continuous at $x_0\in X$ if and only if $x_n\rightarrow x$ implies $Tx_n\rightarrow Tx_0$.

*Proof*. ($\Longrightarrow$) Suppose that $T$ is continuous and $x_n\rightarrow x$ in $X$. Let $\epsilon>0$ be given. Then $\exists\delta>0$ s.t. whenever $d(x,x_0)<\delta$, $d(Tx,Tx_0)<\epsilon$. Since $x_n\rightarrow x$, $\exists$ a positive integer $N$ s.t. $d(x_n,x_0)<\delta$ $\forall n\geq N$. So, $\forall n\geq N$, $d(Tx_n,Tx_0)<\epsilon$. Hence, $Tx_n\rightarrow Tx_0$.

($\Longleftarrow$) Suppose that $T$ is not continuous. Then $\exists\epsilon>0$ s.t. $\forall\delta>0$, $\exists x\ne x_0$ satisfying $d(x,x_0)<\delta$ but $d(Tx,tx_0)\geq\epsilon$. So, $\forall n=1,2,\cdots$, $\exists x_n\ne x_0$ satisfying $d(x_n,x_0)<\frac{1}{n}$ but $d(Tx_n,Tx_0)\geq\epsilon$. This means that $x_n\rightarrow x_0$ but $Tx_n\not\rightarrow Tx_0$.