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Having trouble understanding Open Covers

  1. Apr 7, 2008 #1
    Hey all, i'm hoping someone can help me understand this example in my book. I'm pretty bad at analysis, so explaining things as elementary as possible would be nice.

    Example: Consider the open interval (0,1). For each point x in (0,1) let O_x be the open interval (x/2, 1). Taken together, the infinite collection {O_x : x in (0,1)} forms an open cover for the open interval (0,1). Notice, however, that it is possible to find a finite subcover. Given any proposed finite subcollection
    {O_x_1, O_x_2, ..., O_x_n)
    set x' = min{x_1, x_2, ..., x_n) and observe that any real number y satisfying 0 < y <= x'/2 (that symbol is less than or equal to) is not contained in the union from i=1 to n O_x_i.

    I understand that the infinite collection of open intervals forms an open cover for the open interval (0,1), but i don't understand why you can't come up with a finite amount of open intervals, like for example, (-1,2/3) and (1/3,2) to cover the open interval (0,1). Is it because -1 and 2 aren't contained in (0,1)? The definition of open cover is a little vague in my book.

    Thanks for any help you guys can provide.
  2. jcsd
  3. Apr 7, 2008 #2
    it says it's impossible to find a finite SUBcover. they're saying there is no finite subset of their proposed cover that covers (0,1)
  4. Apr 7, 2008 #3
    ohhh, okay, i get it. so a SUBcover of (a,b) implies it's a collection of SUBsets of (a,b)?
  5. Apr 7, 2008 #4
    no, not quite. given a cover C, a subcover is a subset of that cover that still covers the set. they give an example of a cover ({O_x : x in (0,1)}). they're saying there is no FINITE subset of this cover that still covers (0,1). you can't have a subcover without a cover.
  6. Apr 7, 2008 #5
    does my example of (-1, 2/3) and (1/3, 2) count as an open subcover of the cover (-1, 2) to cover the interval (0,1), then?
  7. Apr 7, 2008 #6


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    If your cover is just
    { (-1, 2) }
    (is this even a cover, wouldn't it have to be { (-1, 2) intersected with (0, 1) } ?
    then no, because (-1, 2/3) and (1/3, 2) are both not in the cover.
    If you'd consider { (-1, 2), (-1, 2/3), (1/3, 2) } a valid cover (despite the fact that the intervals lie outside (0, 1)) then the answer would be "yes", though.
  8. Apr 7, 2008 #7
    i don't know whether this is a cover or not, and that's one of the reasons i posted this thread, because i'm having difficulty understanding what a cover and subcover actually are.
  9. Apr 7, 2008 #8


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    Yes, {(1,2)} is a "cover" of (0, 1) because every point of (0, 1) is in some set in {(-1,2)}. In fact, since there is only one set in that cover, it is true because (0, 1) is a subset of (-1, 2).
    {(1/n, 1)} is a also a cover of (0,1) because, for any x in (0,1), that is, any x such that 0< x< 1, we can find n such that 1/n< x. (Choose integer n> x. Then x< 1/n.) There is no finite subcover because then there would have to be a largest N. choose x< 1/N for that largest N. It is not in any of the intervals.

    No, a SUBcover of {Xi}, where {Xi} is a cover of (a,b) is a SUBset of {Xi} which also covers (a,b).
  10. Apr 7, 2008 #9
    thanks for all the replies, guys. i think i got it now :)
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