I know that in a complete metric space, if $F_1\supseteq F_2\supseteq…$ are closed, and $\text{diam}(F_n)\to0,$ then $\bigcap\{F_n\}\not=\emptyset.$ John B. Conway claims falsely that even if $\text{diam}(F_N)\not=0,$ but is still bounded for all $n$, then $\bigcap\{F_n\}\not=\emptyset.$ I say he claims it falsely as we can conisder
$$F_n=\{n,n+1,n+2,…\}$$
as a subset of $\mathbb{N}$ with the discrete metric, in which case $F_n$ is trivially closed and bounded, since discrete, but $\bigcap\{F_n\}=\emptyset.$
Now, I am curious if there are conditions we can add to Conway's statement to make it correct? What requirements would we need on a metric space to guarantee that for all bounded closed sets $F_1\supseteq F_2\supseteq…$ we would have $\bigcap\{F_n\}\not=\emptyset$?
Best Answer
If the sets were closed and totally bounded in a complete metric space then it would be correct. As you've shown, merely bounded is not enough, but if they're totally bounded then they are compact, and hence the intersection is nonempty.
In the usual metric on $\mathbb R^n$, all bounded subsets are totally bounded. If the metric has this property, then the statement is correct.