# 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 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.
$$(0,1)$$ 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.