Completeness Axiom

charikaar · #1 $Old$ November 7th, 2009, 04:12 AM

A= $\left\{ {x\in \mathbb{R}:x^3 < 2} \right\}$

Prove carefully that A has a least upper bound.

A is non-empty and bounded above therefore by the completeness axiom the supremum of A exists. Can you get me started please?

Also is maximum and least upper bound of a set the same thing?

Thanks

theodds · #2 $Old$ November 7th, 2009, 07:55 AM

Supremum and LUB are two words for the same thing. LUB and maximum are not the same concept, since a maximum of a set must reside inside the set, whereas the LUB is not required to.

My guess is that what they are looking for is a proof that A is non-empty and bounded above, so that the LUB exists. Probably the main point is showing that it is bounded above.

tonio · #3 $Old$ November 7th, 2009, 09:13 AM

Quote:

Originally Posted by charikaar $View Post$

A= $\left\{ {x\in \mathbb{R}:x^3 < 2} \right\}$

Prove carefully that A has a least upper bound.

A is non-empty and bounded above therefore by the completeness axiom the supremum of A exists. Can you get me started please?

Also is maximum and least upper bound of a set the same thing?

Thanks

As already said, you must first prove A is non-empty and bounded above, which seems to be fairly simple. Now take a look at $w:=\sqrt[3]{2}$ and prove this number is the LUB:

1) First, prove w is an upper bound;

2) Prove that $\forall \epsilon>0\,\,\exists\,a_{\epsilon}\in A\,\,s.t.\,\,w-\epsilon<a_{\epsilon}\leq w$

Tonio