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Does there exist an elementary function on some subset $I$ of $R$ such that you can prove that $A = \{\sup(f(x)): x \in I\}$ exists, but you cannot find the value of $A$ ?

If the answer is "no", then replace "elementary function" above by "function" as a further question

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Does something like $A = \sup \{1 - \frac 1x \mid x \in 2\mathbb Z, x \ge 4, \exists p,q \text{ prime }: x = p+q\}$ count for you? – martini Nov 6 '12 at 14:23
Ha. Good points. Will have a think about that and get back to you later. – Adam Rubinson Nov 6 '12 at 14:30
Martini - your comment only makes sense if you are able to prove that A exists. Which you are not. Will - what is meant by "cannot find" is self-evident. And I don't see any problems with doing this. – Adam Rubinson Nov 7 '12 at 13:40
Bumpety bump bump – Adam Rubinson Nov 12 '12 at 13:56
Ah..... A exists because we do know that {x: x is even, x = p+q where p and q are prime} is unbounded. – Adam Rubinson Nov 13 '12 at 13:54

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