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Math Help - Open and closed sets #3

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    Open and closed sets #3

    Let A be bounded above so that s = sup A exists. Show that s Є A bar.
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    Quote Originally Posted by ajj86 View Post
    Let A be bounded above so that s = sup A exists. Show that s Є A bar.
    Let A be a bounded non-empty set. Let  s = \sup A. To show s \in \bar A it is sufficient to show s \in \partial A i.e. is a boundary point. Let \epsilon > 0 then s - \epsilon is not an upper bound which means there is a\in A such that s - \epsilon < a\leq s, and s+\epsilon is above the supremem thus if b>s+\epsilon then b\not \in A. This means (s-\epsilon,s+\epsilon) intersects both A and \mathbb{R} - A non-trivially. It follows that a\in \partial A
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