Nested Set Property I am having trouble understanding the Nested Set Property. For example, say I 
would want to verify the nested set property for the following:
$$F_k=\{x\in \mathbb{R} | \ x\geq0,\  2\leq x^2\leq 2+1/k \}.$$
I know that the Nested Set Property says that if $F_k$ is a sequence of nonempty 
compact sets in a metric space $M$ such that $F_{k+1}\subset F_k \forall k$ 
then there is at least one point in $\bigcap_{k=1}^\infty F_k$. But how would I 
go about proving this for a given set or finding such a point explicitly?
 A: Find a point that is in all the sets!
Notice in this case that $2\leq\sqrt{2}^2\leq2+1/k$ for all positive integers $k$. Thus $\sqrt{2}\in F_k$ for all positive integers $k$. Thus $\sqrt{2}\in\cap F_k$.
A: Take the def'n of compact: $S$ is compact iff every open cover of $S$ has a finite sub-cover ,which means that whenever  $F$ is an open cover of $S$ (which means $F$ is a family of open sets and $\cup F\supset S$)  there is a finite $G\subset F$ such that $\cup G\supset F$. Now the def'n of the F.I.P. (Finite Intersection Property) is that a family $H$ of sets has the F.I.P. iff every finite non-empty $J\subset H$ satisfies $\cap J\ne \phi.$  $\bullet $ We have: A space  $S$ is compact iff every non-empty  family $K$ of closed sets that has the F.I.P. satisfies $\cap K\ne \phi.$ To prove this :(1)If $S$ is compact, let $K$ be a non-empty family of closed subsets of $S$ where $K$ has the F.I.P. Let $[K]^{<\omega}$ be the set of finite non-empty subsets of $K$.(This is standard notation.) Let $F= \{S\backslash \cap J :\phi \ne J\in [K]^{<\omega}\}.$ Now if $S$ is compact then we cannot have $\cap K=\phi$ otherwise $F$ is an open cover of $S$ with no finite sub-cover. (2) If $S$ is not compact then $S$ has an open cover $C$ with no finite sub-cover. Then $K$=$\{S\cup D :\phi \ne D\in [C]^{<\omega}\}$  has the F.I.P. but $\cap K=\phi$.$\bullet$ In your Q, prove that $F_1$ is compact and that every $F_k$ is closed in the subspace $F_1$. And since $\{F_k:K\in N\}$ has the F.I.P. ,therefore $\cap_kF_k\ne \phi.$ But we can also see at a glance that $\sqrt 2\in \cap_kF_k$.
