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Compact subset of open set?

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Let [itex]K \subset \mathbb{R^n}[/itex] be compact and [itex]U[/itex] an open subset containing [itex]K[/itex]. Verify that there exists [itex]r > 0[/itex] such that [itex]B_r{u} \subset U[/itex] for all [itex]u \in K [/itex].


    2. Relevant equations

    Every open cover of compact set has finite subcover.


    3. The attempt at a solution

    I tried to cover my [itex]K[/itex] with open balls therefore there should be finitely many open balls (because [itex]K[/itex] is compact). If I choose [itex]r' = min(r_1,r_2,...,r_n)[/itex] ([itex]r_i[/itex] being the radius of the ball) then every element in [itex]K[/itex] has that required ball-neighbourhood. Because [itex]U[/itex] is open then [itex]B_{r'}{u} \subset U[/itex]. Is this correct?
     
  2. jcsd
  3. Feb 6, 2014 #2

    PeroK

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    That doesn't prove it. Even if you have a finite subcover of open balls, then that subcover might extend beyond U. You need to find a way to make your subcover be subsets of U.

    ... also, you need the r to work for every u in K, not just a finite subset.

    ... maybe using open covers is not the answer? You are working in $$\mathbb{R}^n$$
     
    Last edited: Feb 6, 2014
  4. Feb 6, 2014 #3
    Erh...


    Ok, so my reasoning does not work. How about the fact that since [itex]U[/itex] is open, then every element [itex]x \in U[/itex] has an open ball-neigbourhood [itex]B_r{x} \subset U[/itex]. Then every element in [itex]K[/itex] has that neigbourhood also. Take all these neighbourhoods and then this is an open cover of [itex]K[/itex]. Because [itex]K[/itex] is compact then it has to have finite subcover which has finitely many of these open balls. Then this subcover has to be subset of [itex]U[/itex] and if I take minimum of the [itex]r[/itex]:s, then every element [itex]u \in K[/itex] has that required ball-neighbourhood.

    I do not know why [itex]\mathbb{R^n}[/itex] is relevant here.. Am I supposed to use some metric?
     
  5. Feb 6, 2014 #4

    PeroK

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    Take a ball, radius r, centred on k in U. Let x be in this ball. Then the ball, radius r, centred on x may not be in U.

    In general, any finite subcover will give you a finite set of balls, but that isn't easily going to show that a ball of radius r centred on any point in K is in U.

    Instead, I'd think about K being closed and bounded.
     
  6. Feb 6, 2014 #5
    I feel I'm completely lost now.. Of cource since [itex]K[/itex] is compact then it is closed and bounded. Because it is bounded then for every [itex]u \in K[/itex] [itex]\|u\| \leq M[/itex].

    I really need some hint now..
     
  7. Feb 6, 2014 #6
  8. Feb 6, 2014 #7

    PeroK

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    Here's an outline of a basic proof:

    Assume not, then you get a sequence in K and a sequence not in U that get arbitrarily close.

    As K is compact, the sequence in K has a subsequence which converges to k in K.

    And, it's easy to show that the corresponding subsequence of points not in U also converges to k.

    But, as U is open, this is a contradiction (as k has a neighbourhood in U).

    Can you do your own formal proof based on that?
     
  9. Feb 6, 2014 #8
    Well, only difficulty I see in the proof is that how to show the corresponding subsequence of points not in U also converges to k. Of cource if points get arbitarily close, so the distance between points is less than r > 0.. ?
     
  10. Feb 6, 2014 #9
    I think I got it now.

    So let's assume that there exist no such r. Therefore there is sequence [itex](x_n) \subset \mathbb{R^n}\setminus U[/itex] such that [itex]\|{x_n - k_{n_j}}\| < \frac{\epsilon}{2}[/itex]. Now [itex]k_{n_j}[/itex] is a converging subsequence in [itex]K[/itex].

    Because [itex]K[/itex] is compact there exists subsequence [itex]k_{n_j}[/itex] which converges to some [itex]k[/itex], so [itex]\|{k_{n_j} - k}\| < \frac{\epsilon}{2} [/itex]

    Now consider [itex]\|{x_n - k}\| =\|{x_n - k_{n_j} + k_{n_j} - k}\| \leq \|{x_n - k_{n_j}}\| + \|{k_{n_j} - k}\| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon [/itex]

    So [itex](x_n)[/itex] converges too to [itex]k[/itex]. This is impossible, since [itex]U[/itex] is open and therefore every element in [itex]U[/itex] has open ball-neigbourhood that is a subset of [itex]U[/itex].

    Therefore there is such r.

    Q.E.D ?
     
    Last edited: Feb 6, 2014
  11. Feb 6, 2014 #10

    Dick

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    I don't see how those inequalities with a fixed ##\epsilon## are telling you much. Why don't you start by assuming something like ##\|{x_{n} - k_{n}}\| < \frac{1}{n}##?
     
  12. Feb 7, 2014 #11
    You could also try out the Lebesgue's number lemma . But I think you will have to prove that lemma for sake of completion. Of course , proof is available online and using the Lemma almost solves the question. http://en.wikipedia.org/wiki/Lebesgue's_number_lemma
     
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