1
$\begingroup$

Let $(X,d)$ be a metric space. Let $\mathcal{C}$ be the set of all collections $\{O_i\}_{i=1}^\infty$ of non-empty closed subsets such that \begin{align*} &(a) O_{n+1}\subset O_n \forall n \\ &(b) \lim\operatorname{diam} (O_n) = 0 \ as \ n \to \infty \end{align*}

Prove that $X$ is complete if and only if $\forall C \in \mathcal{C}$ \begin{align*} \bigcap_{A\in C} A \not= \emptyset \end{align*} For the $if$ part: For every $n$, choose $x_n \in O_n$. Then since $O_{n+1}\subset O_n$, the set $\{x_n,x_{n+1},x_{n+2},\cdots\}\subset O_n$. Since $\lim\operatorname{diam}(O_n) = 0$, for any $\epsilon > 0$ choose a natural number $N$ so that $\operatorname{diam}(O_n)<\epsilon$ for $n\geq N$. This means that for any $n,m \geq N$, $|x_n-x_m| \leq\operatorname{diam}(\{x_{N},x_{N+1},\dots\}) \leq\operatorname{diam}(O_N) < \epsilon$. So $\{x_n\}$ is a Cauchy sequence. By completeness of $X$, $\{x_n\}$ converges to a point, $a$. By $O_n$ being closed it must contain $\{x_n,x_{n+1},\dots\}$. Thus, for any given $D\in\mathcal{C}$ we get \begin{align*} a\in \bigcap_{A \in D} A. \end{align*} hence non-empty.

Also, I am not sure how to start the other direction. I think you need to look at the tails of each sequence

$\endgroup$
0
$\begingroup$

Consider the Cauchy sequence $\{x_n\}$. Define $$F_n :=\overline{ \{x_m \vert m \geq n\}}$$.Note that $F_n $ is closed and satisfies the given two conditions. Let $\mathcal{F} = \{F_n\}_{n=1}^\infty$. Then $$F=\bigcap_{F_n\in \mathcal {F} } F_n \not= \emptyset$$. Take $x\in F $ and prove that $x $ indeed is the limit of the sequence $\{x_n\}$. You may have to use the fact that $d(A)= d(\overline {A})$, where $d(A) $ is the diameter of the set $A $ and $\overline {A}$ is the closure of $A $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.