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!
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Sign up to join this communityHi 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!
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).
In constructive mathematics we do not need the axiom of choice to show that the (two-sided) Dedekind cuts form an ordered archimedean field which is complete in the sense that every Cauchy sequence converges.
However, it is possible to define a monotone bounded sequence, known as a Specker sequence, such that it is not provable constructively that the sequence has a limit.
We may ask what is needed to prove that every monotone bounded sequence has a limit. It is sufficient to assume the principle LPO, which states that every binary sequence is either constantly 0 or it contains a 1. In fact, if every monotone bounded sequence has a limit, then LPO holds. Again (as far as I can see), no choice is involved.