Prove that if and if , then there exists a natural number such that for all . My thoughts Intuitively, this makes sense. If as then this means that at some point where . So far I've gotten: Given then,
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Originally Posted by Luxury Prove that if and if , then there exists a natural number such that for all . My thoughts Intuitively, this makes sense. If as then this means that at some point where . So far I've gotten: Given then, Since which is an open set there exists some such that and since there exists some such that . Form you conclusion.
Originally Posted by Drexel28 Could you clarify that part? I'm not sure how you went from left to imply right.
Originally Posted by Luxury Could you clarify that part? I'm not sure how you went from left to imply right. You may not be familiar with the open ball notation. Well, how about letting so there exists some such that and so...
Since x>0, it is a little easier to simply let x=epsilon. Substitute in x for epsilon in you inequality. And since from the definition of the , let K(epsilon) = M.
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