Why is the l.u.b. property equivalent to Cauchy-sequence convergence for $\mathbb{R}$? Math people:
I browsed some questions with similar titles and could not find a duplicate.  I apologize if it is.  If you read my question it will be obvious that I am not a logician, so please be gentle.  From what I've seen, the three most well-known ways to define or "construct" the real numbers are:
(1) As the unique Archimedean ordered field satisfying a certain least-upper-bound property
(2) Using Dedekind cuts
(3) Using equivalence classes of Cauchy sequences of rational numbers.
That's enough definitions for me for now.  I just discovered there are more, but I will focus on these three. I am interested in knowing exactly why, for each definition, the least-upper-bound property (every nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound) is equivalent to the property that $\mathbb{R}$ is complete in the sense that every Cauchy sequence converges.  It seems built in to (1) and intuitively obvious in (2) and (3), but I would like to know if the Axiom of Choice is required using each definition (becomes sometimes one uses AC in "intuitively obvious" proofs without thinking about it). I prefer to avoid the Axiom of Choice if possible.  I consulted Wikipedia, which I usually find to be an excellent reference, and they did not mention AC at all (for (1)-(3)).  I don't know if this means AC is not required or if they just forgot to mention it.  I read somewhere else on the Net that you need AC to establish the equivalence if you use definition (3), so I am not sure.  If indeed AC is required with definition (3) but not with (2), wouldn't that just make definition (2) and possibly (1) simply better than (3)? 
 A: The Axiom of Choice is not needed in any of the definitions and it is not needed to show that the definitions are equivalent.
Let $\mathbb R_2$ be the set of Dedekind cuts and let $\mathbb R_3$ be the set of Cauchy sequences modulo zero sequences.
Here is a map $\mathbb R_2\to\mathbb R_3$: Given a Dedekind cut $\emptyset \ne A\subsetneq\mathbb Q$, let $f(n)=\inf\{\,m\in\mathbb Z\mid \frac mn\notin A\,\}$. Then $(\frac {f(n)}n)_{n\in\mathbb N}$ is a Cauchy sequence.
For a map $\mathbb R_3\to\mathbb R_2$ in the other direction, let $X$ be an equivalence class of Cauchy sequences.
 Then a Dedekind cut is obtained by letting $A$ the set of all rationals being less than almost all sequence members for all elements of $X$.
Now one can show (constructively of course) that these two maps are inverse to each other.
A: The reason you don't need the axiom of choice is that $\Bbb Q$ is linearly ordered, and countable. So we can well-order it as $\{q_n\mid n\in\Bbb N\}$ and use that order for things.
Now, given $r$ which is defined as an equivalence class of Cauchy sequences, we can find a Cauchy sequence which is strictly increasing $r_n$, and define the cut $\{q\in\Bbb Q\mid\exists k q<r_k\}$, and its complement of course.
Given a Dedekind cut and $A$ is the lower part, we know that $A=\{q_{n_i}\mid i\in\Bbb N\}$ and we can define an increasing sequence $r_0=q_{n_0}$ and $r_{n+1}=q_k$ where $k=\min\{n_i\mid q_{n_i}>r_n\}$. This sequence is strictly increasing, clearly, and cofinal in $A$. But it is also true that it is bounded, because $A$ is bounded as the lower half of a Dedekind cut.
It is not hard to show that $r_n$ is Cauchy, if there exists $\varepsilon>0$, suppose that for all $k$ there is some $n,m>k$ such that $|r_m-r_n|>\varepsilon$, then it is easy to see that $r_n$ is not bounded, in contradiction to the fact that $A$ is.
No choice was used, as you can verify. Whenever it was used, then we used the choice given from the countability of $\Bbb Q$ which is provable in $\sf ZF$ itself of course.
