Find an example of a family of intervals A={A sub n } such that n belongs to N s
ID: 2943104 • Letter: F
Question
Find an example of a family of intervals A={A sub n } such that n belongs to N such that for all n in the natural numbers, A sub n+1 is a subset of A sub n and the intersection of A sub n is the empty set, such that The sets A sub n are all closed and the sets A sub n are all bounded. I already know the answers but I need a explanation of why. For the sets A sub n are all closed my answer is A ={A sub n} such that n belongs to N of sets A sub n = [n,infinity) where n belongs to N. For the sets A sub n are all bounded my answer is A = {A sub n} such that n belongs to N of sets A sub n = (0,(1/n)) where n belongs to N.Explanation / Answer
From your desription it looks like a question with 2 parts: Find a sequence {An} s.t. An+1 is a subset of An for each n and
1. the sets are all closed. In this case you may take An = [n,). Note that these sets are all closed. Also clearly [n,) contains [n+1,). Also if say x lies in all these intervals, i.e. in the intersection of these sets then suppose x lies in the interval [n,n+1) for some integer n. Note that x does not lie in An+1 then and that contradicts that x is in all the Ai.
2. The sets are all bounded. In this case, you may take An = (0,1/n). Again, the only thing that needs to be proved is that these sets have empty intersection. Again if x is in the intersection, then let n be the unique integer such that 1/(n+1) < x <=1/n. Then again, as before x is not in the set An+1 and that contradicts the assumption.
An intersting side point here is this: If you have both closed and bounded then any such sequence cannot have nonempty intersection. THis is so since any such set is necessarily compact and compact sets have the finite intersection property - i.e., if finite subcollections have non empty intersection then the entire collection also has nonempty intersection.
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