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Maybe this problem isn't so strange to someone with better understanding. Here it is:

Prove that if sup$A$ $\lt$ $\infty$, then for each natural number $n$ there is an elements $a_n$$\epsilon$ A such that sup$A$ - $\frac1n$ $\lt$ $a_n$ $\le$ sup$A$.

I have tried to apply the Archemedean property here, but didn't get anywhere useful. I also have tried to say sup$A$ - $\frac1n$ is a lower bound of A so inf$A$ is greater than it, but I am not sure if that is useful. If you have any insight on where to move with this problem, please share your knowledge!

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  • $\begingroup$ Can you say what you mean by $A$? Is it a set of real numbers? $\endgroup$
    – Bernard W
    Commented Jun 2, 2016 at 1:36
  • $\begingroup$ This is of no help but I have no idea based on any context whether it is a subset of the reals or not. But I feel that if it was a subset of reals, this property wouldn't hold (?). $\endgroup$
    – rlh282
    Commented Jun 2, 2016 at 1:42
  • $\begingroup$ Would it work for naturals? $\endgroup$
    – rlh282
    Commented Jun 2, 2016 at 1:43
  • $\begingroup$ The supremum (if it exists) of a set $S$ of natural numbers is just the largest element in the set. Suppose that $\sup S=k$ and note that for each $n$ it holds that $k-1/n < k\le k$ so set $a_n=k$ and you are done. $\endgroup$
    – Bernard W
    Commented Jun 2, 2016 at 1:48

1 Answer 1

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Let $\alpha = \sup A$. Then, by definition, $\alpha$ is the smallest upper bound for $A$. Since $\alpha - \frac1n < \alpha$, we conclude that $\alpha - \frac1n$ cannot be an upper bound for $A$. Therefore, there must exist an element $a_n \in A$ such that $\alpha - \frac1n < a_n $. Since $a_n \le \alpha$, we're done.

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  • $\begingroup$ Thank you, I was obviously making this much more complicated. $\endgroup$
    – rlh282
    Commented Jun 2, 2016 at 2:21

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