# Thread: proof that a empty set is bounded

1. ## 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

2. 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: $\displaystyle \left\{ {x \in \mathbb{Q}^ + :x^2 < 2} \right\}$.

3. 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 $\displaystyle \pi$ 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?

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# empty set bounded

Click on a term to search for related topics.