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Liminf subset of limsup

  1. Sep 18, 2011 #1
    I have been told that given a sequence of subsets, liminf [itex] A_n\subseteq [/itex] limsup [itex] A_n [/itex]. I have spent some time trying to cook up such a sequence, but I can only ever show they are equal. Can someone point me in the right direction?
  2. jcsd
  3. Sep 18, 2011 #2


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    If there is no restriction on An, then you could set up a sequence:
    An = empty set for even n and An = whole space for odd n.
  4. Sep 18, 2011 #3
    I don't know if that one works. Note that

    liminf [itex] A_n = A_1 \cup [A_1 \cap A_2] \cup [A_1 \cap A_2 \cap A_3] \dots = X \cup [X \cap \emptyset] \cup [X \cap \emptyset \cap X] \dots = X \cup \emptyset \cup \emptyset \dots = X, [/itex]


    limsup[itex] A_n = A_1 \cap [A_1 \cup A_2] \cap [A_1 \cup A_2 \cup A_3] \dots = X\cap [X \cup \emptyset] \cap [X \cup \emptyset \cup X] \dots = X \cap X \cap X \dots = X. [/itex]

    Hence they are equal.
  5. Sep 19, 2011 #4


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    I am a little confused by your definitions. My understanding is that limsup consists of all points which are in an infinite number of sets of the sequence, which would be the whole space. Liminf according to my understanding are those points which belong to all but a finite number of sets in the sequence, which would be none.

    To use your expression I think you need to start with An (not A1) and then let n become infinite.
    Last edited: Sep 19, 2011
  6. Sep 20, 2011 #5
    Yeah, I had the definition wrong. UGH! Thanks!
  7. Sep 22, 2011 #6
    How can I prove that liminf of a sequence of events An is a subset of lim sup of An given n goes to infinity? I'm inexperienced in constructing proofs, so please do not be strict.
  8. Sep 23, 2011 #7


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    One way of getting there is to look at
    supn = An∪An+1∪...
    infn = An∩An+1∩...
    supn contains infn

    So let n become infinite and the relationship still holds.
  9. Sep 24, 2011 #8
    Thank you, it's clear now.
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