I am a bit confused on whether completeness of real numbers allows for their Dedekind construction or whether Dedekind construction dictates the completeness of real numbers.
In other words, is completeness an inherent property of real numbers, or are they complete because we construct them this way.
The answer is both. No matter how you construct real numbers they are complete. If a construction does not produce a complete set then it is not producing real numbers. The goal of Dedekind was to make precise the completeness of reals by constructing them with his cuts.
You can define the irrational reals as the Dedekind cuts on $\mathbb Q$ and prove that $\mathbb R$ is order-complete. Or you can define the irrational reals as equivalence classes of Cauchy sequences in $\mathbb Q$ that do not converge to rationals, and prove that $\mathbb R$ is order-complete, and also prove that the Dedekind-cuts on $\mathbb Q$ also produces this "version" of $\mathbb R.$ Or you can define the rules of arithmetic on infinitely long decimal-sequences, and prove that this "version" of $\mathbb R$ is isomorphic to each (either) of the two above.
You can say that $\mathbb R$ is complete because we made it so: Any two complete ordered fields are isomorphic..... BTW, we can extend any ordered field $F$ to a larger ordered field $G,$ where $G$ has "infinitesimals", that is, positive members that are smaller than any positive member of $F$. But any such $G$ cannot be order-complete nor Dedekind-complete... And we cannot extend $\mathbb R$ to a strictly larger ordered field without losing order-completeness and Dedekind-completeness.
