Method of Dedekind cut is applicable for some other ordered subsets? There are two ways of construction of reals from rationals - by Dedekind cut and by Cauchy sequences. In Dedekind cut, I think, ordering on the set $\mathbb{Q}$ is required to be considered. Whereas, Cantor's method considers Cauchy sequences. I was thinking, are there some ordered subsets in $\mathbb{R}$, which are not complete (as metric spaces), but, for the completion of them, Cantor's method is applicable but not of Dedekind cuts.

If we take $(0,1)$ a subset of $\mathbb{R}$, can we make its completion to $[0,1]$ using the Dedekind cuts philosophy?

 A: A linearly ordered set $(X,<)$ is called (order-)complete iff every non-empty subset that has an upper bound has a supremum. $(\Bbb R, <)$ is an example, constructed from the incomplete linear order $(\Bbb Q, <)$ by Dedekind cuts.
It has been generalised, first to all linearly ordered sets then the partially ordered ones (which a slighly modified version of completeness) in Dedekind-MacNeille completion.
$(0,1)$ is the same as an ordered set as $\Bbb R$. A version of the tangent function can work as an order isomorphism, e.g. So its completion (as an order) is itself, not $[0,1]$. We can see that space as the ordered compactification of $(0,1)$ (a linearly ordered set has a natural order topology, which is Tychonoff, and so they can be compactified as a topological space, but better still, they can be compactified as ordered topological spaces: take the completion first and add a minimum if the completion doesn't have one, and the same for a maximum).
So metrically (with the inherited metric), $[0,1]$ is the Cauchy-completion of $(0,1)$ (which is also compact as $(0,1)$ is totally bounded), and topologically it is its ordered compactification. It's not its order-completion.
