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Math Help - Prove Compactness - Directly from definition

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    [SOLVED] Prove Compactness - Directly from definition

    Let K := \{1/n : n \in \mathbb{Z}^+\} \cup \{0\} .Prove that K is compact directly from the definition(without using Heine-Borel theorem).

    The definition of compactness I'm using is: A set K \subseteq \mathbb{R}^n is said to be compact if every open cover of K has a finite subcover.

    But the problem I'm having is, how do I show that universal statement is true. I can think of open covers which have finite subcovers, but how can I show that it will be the case for any and all open covers without referencing the Heine-Borel theorem in any way.
    Last edited by chrischen88; October 13th 2009 at 07:27 PM.
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    Let (U_a)_{a \in A} be an open cover for K. There exists U_{a_0} such that 0 \in U_{a_0}. Since the sequence 1/n converges to 0 U_{a_0} contains all but finitely many elements of K, and the rest can obviously be covered by a finite subcover.
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    Junior Member platinumpimp68plus1's Avatar
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    You can show that the subcovers are finite by finding an arbitrary one.


    You can argue:
    At the point 0 for some large N, we have that 1/N<L, L some element of K. ie. 0 is a limit point of K. Pick a subcover that covers 0 and 1/N. Note that this covers infinitely many points between 0 and 1/N. That means that you only have finitely many points remaining: ie. 1, 1/2, 1/3, 1/4, 1/5,....1/(N-1). Pick one subcover to cover each of these points. Now you have a finite set of subcovers.
    Last edited by platinumpimp68plus1; October 13th 2009 at 07:22 PM.
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    Quote Originally Posted by Jose27 View Post
    Let (U_a)_{a \in A} be an open cover for K. There exists U_{a_0} such that 0 \in U_{a_0}. Since the sequence 1/n converges to 0 U_{a_0} contains all but finitely many elements of K, and the rest can obviously be covered by a finite subcover.
    I'm not sure I understand this step: 1/n converges to 0 so U_{a_0} contains all but finitely many elements of K.

    0 \in U_{a_0} and U_{a_0} contains all the values that are infinitely close to 0? So are we trying to say that all the values that are infinitely close to 0 have a finite subcover?
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    Quote Originally Posted by chrischen88 View Post
    I'm not sure I understand this step: 1/n converges to 0 so U_{a_0} contains all but finitely many elements of K.

    0 \in U_{a_0} and U_{a_0} contains all the values that are infinitely close to 0? So are we trying to say that all the values that are infinitely close to 0 have a finite subcover?
    Since U_{a_0} is open there is an r>0 such that (-r,r) \subset U_{a_0}. Take N such that 1/N<r then for all n>N we have 0<1/n<1/N<r and so for all n \geq N we have 1/n \in U_{a_0}. Now just pick N-1 open sets from the original cover that enclose 1, 1/2,...,1/(N-1).
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    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by chrischen88 View Post
    I'm not sure I understand this step: 1/n converges to 0 so U_{a_0} contains all but finitely many elements of K.

    0 \in U_{a_0} and U_{a_0} contains all the values that are infinitely close to 0? So are we trying to say that all the values that are infinitely close to 0 have a finite subcover?
    Because the sequence x_n=\frac{1}{n} converges to 0, then \forall~\epsilon>0, \exists~N such that n>N implies \frac{1}{n}<\epsilon.

    Thus, the open set U_{a_0} that contains 0 has some radius \epsilon, and therefore all elements with n>N will be contained in this set. But the set of all natural numbers larger than some fixed constant N is clearly infinite.

    The remaining points that need to be covered are all the natural numbers less than N, which is clearly a finite set and will be covered by at most N open sets.
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    Quote Originally Posted by platinumpimp68plus1 View Post
    You can show that the subcovers are finite by finding an arbitrary one.


    You can argue:
    At the point 0 for some large N, we have that 1/N<L, L some element of K. ie. 0 is a limit point of K. Pick a subcover that covers 0 and 1/N. Note that this covers infinitely many points between 0 and 1/N. That means that you only have finitely many points remaining: ie. 1, 1/2, 1/3, 1/4, 1/5,....1/(N+1). Pick one subcover to cover each of these points. Now you have a finite set of subcovers.
    For this part " ie. 1, 1/2, 1/3, 1/4, 1/5,....1/(N+1)", did you mean " ie. 1, 1/2, 1/3, 1/4, 1/5,....1/(N-1)"?

    Then what you said makes sense to me. Ie. we can pick a subcover to cover the interval from 0 to 1/N, which contains infinitely many points, and then we only have finitely many " ie. 1, 1/2, 1/3, 1/4, 1/5,....1/(N-1)" left over to cover.

    So to keep this subcover arbitrary in my argument I should never mention any specific subcover, I mean I should never say choose the collection \{(-1,1), (0,1)\} as a subcover for [0,1/N] etc. Correct? I should just say "a subcover can be chosen to cover the points in this interval: [0,1/N] " etc.
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  8. #8
    Super Member redsoxfan325's Avatar
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    Right. You can't specify a specific open cover, because then you're only proving that it works for that open cover. You should choose an arbitrary open cover and then use general properties of that open cover to prove there is a finite subcover.
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