# proof that a empty set is bounded

• Sep 17th 2009, 09:02 AM
jmheaney
proof 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, 09:27 AM
Plato
Quote:

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

Consider the set: $\left\{ {x \in \mathbb{Q}^ + :x^2 < 2} \right\}$.
• Sep 17th 2009, 09:31 AM
HallsofIvy
Quote:

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

??? 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 $\pi$ 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?