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Thread: Open sets

  1. #1
    Aug 2007

    Open sets

    Theorem: Every open set $\displaystyle U \subset \mathbb{R} $ can be uniquely expressed as a countable union of disjoint open intervals. The endpoints of $\displaystyle U $ do not belong to $\displaystyle U $ (Pugh, p. 63).

    The book goes on to prove this by first defining $\displaystyle a_x = \inf\{ a : \exists(a,x) \subset U \} $ and $\displaystyle b_x = \sup \{b: \exists (x,b) \subset U \} $ and then an interval $\displaystyle I_x = (a_x, b_x) $.

    I don't understand why they define $\displaystyle \sup $ and $\displaystyle \inf $ for those particular sets. What does $\displaystyle I_x $ represent? I know that its the largest interval, because the endpoints are the glb and lub.
    Last edited by shilz222; Sep 10th 2007 at 12:56 PM.
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  2. #2
    MHF Contributor

    Aug 2006
    Once again, I am not quite sure about your misunderstanding.
    What the text is doing is constructing what are know as components in topology. A component is a maximal connected set. Components are pair-wise disjoint. In $\displaystyle \Re^1$ the components of an open subspace are open intervals. You are correct to worry about glbís and lubís. In the example you gave, unless $\displaystyle U$ is bounded, it is possible that $\displaystyle a_x = - \infty \mbox{ or } b_x = \infty $.
    Does this help you?
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