prove/ disprove that every non empty subset of Q that is bounded above has a supremum in Q

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- Sep 17th 2009, 08:02 AMjmheaneyproof that a empty set is bounded
prove/ disprove that every non empty subset of Q that is bounded above has a supremum in Q

- Sep 17th 2009, 08:27 AMPlato
- Sep 17th 2009, 08:31 AMHallsofIvy
??? You titled this "proof that an empty set is bounded"! This has nothing to do with the empty set!

Since that is false, a counter-example suffices. Let A= {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...} where each number in the sequence is $\displaystyle \pi$ to one more decimal place. Since the**real numbers**are "complete", this statement is true for**R**- there is a supremum in**R**. If this set had a supremum in**Q**, it would have to be the same as the supremum in**R**. What**is**the supremum of this set (in**R**). Is that a rational number?