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Give examples of set following the given condition | if not, why not?
(a) A sequence of nested unbounded closed intervals ${L_1} \supseteq {L_2} \supseteq {L_3} \supseteq {L_4} \supseteq ...$ with $ \cap _{n = 1}^\infty {L_n} = \phi $.
(b) A sequence of closed (not necessarily nested) intervals ${I_1},{I_2},{I_3},...$ with the property that $ \cap _{n = 1}^N{I_n} \ne \phi $ for all $N \in N$, but $ \cap _{n = 1}^\infty {I_n} = \phi $.
For (a) if I chose ${L_n} = {1 \over {n - 1}}$ then the set would be $[0,\infty)$ so I have unbounded closed interval type set but certainly I won't get $ \cap _{n = 1}^\infty {L_n} = \phi $. Can there be any set following the given condition be obtained? if not, why ?
For (b) No clue at all.

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    $\begingroup$ For (a), consider $L_n:=[n,\infty)$. Actually consider that for (b) as well. $\endgroup$ – NeedForHelp Jan 24 '17 at 9:04
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    $\begingroup$ Doesn't $L_n=[n,\infty)$ work for both (a) and (b)? $\endgroup$ – MPW Jan 24 '17 at 9:05
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For a) and b), consider the colection of intervals $[n,\infty)$ with $n\in\mathbb{N}$. Clearly, there are closed and if $m>n$, then $[n,\infty)\supset [m,\infty)$. Thus, there are nested. The interseccion of all of them is the empty set and the intersection of $N$ of them is not-empty.

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