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Open coverings and Heine-Borel theorem

  1. Sep 17, 2009 #1
    Okay my book says that a collection H of open sets is an open covering of a set S if every point in S is contained in a set F belonging to H.

    Then it says that the set S = [0,1] is covered by the family of open intervals

    F1 = {(x-(1/N), x +(1/N))|0< x <1}
    N = positive integer

    My first question is that.. Does this mean that if I pick an N and make it constant I can find an opencovering by picking arbitrary x ?

    In this case how to I use the set F1 to construct the open cover of S ?

    How exactly does this open covering work in this case ? According to Heine-Borel a compact set has finitely many open sets that make up it's open covering.



    My second question is that if [0,1] has finitely many open sets that make up it's open covering then so should (0,1), right ? Since (0,1) is a smaller set than [0,1]. But (0,1) is not closed, however it has a supremum and infimum that are not in the open interval.


    Basically I want to understand how to use F1 to cover [0,1]; like what do I chose, any arbitrary N or a bunch of arbitrary x.


    I'm studying this stuff on my own because the Analysis course I want to take isn't going to be offered at my school this year, but I can't wait. So please have some patience .

    ~Regards,
    Elmer
     
  2. jcsd
  3. Sep 18, 2009 #2

    LCKurtz

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    You said:

    My first question is that.. Does this mean that if I pick an N and make it constant I can find an opencovering by picking arbitrary x ?

    I think the statement is a bit ambiguous. My interpretation is that N can be any positive integer, but it is chosen and "fixed". But you don't "find an open covering by picking x". What is given is and open covering. For example, if N = 10 every x in the interval is covered by its own private open interval of length 2/10. There are zillions [infinitely many] of them, one for each x, and they greatly overlap each other and cover the interval many times over.

    Then you asked:

    How exactly does this open covering work in this case? According to Heine-Borel a compact set has finitely many open sets that make up it's open covering.

    Well, in our example where the given intervals are 1/5 long, it shouldn't be hard to find just a few of them that do the job. And the same idea works for any fixed N.

    Regarding your question about (0,1), no. Not every open cover of it can be reduced to finitely many. Consider the collection on intervals (1/n, 1 - 1/n) for n from 3 to infinity. Can you see the union of those covers (0,1) but finitely many is never enough?
     
  4. Sep 18, 2009 #3

    LCKurtz

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    I just noticed this, which I should have commented on. You said:

    "According to Heine-Borel a compact set has finitely many open sets that make up it's open covering."

    No, that is not what Heine-Borel theorem states. It says that if you are given any open covering (which may have infinitely many open sets in it), you can always find a finite subset of the given covering that still covers the set. In other words every open covering of a compact set has a finite sub-covering.
     
  5. Sep 18, 2009 #4
    Thanks a lot! You addressed all my misconceptions in a clear and concise manner.

    You hit the right spot. Thanks again.
     
  6. Sep 19, 2009 #5

    HallsofIvy

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    And, strictly speaking, that is not what the "Heine-Borel theorem" says! That is the definition of "compact set". The "Heine-Borel theorem" says that every closed and bounded subset of the real numbers is compact.

    And, getting back to your first post,
    That set consists of all different open intervals for all real numbers x between 0 and 1 and all positive integers N- a lot of sets! And it is, strictly speaking, "overkill" since taking x= 1/2, N= 1 is sufficient- every real number in [0, 1] is contained in either (1/2- 1, 1/2+ 1)= (-1/2, 3/2).

    I believe that answers your second question, "My second question is that if [0,1] has finitely many open sets that make up it's open covering then so should (0,1), right ? Since (0,1) is a smaller set than [0,1]. But (0,1) is not closed, however it has a supremum and infimum that are not in the open interval", in the negative since that "open cover" consists only of a single set.

    Remember that just finding a finite open cover for a set does NOT prove it is compact, nor does finding a finite subcover for a given open cover prove that it is finite. The original, infinite, F1, and my finite subcover could apply to (0, 1) which is NOT compact (every compact set is closed). To show that a set is compact, you must show that every possible open cover has a finite subcover.

    Consider the open cover of (0, 1) defined by {(0, 1/2), (0, 3/4), (0, 7/8), (0, 15/16), ..., (0, [itex]2^{n-1}/2^n[/itex]}. That covers (0,1) but no finite subcover does.
     
    Last edited: Sep 19, 2009
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