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Math Help - Proving compactness without using Heine-Borel Thm

  1. #1
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    Proving compactness without using Heine-Borel Thm

    Q: Let (a_n) be a sequence of reals so that a = lim(a_n) exists as n→∞. Let S = {a_1, a_2, a_3, · · · } ∪ {a}. Prove that any open cover of S has a finite subcover.

    I am trying to prove this without using the Heine-Borel Theorem (Heine?Borel theorem - Wikipedia, the free encyclopedia), which is why it is giving me a little trouble. I'm guessing the goal is basically to arrive at a conclusion showing that S is compact, without using the fact that S is bounded and closed (because that's the Heine-Borel Thm).

    My intuition would be to show that S is sequentially compact, so that any sequence in S has a subsequence converging to a limit point in S? I'm not sure where to go from here though... Any hint/help would be appreciated.
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  2. #2
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    Use the definition of compactness. Take (U_i)_i an open covering of S. Choose i_0 such that a\in U_{i_0}. By the definition of convergence of a sequence, outside U_{i_0} there are only a finite number of terms of the sequence. Now the conclusion is almost immediate.
    Last edited by Enrique2; November 11th 2009 at 03:43 PM.
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  3. #3
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    Quote Originally Posted by Enrique2 View Post
    Use the definition of compactness. Take (U_i)_i an open covering of S. Choose i_0 such that a\in U_{i_0}. By the definition of convergence of a sequence, outside U_{i_0} there are only a finite number of terms of the sequence. Now the conclusion is almost immediate.
    Just to clarify, so within U_{i_0} there would be an infinite # of a_{n} terms converging to a. If so, then S satisfies the definition of compactness that I had on the first post: "any sequence in S has a subsequence converging to a limit point in S" because a_{n} converges to a and any subsequences of a_{n} converges to a. Is this right?
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  4. #4
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    Quote Originally Posted by DPMachine View Post
    Just to clarify, so within U_{i_0} there would be an infinite # of a_{n} terms converging to a. If so, then S satisfies the definition of compactness that I had on the first post: "any sequence in S has a subsequence converging to a limit point in S" because a_{n} converges to a and any subsequences of a_{n} converges to a. Is this right?
    Within U_{i_0}, there are infinitely many points of S. What Enrique is saying is that since this one open set covers all but finitely many x_n\in S (say all n>N, for some N), then you can cover remaining points a_1,...,a_N individually.

    The "every sequence has a convergent subsequence" (Bolzano-Weierstrass Theorem) is not used in this particular proof; however, it can be used. Since \{a_n\} converges to a, every subsequence also converges to a. Since S=\{a_n\}\cup a every sequence has a convergent subsequence, so S is compact.
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