# Prove that metric Space $X$ must be complete

If every nest $F$ of non empty closed subsets of a metric spaces $X$ satisfying $\inf\{diam(A)~|~A \in F\}=0$ has singleton intersection,then $X$ must be complete.

Attempt:

My textbook gives the following approach : Suppose $Y$ is a metric superspace of $X$ in which $X$ is not closed and let $z \in Cl_Y(X) - X$.

( Note that $Cl_A(B)$ implies the closure of set $B$ in the metric space $A$.)

For each $r>0,$ let $B_r=Cl_X(\{x \in X~|~d(x,z)<r\}).$ Then, it can be easily checked that :

$(i)~\{B_r~|~r \in \mathbb R^+ \}$ is a nest of non empty closed subsets of $X$.

Proof: if $r_1<r_2,$ then, $\{x \in X~|~d(x,z)<r_1\} \subseteq \{x \in X~|~d(x,z)<r_2\}$

$(ii)~diam(B_r) \le 2r ~\forall~ r \in \mathbb R^+$

Proof: Let $x,y \in B_r = Cl_X(\{x \in X~|~d(x,z)<r\}).$.

Then, $d(x,y) \le d(x,z) + d(z,y) \le 2r$

$\implies diam(B_r) \le 2r \le \infty$.

So, everything is well defined here.

$(iii)~ \bigcap \{B_r~|~r \in \mathbb R^+ \} = \emptyset$

we already know that if $(X,d)$ is a metric space and $F$ is the nest of non empty subsets of $X$ for which $\inf\{diam(A)~|~A \in F\}=0$, then $\bigcap F= \emptyset$ or a singleton set.

So, to prove that $X$ is complete, we must being a contradiction that $\bigcap \{B_r~|~r \in \mathbb R^+ \} = \emptyset$ contradicting the initial given statement that the intersection is a singleton.

Please guide me on how to move forward. Thank you very much!

• – Siminore Jul 16 '15 at 11:59
• Note that there is a much simpler approach. Take a Cauchy sequence $\{u_n\}$, and let $F_k$ be the closure of $\{ u_n \mid n \ge k \}$. Then $F_0 \supset F_1 \supset \dots$ is a nested sequence of closed sets, and because $u_n$ is Cauchy, $\inf \operatorname{diam} F_k = 0$. Then $x \in \bigcap F_k$ is the limit of $\{u_n\}$. (There are some details to fill in but the idea is here.) – Najib Idrissi Jul 16 '15 at 12:09

"we already know that if $(X,d)$ is a metric space and $F$ is the nest of non empty subsets of $X$ for which $\inf\{diam(A)~|~A \in F\}=0$, then $\bigcap F= \emptyset$ or a singleton set."
If you set $\overline{B_r}= Cl_Y(B_r)$ to be the closure of $B_r$ in Y, then clearly $$z\in \bigcap \overline{B_r}\neq \emptyset.$$
Applying your statement again, we conclude that $\bigcap \overline{B_r}$ is a singleton. Clearly, $\bigcap B_r\subset\bigcap \overline{B_r}$, and $z\notin \bigcap B_r$, since $z\notin X$.
Hence, $\bigcap B_r = \emptyset$ as it is the subset of a singlton and does not contain that element.