# Completeness of real numbers as Dedekind cuts

(Title should be Connectedness of...)
Hi. I'm trying to prove that if a set of Dedekind cuts is bounded, it has a least upper bound. We've defined a Dedekind cut, called E, to be a nonempty subset of Q (i) with no last point, (ii) an upper bound in Q, and (iii) the property that if x belongs to Q and y belongs to E, then x < y implies that x belongs to E. And we've defined A < B for Dedekind cuts to mean that A is a subset of B.
I have made some tentative observations without any idea of how the whole proof will look. Dedekind cuts are open because of properties (i) and (iii). The union of a collection of Dedekind cuts will also be open then. In fact, it seems that the union of a bounded collection of Dedekind cuts would itself be a Dedekind cut. Furthermore, no upper bound of a Dedekind cut belongs to it (because of its openness). The same is true for a collection of Dedekind cuts. The complement of any Dedekind cut (the set of upper bounds of that cut) will be closed.
I'm not sure how to proceed. Any advice would be much appreciated. Thanks!

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It would seem that the union of a bounded collection of Dedekind cuts is the least upper bound of that collection. Obviously this union is an upper bound of the collection, and it seems to be the smallest upper bound (maybe this point deserves a more rigorous proof). Any help?

The union of the cuts is the least upper bound. Since it is the union, every cut in the collection is a subset of it, hence less than it. Therefore it is an upper bound. Now we only have to prove that it is the least upper bound. It can be proved as follows:

Proof (by contradiction). Assume the union, which we denote by U isn't the least upper bound. Hence there is another upper bound, which we denote by $\gamma$, which is less than U, or it is a proper subset of U. The set $U\smallsetminus\gamma$ is not empty, i.e. contains at least one element. We call one of the elements a. Since a is in $U\smallsetminus\gamma$, which is a subset of U, it is also in U. If it is in U, it is in one of the cuts but not in $\gamma$. Hence $\gamma$ doesn't include everything in the collection of cuts, which is a contradiction, since it is supposed to be an upper bound and every cut in the collection should be a subset of $\gamma$, which isn't the case here. Since we have reached a contradiction by assuming the union isn't the least upper bound, it is true that the union is the least upper bound.

Hence every bounded collection of cuts contains a least upper bound since we can construct it by taking the union of the cuts.

HallsofIvy