1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Real analysis

  1. Feb 16, 2006 #1
    I'm trying to show that this set is compact {0} union {2^-m + 2^-n} where n,m are integers and m is less than n

    I'm thinking that since it's bounded below by 0 and the set contains it's bound that if I can show that it's closed and bounded above that will be a proof but i'm not really sure how i would do that
    or
    Say there's an collection of open sets who's union is (-2, 2^-m + 2^-n + 2) that since that's an open cover then and you can have a finite subcover of something like (-1, 2^-m + 2^-n +1) that'll still contain the set therefore the set is compact.

    I don't know if either of those are correct or if i'm heading in the right direction. Can anyone help?
     
  2. jcsd
  3. Feb 17, 2006 #2
    Zero is in the set, so any cover of the set must cover zero. Can you show that all but a finite number of points in the set are covered by this cover on {0}?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Real analysis
  1. Real analysis (Replies: 3)

  2. Real analysis (Replies: 0)

  3. Real analysis (Replies: 16)

  4. Real Analysis (Replies: 1)

  5. Real Analysis (Replies: 5)

Loading...