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Is it true that every nested sequence of non-empty closed sets $(I_n)$ (one such that $I_{n+1} subset I_n$) has a non-empty intersection?

If $X$ is a compact space, this is true. $mathcal{F} = {I_alpha}_{alpha in I}$ is an arbitrary collection of nonempty closed sets such that any intersection of finitely many sets from $mathcal{F}$ is nonempty, then $bigcap_{alpha in I} I_alpha$ is nonempty.

To prove this, suppose $bigcap_{alpha in I} I_alpha$ is empty. Then ${X - I_alpha}$ is an open cover of $X$. Since $X$ is compact, there exists $alpha_1, ..., alpha_n$ such that $bigcup_{j = 1}^n (X - I_{alpha_j}) = X$. Then $bigcap_{j = 1}^n I_{alpha_j} = emptyset$. This contradicts the assumption that any finite intersection of elements of $mathcal{F}$ is nonempty.

For example (in your case) if $mathcal{F}$ consists of nonempty nested intervals, then $mathcal{F}$ has the finite intersection property and the above proof assured that the infinite intersection is nonempty.

No: take $I_n=[n,to)={xinBbb R:xge n}$. If, however, some $I_m$ is compact, then $I_n$ is compact for all $nge m$, and the intersection will be non-empty.

I'm studying intro to analysis and I got that a nested sequence of closed and nonempty subsets of $mathbb{R}$ do not necessarily have a nonempty infinite intersection:

$textbf{Claim 1}$: The set $A_{n} = (-infty, -n]$ is closed, $nin mathbb{N}$.

$textit{proof}$: Let $x$ be a limit point for $A_{n}Rightarrow$ if we had $x>-n,$ choose $epsilon = x+nRightarrow x-epsilon=-n<displaystyle x-frac{epsilon}{2}Rightarrow
V_{epsilon/2}(x)cap A_{n}=emptyset Rightarrow $ contradiction $Rightarrow xin A_{n}$.

$textbf{Claim 2}$: $A_{n+1}subseteq A_{n}forall n$.

$textit{Proof}$: $xin A_{n+1}Rightarrow xleq-n-1<-nRightarrow xin A_{n}$.

$textbf{Claim 3}$: The sequence $A_{n}supset A_{n+1}, A_{n}=(-infty,-n] forall n$ has
empty infinite intersection.

$textit{Proof}: bigcap_{n=1}^{infty}A_{n}neq emptyset Rightarrow$ choose $xin
bigcap_{n=1}^{infty}A_{n}Rightarrow xleq -nhspace{.5pc}forall nRightarrow -xgeq nhspace{.5pc}forall nRightarrow$ contradiction ($mathbb{N}$ is not bounded above)

$textbf{Conclusion}$: By claims 1,2, and 3, we have a nested sequence of closed sets with empty infinite intersection.

$underline{Legend}$:

1. $V_{epsilon}(x)=(x-epsilon,x+epsilon)$


2. Infinite intersection of $A_{n}=bigcap_{n=1}^{infty}A_{n}$