Suppose $(X,d)$ is a metric space and that $F$ is a nest of non empty subsets of $X$ for which $\inf\{\operatorname{diam} ~(A)~|~A \in F\} =0.$ Show that $\bigcap F = \emptyset$ or $\bigcap F$ is a singleton set.
Attempt:
Let $A_i \in F$ where $F$ is a nest of non empty subsets of $X$ such that
$$\cdots \subseteq A_0 \subseteq A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n \subseteq \cdots .$$
Then, $~~ \cdots \le \operatorname{diam}(A_0) \le \operatorname{diam}(A_1) \le \operatorname{diam}(A_2) \le\cdots \le \operatorname{diam}(A_n) \le \cdots $
$$\inf\{\operatorname{diam} ~(A)~|~A \in F\} =0 \implies ~\forall~\epsilon_i >0,~\exists~A_i~|~\operatorname{diam}(A_i)<\epsilon_i$$
$\sup ~d(a,b) <\epsilon_i$ where $~a,b \in A_i.$
Since, this is valid for all $\epsilon_i >0$, this means, $d(a,b)=0~\forall~a,b \in A_i~~\implies a=b.$
Since, $F$ is a nest of non-empty subset of $X\implies \bigcap F = \{a\},$ a singleton set.
However, I am not able to prove the possibility for $\bigcap F = \emptyset$.
If $\bigcap F = \emptyset$, then, will the condition $\inf\operatorname{diam} ~(A)~|~A \in F\} =0$ be satisfied?
Could someone please check if my above attempt is correct . Also please tell me how to prove the possibility for $\bigcap F = \emptyset$.
Thank you very much for your help in this regard.