Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Question about the set (a,a]

  1. Sep 22, 2004 #1
    question about the set [a,a)

    If S=[a,a) Can such a set exist? It implies that a is in S and not in S, which doesn't make sense, but it seems a problem I'm trying to do requires it to be considered empty.

    The question is:

    Let In = [an,bn) where

    In+1 < In for all natural numbers n. [< denotes subset]

    Give an example of those In for which the intersection of In (for all n) is empty.

    I can't see any other way to construct an empty set.
    Last edited: Sep 22, 2004
  2. jcsd
  3. Sep 22, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    (a, a] would be the set of all x such that a < x <= a.

    However, I think the question intends you to limit yourself to intervals whose endpoints are distinct.

    Recall that when you take a nested intersection of nonempty closed sets, you never get the empty set. Since you're searching for behavior that is not demonstrated by closed sets, I would suggest focusing at the open end of your intervals, where they resemble closed sets the least.
  4. Sep 22, 2004 #3
    Hi. Thanks for the response. Based on your definition [a,a) is indeed empty. Although I'm not sure why you think the question requires the endpoints to be distinct.

    If every set In+1 is a subset of In, then the only way the intersection of In for all n is empty is if one of those sets is empty. To my eyes (which admittedly are stupid at times), the only set of the form [a,b) that is empty is when a=b (or b<a)
    Last edited: Sep 22, 2004
  5. Sep 22, 2004 #4

    I just learned that the set [a,a) is not a proper subset of itself. So if In=[a,a), In+1 can't satisfy In+1 < In.

    Now I have no idea what's going on. If no In can be empty, how can the intersection of all In be empty in any case! HELP
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook