Open and closed sets #3

**ajj86** · November 2nd 2008, 05:01 PM

Let A be bounded above so that s = sup A exists. Show that s Є A bar.

**ThePerfectHacker** · November 2nd 2008, 05:25 PM

Originally Posted by ajj86

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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