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#### conscipost

##### Member

- Jan 26, 2012

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In a book I am currently reading, the statement "every limited hyperreal is infinitely close to a real #" is shown to imply the completeness of R, that is that any subset A of R bounded above has a least upper bound. What the author offers to do is introduce this construction: for each natural n, let s

Without completeness I'm not sure why s

Thanks,

_{n}be the least k in the integers so that k/n is an upper bound of A. Then we are to take an unlimited N and let L, an element of R, be infinitely close to s_{N}/N.Without completeness I'm not sure why s

_{n}necessarily exists, can anyone give me some hints? Is it just because once I know the set is bounded above, I can start with an integer greater than this upper bound multiplied by n and "count down" so to say, checking whether each integer less than the last is an upper bound until I find one that is not?Thanks,

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