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

Any closed interval [a,b] is compact ?

  1. Aug 8, 2010 #1
    Hi All,

    So all closed interval [a,b] is compact
    (see Theorem 2.2.1 in Real Analysis and Probability by RM Dudley)

    Now, Let's say I have [0,10] as my closed interval.
    Let My Open Cover be
    (0, 5)
    (5, 7.5)
    (7.5, 8.75)
    (8.75, 9.375)

    ....

    Essentially, The length of each open interval is cut by half, i.e.
    length (0,5) = 5
    length (5,7.5) = 2.5 = 5/2
    length (7.5,8.75) = 1.25 = 5/(2*2)
    length (8.75, 9.375) = 0.625 = 5/(2*2*2)

    ....
    the Union of all these interval gives us [0.10] \ ({0} U {5} U {7.5} U {8.75} U {9.375} ... )

    Therefore, we add
    (0-[tex]\epsilon[/tex], 0+[tex]\epsilon[/tex])
    (5-[tex]\epsilon[/tex], 5+[tex]\epsilon[/tex])
    (7.5-[tex]\epsilon[/tex], 7.5+[tex]\epsilon[/tex])
    (8.75-[tex]\epsilon[/tex], 8.75+[tex]\epsilon[/tex])

    .... to cover all these missing single points.

    The choice of [tex]\epsilon[/tex] is arbitrary, as long as it doesn't touch the middle-point of the next interval.

    for example,
    to cover {5}, we can add a (5-0.1, 5+0.1)
    to cover {7.5} we can add a (7.5-0.01, 7.5+0.01)
    etc etc


    So we have create ourselves an Open cover for [0, 10]
    but i can't see any finite members of such cover become yet another cover for [0, 10]

    I know [0, 10] is compact, I simply can't find a finite subcover.
    Where is the loophole ?

    Thanks
    J
     
  2. jcsd
  3. Aug 8, 2010 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Don't you get a cover by taking all of your intervals up to and including the first one that contains 10?
     
  4. Aug 8, 2010 #3
    Yes but that won't be finite. It takes infinitely many of such open interval to cover [0,10] because I cut the length of each interval by half, i.e.

    length of [0,10] = 10
    10 = 5 + 5/2 + 5/4 + 5/8 + 5/16 + .....

    To show [0,10] is compact we need to find a finite subcover under such construction.
     
  5. Aug 8, 2010 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It only takes one interval to contain 10. Which one does it?

    If none of them, then you don't have a cover of [0,10].
     
  6. Aug 8, 2010 #5

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Okay. Then just take this, and all of your intervals up to, and including the first one that contains 10-0.0000000001.
     
  7. Aug 8, 2010 #6
    Let's say, for (0, 5)

    we take (0 - 0.1, 0 + 0.1) to cover the point 0
    we take (5 - 0.01, 5 + 0.01) to cover the point 5
    we take (7.5 - 0.001, 7.5 + 0.001) to cover the point 7.5

    etc etc

    how can we get a FINITE subcover ?
     
  8. Aug 8, 2010 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You can't find a finite subcover cover of [0, 10 - 1010]?
     
  9. Aug 8, 2010 #8
    The point is, each mid-point of the open interval is not covered by any other interval and we have infinitely many of such mid-point.

    For example,
    only (0,5) cover {2.5}
    only (5, 7.5) cover {6.25}
    only (7.5, 8.75) cover {8.125}
    only (8.75, 9.375) cover {9.0625}
    etc etc

    we have infinitely many of such mid-point which therefore takes infinitely many open interval to cover them all. I simply can't find a FINITE subcover for [0,10]
     
  10. Aug 8, 2010 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You said that you had the interval (10 - 1010, 10 + 1010) as one that contains 10.

    So you just need to find a finite cover of [0, 10 - 1010] to go along with it, and you have a cover of the entire interval [0,10].
     
  11. Aug 8, 2010 #10
    Not if I construct my Open interval the way I start my post.
    Compactness require EVERY open cover U possesses a FINITE subcover V which is the subset of the U

    With U being the way we construct it, i simply can't find a finite V
     
  12. Aug 8, 2010 #11
    Sorry if my writing is a little bit rough.
    The (10-[tex]\epsilon[/tex], 10 + [tex]\epsilon[/tex]) covers {10} and does not cover mid-point of any other interval. the way we choose [tex]\epsilon[/tex] is arbitrary as long as it doesn't touch the mid-point of other intervals
     
  13. Aug 8, 2010 #12

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What's wrong with

    (0.00000000000, 0.50000000000)
    (0.50000000000, 0.75000000000)
    (0.75000000000, 0.87500000000)
    (0.87500000000, 0.93750000000)
    (0.93750000000, 0.96875000000)
    (0.96875000000, 0.98437500000)
    (0.98437500000, 0.99218750000)
    (0.99218750000, 0.99609375000)
    (0.99609375000, 0.99804687500)
    (0.99804687500, 0.99902343750)
    (0.99902343750, 0.99951171875)
    (0.99951171875, 0.99975585938)
    (0.99975585938, 0.99987792969)
    (0.99987792969, 0.99993896484)
    (0.99993896484, 0.99996948242)
    (0.99996948242, 0.99998474121)
    (0.99998474121, 0.99999237061)
    (0.99999237061, 0.99999618530)
    (0.99999618530, 0.99999809265)
    (0.99999809265, 0.99999904633)
    (0.99999904633, 0.99999952316)
    (0.99999952316, 0.99999976158)
    (0.99999976158, 0.99999988079)
    (0.99999988079, 0.99999994040)
    (0.99999994040, 0.99999997020)
    (0.99999997020, 0.99999998510)
    (0.99999998510, 0.99999999255)
    (0.99999999255, 0.99999999627)
    (0.99999999627, 0.99999999814)
    (0.99999999814, 0.99999999907)
    (0.99999999907, 0.99999999953)
    (0.99999999953, 0.99999999977)
    (0.99999999977, 0.99999999988)
    (0.99999999988, 0.99999999994)
    (0.9999999999, 1.0000000001)

    Along with all of the 30 or so extra intervals to cover the point gaps?
     
  14. Aug 8, 2010 #13

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No matter what positive value epsilon has, it covers the mid-point of some other interval.
     
  15. Aug 8, 2010 #14
    I agree, and there goes the loophole

    Thanks !
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Any closed interval [a,b] is compact ?
  1. Compact manifold (Replies: 9)

  2. Compact spaces (Replies: 2)

Loading...