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Thread: covering

  1. #1
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    covering

    Let $\displaystyle S = (0,1) $ and $\displaystyle \mathcal{O} = \{(1/n,1): n \in \bold{N} \} $. Show that $\displaystyle \mathcal{O} $ is an open cover of $\displaystyle S $.

    So choose $\displaystyle x \in (0,1) $. Pick $\displaystyle n_0 \in \bold{N} $ such that $\displaystyle 1/n_0 < x $. This implies that $\displaystyle x \in (1/n_0, 1) \subset \bigcup_{n \in \bold{N}} (1/n,1) = \bigcup_{G \in \mathcal{O}} G $. Is this correct? Could I pick $\displaystyle n_0 $ in a different way?

    What if we had a finite subset $\displaystyle \mathcal{O'} \subset \mathcal{O} $?
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  2. #2
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    If you had a finite subset of the collection, then the smallest left-hand endpoint would a number $\displaystyle 0 < t < 1$.
    But $\displaystyle \frac{t}{2} \in \left( {0,1} \right)$ and $\displaystyle \frac{t}{2} < t$.
    Therefore, the subcollection does not cover $\displaystyle \left( {0,1} \right)$.
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  3. #3
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    The statement that set is compact means every open covering has a finite subcover.
    Having one open cover which has a finite subcover does not imply that the set is compact.

    Consider: $\displaystyle \left( { - 1,.000009} \right) \cup \left[ {\bigcup\limits_{n \in N} {\left( { n^{ - 1} ,1} \right)} } \right]$
    That is clearly an open covering of $\displaystyle \left( {0,1} \right)$ and the collection contains a finite subcover.
    But $\displaystyle \left( {0,1} \right)$ is still not compact.
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