$\begingroup$

The following statement is the axiom of completeness:

Every non empty subset of $\mathbb R$ that is bounded above has a least upper bound.

So I was wondering: is it an axiom or can it be proven? One can construct $\mathbb R$ as the set of equivalence classes of Cauchy sequence in $\mathbb Q$. Two sequences are equivalent iff the have the same limit. Then, I tried to prove that a non empty set bounded above has a least upper bound but couldn't find a proof. But I think it is possible because $\mathbb R$ is defined by construction and then there can not be axioms in it?

$\endgroup$ 2

1 Answer

$\begingroup$

It depends on your choice of axioms. Many introductory texts define the real numbers axiomatically, as a set of objects that obey certain axioms. One of the required axioms is the least upper bound axiom you mention. However, using more fundamental (and abstract) axioms, one can construct the real number system from even more primitive notions. In that case, it would be a theorem. In that scenario it is often called the least upper bound property of the reals.

I see you have edited your question. Yes, you can prove the least upper bound property from the construction you mention. You can find a thorough treatment of that construction in many analysis textbooks. There is also a brief treatment on Wikipedia.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy