proof that a empty set is bounded

**jmheaney** · September 17th 2009, 08:02 AM

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

**Plato** · September 17th 2009, 08:27 AM

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

**HallsofIvy** · September 17th 2009, 08:31 AM

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?

Thread: proof that a empty set is bounded

LinkBack

Thread Tools

Display

proof that a empty set is bounded

Similar Math Help Forum Discussions

Let B={-x : x ϵ A}, A a non-empty subset of R. Show B is bounded below

Proof: Intersection of infinitly many subsets empty?

Closed, non-empty and bounded sets.

Proof of non-empty subsets, glb and lub

How to proof that a set is not empty?

Search Tags