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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?

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  • $\begingroup$ Closed sets where? In what topological space? $\endgroup$ – Asaf Karagila Sep 10 '13 at 7:55
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    $\begingroup$ I added 'non-empty' to the question, because otherwise there is the trivial counter-example where $I_n =\emptyset$ for large enough $n$. $\endgroup$ – Rhys Sep 10 '13 at 7:56
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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.

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

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  • $\begingroup$ How do you use compactness to prove that? $\endgroup$ – user94055 Sep 10 '13 at 7:54
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    $\begingroup$ @user94055: For $n\ge m$ let $U_n=\Bbb R\setminus I_n$. Show that if $\bigcap_{n\ge m}I_n=\varnothing$, then $\{U_n:n\ge m\}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful. $\endgroup$ – Brian M. Scott Sep 10 '13 at 7:57
  • $\begingroup$ @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks. $\endgroup$ – M.Sina Sep 10 '13 at 8:08
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    $\begingroup$ @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow. $\endgroup$ – Brian M. Scott Sep 10 '13 at 8:20
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    $\begingroup$ @BrianM.Scott: thanks. good night :). $\endgroup$ – M.Sina Sep 10 '13 at 8:27
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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, $n\in \mathbb{N}$.

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

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

$\textit{Proof}$: $x\in A_{n+1}\Rightarrow x\leq-n-1<-n\Rightarrow x\in A_{n}$.

$\textbf{Claim 3}$: 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 $x\in \bigcap_{n=1}^{\infty}A_{n}\Rightarrow x\leq -n\hspace{.5pc}\forall n\Rightarrow -x\geq n\hspace{.5pc}\forall n\Rightarrow$ 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 $A_{n}=\bigcap_{n=1}^{\infty}A_{n}$
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In $\mathbb Q^c$, take the sequence $\{(-1/n,1/n)\}_{n=1}^\infty$, where $(-1/n,1/n)=\{x \in \mathbb Q^c:-1/n<x<1/n \}$

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