LEC CAL1 6

Cantor intersection theorem

For a non empty subset $E$ of a metric space $(X,d)$, we say $E$ is bounded provided it has finite diameter. A descending sequence $(E_n)$ of non empty subsets of $X$ is called a contracting sequence if ${\lim }_{n\to \infty }\text{diam}(E_n)=0$.

Definition 175.1(Definition).

A metric space $X$ is said to have the Cantor intersection property if whenever $(F_n)$ is a contracting sequence of non empty closed subsets of $X$, ${\bigcap }_{n=1}^{\infty }{F}_n=\{x\}$, $x\in X$.

Note the similarity between the Cantor intersection property (of a metric space) and the nested interval property (of $\mathbb{R}$).

Theorem 175.2(Cantor Intersection Theorem).

A metric space $X$ has the Cantor intersection property if and only if it is complete.

Proof.

$(⟹)$ Let $(s_n)$ be a Cauchy sequence. For each index $n$ define $F_n$ to be $\{s_i|i\ge n\}$. For all $ϵ>0$, there exists $N$ such that for all $n,m>N$, $|s_n-s_m|<ϵ$. Thus, $\text{diam}(F_N)\le ϵ$, from which it follows that $\text{diam}(\overline{F_N})\le ϵ$ (for all $N^{\prime }>N$ too, since the diameters are decreasing). Thus, ${\lim }_{n\to \infty }\text{diam}(\overline{F_n})=0$, and $(\overline{F_n})$ is a contracting sequence. Thus, ${\bigcap }_{n=1}^{\infty }\overline{F_n}=\{s\}$, $s\in X$. For all $F_n$, $s\in F_n$ or $s$ is a limit point of $F_n$ (or both). In any case, $B_{\frac{1}{k}}(s)\cap F_n\ne \varnothing$ for all $k,n$. This allows us to select a subsequence of $(s_n)$ which converges to $s$. If a subsequence of a Cauchy sequence converges to a point, the entire sequence converges to said point. Thus, $(s_n)\to s$, every Cauchy sequence converges, and $X$ is complete.

$(⟸)$ Let $(F_n)$ be a contracting sequence of nonempty closed subsets of $X$. Pick $s_n\in F_n$ for each $n\in \mathbb{N}$. For any $ϵ>0$, there exists $F_N$ such that $\text{diam}(F_N)<ϵ$, $i$.$e$, $d(s_n,s_m)<ϵ$ for all $n,m>N$. This makes $(s_n)$ a Cauchy sequence. Since $X$ is complete, it must converge, say to a point $s$. Then, $s$ is a limit point of each $F_n$. Since $F_n$ is closed for all $n$, $s\in F_n$ for all $n$. Thus, $s\in {\bigcap }_{n=1}^{\infty }{F}_n$. The intersection does not contain more than one point, since if it did, then ${\lim }_{n\to \infty }\text{diam}(F_n)\ne 0$.□

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Completion of a metric space

Definition 175.3(Definition).

Let $(X,d)$ be a metric space. Then there is a complete metric space $(\tilde{X},\tilde{d})$ called the completion of $(X,d)$ for which $X$ is a dense subset of $\tilde{X}$ and $d(u,v)=\tilde{d}(u,v)$ for all $u,v\in X$.

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Constructing the completion of a metric space

Let $(X,d)$ be a metric space. We will assume that $\mathbb{R}$ is a complete metric space, so remember that what follows hinges on a separate proof of the completeness of $\mathbb{R}$.

Suppose $(x_n)$ and $(y_n)$ are two Cauchy sequences in $X$. Then, $(d(x_n,y_n))$ is a Cauchy sequence in $\mathbb{R}$, and hence must converge. Let $S$ be the space of all Cauchy sequences in $X$. Define $\rho ((x_n),(y_n))\equiv {\lim }_{n\to \infty }d(x_n,y_n)$. It is easy to see that

• $\rho \ge 0$, and $(x_n)=(y_n)$ implies $\rho ((x_n),(y_n))=0$.
• $\rho$ is symmetric
• triangle inequality holds

However, $\rho ((x_n),(y_n))=0$ does not imply $(x_n)=(y_n)$. $\rho$ is what is called a pseudometric, and $(S,\rho )$ is called a pseudometric space. On such a space, one can define the relation $(x_n)\sim (y_n)$ if $\rho ((x_n),(y_n))=0$. This is an equivalence relation, and partitions $S$ into equivalence classes $S/\sim =:\tilde{X}$. Define $\tilde{d}([(x_n)],[(y_n)])=\rho ((x_n),(y_n))$. This is well defined¹. That makes $(\tilde{X},\tilde{d})$ a metric space.

$(\tilde{X},\tilde{d})$ is complete

Consider a Cauchy sequence in $\tilde{X}$: $([(x_{n,1})],[(x_{n,2})],\dots )$. By passing to subsequences if necessary, we can assume that given $m$, $\tilde{d}([(x_{n,k})],[(x_{n,l})])<2^{-m}$ for all $l,k\ge m$. Similarly, by passing to subsequences if necessary, we can assume that for all $k$, given $c$, $d(x_{a,k},x_{b,k})<2^{-c}$ for all $a,b\ge c$.

Let $(z_n)\equiv (x_{n,n})$. We will first show that $(z_n)$ is a Cauchy sequence in $X$, and then that $([(x_{\ast ,n})])\to [(z_n)]$.

Let $ϵ>0$. Choose $N$ such that $2^{-N}<ϵ/3$. Now, for all $n,m\ge N$, $d(z_n,z_m)=d(x_{n,n},x_{m,m})$. Since $\tilde{d}([(x_{\ast ,n})],[(x_{\ast ,m})])<2^{-N}$, ${\lim }_{k\to \infty }d(x_{k,n},x_{k,m})<2^{-N}$, so there exists $M$ such that $d(x_{k,n},x_{k,m})<2^{-N}$ when $k\ge M$. Take $M^{\prime }=\max \{M,n,m\}$. Now,

$$\begin{array}{rl}d(x_{n,n},x_{m,m})& =d(x_{n,n},x_{M^{\prime },n})+d(x_{M^{\prime },n},x_{M^{\prime },m})+d(x_{M^{\prime },m},x_{m,m})\\ & <2^{-n}+2^{-N}+2^{-m}\\ & <3\cdot 2^{-N}\\ & <ϵ.\end{array}$$

Thus, $d(z_n,z_m)<ϵ$ for all $n,m\ge N$, and $(z_n)$ is a Cauchy sequence in $X$.

Let $ϵ>0$. Choose $N$ such that $2^{-N}<ϵ/3$. Let $k>N$. $\tilde{d}([(x_{n,k})],[(z_n)])={\lim }_{n\to \infty }d(x_{n,k},x_{n,n})$. For $n>k$, we can always find $M\ge n$ such that $d(x_{M,k},x_{M,n})<2^{-N}$. So, we have $d(x_{n,k},x_{n,n})\le d(x_{n,k},x_{M,k})+d(x_{M,k},x_{M,n})+d(x_{M,n},x_{n,n})$ $<2^{-k}+2^{-N}+2^{-k}<3\cdot 2^{-N}<ϵ$. Thus,

$$\begin{array}{rl}\tilde{d}([(x_{n,k})],[(z_n)])& =\underset{n\to \infty }{\lim }d(x_{n,k},x_{n,n})\\ & <ϵ.\end{array}$$

Thus, $([(x_{\ast ,n})])\to [(z_n)]$ and $(\tilde{X},\tilde{d})$ is complete.

$X$ is dense in $(\tilde{X},\tilde{d})$

For each $p\in X$, there is a Cauchy sequence all of whose terms are $p$. Let $P_p$ be the element of $\tilde{X}$ which contains this sequence. Define the map $\varphi :X\to \tilde{X}$ by $\varphi (p)=P_p$. It is easy to see that $\varphi$ is an isometry, $i$.$e$, $\tilde{d}(P_p,P_q)=d(p,q)$ for all $p,q\in X$.

We will now show that $\varphi (X)$ is dense in $\tilde{X}$. Let $[(x_n)]\in \tilde{X}$ such that $[(x_n)]\notin \varphi (X)$. Consider the sequence $(P_{x_1},P_{x_2},\dots )$ in $\tilde{X}$. Since $(x_n)$ is a Cauchy sequence in $X$ and $\tilde{d}(P_{x_k},P_{x_m})=d(x_k,x_m)$, $(P_{x_n})$ is a Cauchy sequence in $\tilde{X}$. Let $ϵ>0$. There exists $N$ such that for all $k,m>N$, $d(x_k,x_m)<ϵ$. Thus, for $k>N$, $\tilde{d}(P_{x_k},[(x_n)])={\lim }_{m\to \infty }d(x_k,x_m)<ϵ$. Therefore, $(P_{x_n})\to [(x_n)]$, proving that $\varphi (X)$ is dense in $\tilde{X}$.

Footnotes

1. Let $(x_n)\sim (a_n)$ and $(y_n)\sim (b_n)$. Write $d(x_n,y_n)\le d(x_n,a_n)+d(a_n,b_n)+d(b_n,y_n)$ and its dual equation bounding $d(a_n,b_n)$. Combine them to get $|d(x_n,y_n)-d(a_n,b_n)|\le d(x_n,a_n)+d(y_n,b_n)$. It follows that ${\lim }_{n\to \infty }d(x_n,y_n)={\lim }_{n\to \infty }d(a_n,b_n)$, or $\tilde{d}([(x_n)],[(y_n)])=\tilde{d}([(a_n)],[(b_n)])$. ↩