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I Question about compactness

  1. Jul 13, 2016 #1
    The question is stated in the attatched file.
     

    Attached Files:

  2. jcsd
  3. Jul 13, 2016 #2

    fresh_42

    Staff: Mentor

    How are the ##H_i## defined?
     
  4. Jul 13, 2016 #3
    It is just an open set because it is just a part of open covering.
     
  5. Jul 13, 2016 #4

    fresh_42

    Staff: Mentor

    Ok. Now, if you take ##ε = 2(β-α)##, how many ##ε##-intervals around a fixed ##x_0## would you need for coverage? How many if ##ε = \frac{β-α}{2}##, how many for ##ε = \frac{β-α}{3}## or ##ε = \frac{β-α}{n}## with a fixed ##n##? The limit is essentially the union of all points of ##[α,β]## but for a given open coverage, finitely many will be enough.
     
  6. Jul 13, 2016 #5
    Thanks for your reply.

    I wanted to confirm the theorem, "if a set is compact then its open covering will contain a subcovering that consists of finitely many open sets." I believe H is an open covering, since it contains S and consists of open sets. However, because epsilon is converging to 0, it would require infinitely many sets to fully cover the set S, which contradicts to the theorem.

    Each open set overlaps as much as epsilon, but it is shrinking to 0. Then would we need finitely many open sets Hi to cover the closed set? Somehow I feel there has to be infinitely many.
     
    Last edited: Jul 13, 2016
  7. Jul 13, 2016 #6

    fresh_42

    Staff: Mentor

    The point is: For any given open coverage of a compact set (in a metric space - although I think Hausdorff is enough) there is a finite sub-coverage. Again: a given coverage!

    If you take ##ε=0## then you simply have ##S= \{x | x \in S\} = \bigcup_{x \in S} \{x\}##. This is no open coverage anymore! But as soon as you chose a ##ε > 0##, how small it ever maybe, then finitely many will be sufficient. However, the number will depend on your choice of ##\epsilon##.
     
  8. Jul 13, 2016 #7

    Hmm, can I understand this way.

    no matter how small ε is, we can find ε'< ε. which means ε is bounded below by some ε'. Then we can find a sufficiently small integer n which satisfies 1/n < ε.

    Also 1/n'<ε'.

    since 1/n is bounded by 1/n' and countable therefore finite number of open sets will be enough to cover the set S?
     
  9. Jul 13, 2016 #8

    fresh_42

    Staff: Mentor

    I don't see why you need a ##ε'##, but yes, this is correct. Perhaps you shouldn't say countable because this usually refers to an infinite number of things like the amount of natural numbers, since a finite number of things are always countably many.
     
  10. Jul 13, 2016 #9
    Thanks. My approach and terminology might be coarse because I am an engineering student. Thank you! It helepd a lot
     
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