Hi fellows,


I was wondering. Is the axiom of choice used to show that $\mathbb{R}$ is complete? If yes, is there a way to construct monotonic bounded sequences that do not converge?


Thanks in advance!



Axiom of choice and convergence
The standard construction(s) of $\mathbb R$ do not use the Axiom of Choice. Therefore one cannot construct a bounded monotone sequence that does not converge. 


Maybe, it worths to say that there are also constructions of $\mathbb R$ that makes use of AC. See for instance <http://en.wikipedia.org/wiki/Construction_of_the_real_numbers>, where it is proposed the following construction: take the ring of bounded hyperrationals and quotient out the maximal ideal of infinitesimals. One gets the real numbers. Here, AC is used in actually the weaker form of the ultrafilter lemma (existence of free ultrafilters).


