# 1.1 Convergent sequences, cauchy sequences, and complete spaces  (Page 2/2)

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Proof: Assume that a sequence ${x}_{n}\to x$ in $\left(X,d\right)$ . Fix $ϵ>0$ . Since $\left\{{x}_{n}\right\}$ is convergent, there must exist an ${n}_{0}\in {\mathbb{Z}}^{+}$ such that $d\left({x}_{n},x\right)<ϵ/2$ for all $n\ge {n}_{0}$ . Now, pick $j,k\ge {n}_{0}$ . Then, using the triangle inequality, we have $d\left({x}_{j},{x}_{k}\right)\le d\left({x}_{j},x\right)+d\left(x,{x}_{k}\right)<ϵ/2+ϵ/2$ (since both $j$ and $k$ are greater or equal to ${n}_{0}$ ) and so $d\left({x}_{j},{x}_{k}\right)<ϵ$ . Therefore, the sequence $\left\{{x}_{n}\right\}$ is Cauchy.

One may wonder if the opposite is true: is every Cauchy sequence a convergent sequence?

Example 3 Focus on the metric space $\left(X,{d}_{0}\right)$ with $X=\left(0,1\right]=\left\{x\in \mathbb{R}:0 . Now consider the sequence ${x}_{n}=1/n$ in $X$ . This is a Cauchy sequence: one can show this by picking ${n}_{0}\left(ϵ\right)>⌈2/ϵ⌉$ and using the triangle inequality to get

${d}_{0}\left({x}_{j},{x}_{k}\right)\le {d}_{0}\left({x}_{j},0\right)+{d}_{0}\left({x}_{k},0\right)=|{x}_{j}|+|{x}_{k}|=\frac{1}{j}+\frac{1}{k}\le \frac{1}{{n}_{0}}+\frac{1}{{n}_{0}}=\frac{2}{{n}_{0}}<\frac{2}{⌈2/ϵ⌉}\le \frac{2}{2/ϵ}=ϵ.$

However, this is not a convergent sequence: we've shown earlier that ${x}_{n}\to 0$ . Since $0\notin \left(0,1\right]$ and a sequence has a unique limit, then there is no $x\in \left(0,1\right]$ such that ${x}_{n}\to x$ .

## Complete metric spaces

Whether Cauchy sequences converge or not underlies the concept of completeness of a space.

Definition 4 A complete metric space is a metric space in which all Cauchy sequences are convergent sequences.

Example 5 The metric space $\left(X,{d}_{0}\right)$ with $X=\left(0,1\right]$ from earlier is not complete, since we found a Cauchy sequence that converges to a point outside of $X$ . We can make it complete by adding the convergence point, i.e., if ${X}^{\text{'}}=\left[0,1\right]$ then $\left({X}^{\text{'}},{d}_{0}\right)$ is a complete metric space.

Example 6 Let $C\left[T\right]$ denote the space of all continuous functions with support $T$ . If we pick the metric ${d}_{2}\left(x,y\right)={\left({\int }_{T},{|x\left(t\right)-y\left(t\right)|}^{2},d,t\right)}^{1/2}$ , then $\left(X,{d}_{2}\right)$ is a metric space; however, it is not a complete metric space.

To show that the space is not complete, all we have to do is find a Cauchy sequence of signals within $C\left[T\right]$ that does not converge to any signal in $C\left[T\right]$ . For simplicity, fix $T=\left[-1,1\right]$ . Fortuitously, we find the sequence illustrated in [link] , which can be written as

${x}_{n}\left(t\right)=\left\{\begin{array}{cc}-1\hfill & \mathrm{if}\phantom{\rule{3.33333pt}{0ex}}t<-1/n,\hfill \\ 1\hfill & \mathrm{if}\phantom{\rule{3.33333pt}{0ex}}t>1/n,\hfill \\ nt\hfill & \mathrm{if}\phantom{\rule{3.33333pt}{0ex}}-1/n\le t\le 1/n.\hfill \end{array}\right)$

Since all these functions are continuous and defined over $T$ , then $\left\{{x}_{n}\right\}$ is a sequence in $C\left[T\right]$ . We can show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence: let ${n}_{0}$ be an integer and pick $j\ge k\ge {n}_{0}$ . Then,

$\begin{array}{cc}\hfill {d}_{2}\left({x}_{j},{x}_{k}\right)& ={\left({\int }_{-1}^{1},{|{x}_{k}\left(t\right)-{x}_{j}\left(t\right)|}^{2},d,t\right)}^{1/2}={\left({\int }_{-1/j}^{1/j},{|{x}_{k}\left(t\right)-{x}_{j}\left(t\right)|}^{2},d,t\right)}^{1/2}\le {\left({\int }_{-1/j}^{1/j},{1}^{2},d,t\right)}^{1/2}\hfill \\ & \le {\left(2/j\right)}^{1/2}\le \sqrt{2/{n}_{0}}.\hfill \end{array}$

So for a given $ϵ>0$ , by picking ${n}_{0}$ such that $\sqrt{2/{n}_{0}}<ϵ$ (say, for example, ${n}_{0}>⌈2/{ϵ}^{2}⌉$ ), we will have that ${d}_{2}\left({x}_{j},{x}_{k}\right)<ϵ$ for all $j,k>{n}_{0}$ ; thus, the sequence is Cauchy. Now, we must show that the sequence does not converge within $C\left[T\right]$ : we will find a point ${x}^{*}\notin C\left[T\right]$ such that for ${X}^{\text{'}}=C\left[T\right]\cup \left\{{x}^{*}\right\}$ the sequence ${x}_{n}\to {x}^{*}$ in $\left({X}^{\text{'}},{d}_{2}\right)$ . By inspecting the sequence of signals, we venture the guess

${x}^{*}\left(t\right)=\left\{\begin{array}{cc}-1\hfill & \mathrm{if}\phantom{\rule{3.33333pt}{0ex}}t<0,\hfill \\ 1\hfill & \mathrm{if}\phantom{\rule{3.33333pt}{0ex}}t>0,\hfill \\ 0\hfill & \mathrm{if}\phantom{\rule{3.33333pt}{0ex}}t=0.\hfill \end{array}\right),$

For this signal, we will have

$\begin{array}{cc}\hfill {d}_{2}\left({x}_{n},{x}^{*}\right)& ={\left({\int }_{-1}^{1},{|{x}_{n}\left(t\right)-{x}^{*}\left(t\right)|}^{2},d,t\right)}^{1/2}={\left({\int }_{-1}^{0},|,{x}_{n},\left(t\right),-,{x}^{*},{\left(t\right)|}^{2},d,t,+,{\int }_{0}^{1},{|{x}_{n}\left(t\right)-{x}^{*}\left(t\right)|}^{2},d,t\right)}^{1/2}\hfill \\ & ={\left({\int }_{-1/n}^{0},{|nt-\left(-1\right)|}^{2},d,t,+,{\int }_{0}^{1/n},{|nt-1|}^{2},d,t\right)}^{1/2}\hfill \\ & ={\left({\int }_{-1/n}^{0},{|nt+1|}^{2},d,t,+,{\int }_{0}^{1/n},{|1-nt|}^{2},d,t\right)}^{1/2}\hfill \\ & ={\left({\int }_{0}^{1/n},{|1-nt|}^{2},d,t,+,{\int }_{0}^{1/n},{|1-nt|}^{2},d,t\right)}^{1/2}={\left(2,{\int }_{0}^{1/n},{\left(1-nt\right)}^{2},d,t\right)}^{1/2}\hfill \\ & ={\left(\frac{2}{3n}\right)}^{1/2}.\hfill \end{array}$

So if we select ${n}_{0}$ such that ${\left(\frac{2}{3{n}_{0}}\right)}^{1/2}<ϵ$ , i.e., ${n}_{0}>\frac{2}{3{ϵ}^{2}}$ , then we have that ${d}_{2}\left({x}_{n},{x}^{*}\right)<ϵ$ for $n>{n}_{0}$ , and so we have shown that ${x}_{n}\to {x}^{*}$ . Now, since a convergent sequence has a unique limit and ${x}^{*}\notin C\left[T\right]$ , then $\left\{{x}_{n}\right\}$ does not converge in $\left(C\left[T\right],{d}_{2}\right)$ and this is not a complete metric space.

The property of equivalence between Cauchy sequences and convergent sequences often compels us to define metric spaces that are complete by choosing the metric appropriate to the signal space. For example, by switching the distance metric to ${d}_{\infty }\left(x,y\right)={sup}_{t\in T}|x\left(t\right)-y\left(t\right)|$ , the metric space $\left(C\left[T\right],{d}_{\infty }\right)$ becomes complete.

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