# Completeness axiom

1. Oct 28, 2010

### annoymage

1. The problem statement, all variables and given/known data

let A be a set of all positive rational number such that $p^2<2$

B be a set of all positive rational number such that $p^2>2$

2. Relevant equations

n/a

3. The attempt at a solution

Set A is clearly non empty, and is a subset of real number, anyway i can choose 3 is upperbound, therefore upperbound exist, so by completeness axiom, supremum exist.

But the book here said

"Set A is bounded above, in fact every element in B a the upperbound of A. Since B has no smallest element, A has no least upper bound/ supremum in Q."

i'm really sure i'm not wrong. But am i wrong?

p/s; i just realised that this book define least-upper-bound property(more general case from completeness axiom), and also above example are the counterexample that proves Q does not have least-upper-bound property(follows from what the book have shown, not mine).

But aren't this contradicting the completeness axiom?

since Q is a subset of R, and any non-empty subset of Q that bounded from above has supremum(from completeness axiom), therefore Q has the least-upper-bound property.

help, where i gone wrong T_T

Last edited: Oct 28, 2010
2. Oct 28, 2010

### fzero

The completeness axiom applies to the reals, there is no contradiction. As you say, the set A demonstrates that Q does not have the least-upper-bound property, so Q is not complete. The completeness axiom guarantees that a subset of Q has a least-upper-bound in R. However that least-upper-bound is not in Q.

3. Oct 28, 2010

### annoymage

AAAAAAAAAAAAAAAAAAA in R, not Q

Thank YOUUUUUUU, i'm so stupid