# Question about Least Upper Bound Property in Dedekind cut construction of $R$

While reading Rudin's textbook Principles of Mathematical Analysis, the book introduced the idea that a set of rational numbers does not always have the least upper bound property because you can always find a smaller upper bound for a set such as $$\{p \in Q : p^2 < 2\}$$.

If the members of $$R$$ are subsets of $$Q$$ satisfying the conditions for a cut, then every real number is actually a set.

For instance, if I want to think about the number $$\sqrt2$$, I define it as a cut $$\{p \in Q : p^2<2$$ or $$p<0\}$$.

Here's where I have a question.

If you think about it, the reason that the least upper bound property applies for the set of real numbers is because order is redefined to mean "$$a$$ is a proper subset of $$b$$" for $$a,b \in R$$.

What I mean by the above is, suppose there is a nonempty set of real numbers $$A$$. Then this set of real numbers consists of cuts.

The cut that contains all the other cuts is the least upper bound of $$A$$ because we defined order to mean that larger sets contain smaller ones.

So really, the least upper bound property is saying that in a set of cuts, there exists a cut $$A$$ such that every other cut is a proper subset of $$A$$?

If we consider the definition of a least upper bound that:

(1) $$A$$ must be an upper bound and

(2) if $$B$$ < $$A$$, then $$B$$ is not an upper bound.

Then (2) automatically follows from (1) because if $$B \subset A$$ then $$A \not\subset B$$ (by the trichotomy of order) and $$B$$ is not an upper bound.

For a nonempty set of real numbers with its elements as cuts, does the least upper bound always exist as an element in the set?

I've been thinking to an analogous set in the rationals $$\{p \in Q: p<0\}$$ where 0 is the least upper bound and not in the set, but I am not sure how I would construct a similar set of real numbers.

• Are you asking if a set of real numbers always contains its least upper bound? It is false, $(0,1)$ has supremum $1$ not in the set.
– user737563
Commented Dec 31, 2019 at 19:04
• Not necessarily in the given set, but in $\Bbb R$ (if the set is bounded above), and simply take the union of the cuts. Commented Dec 31, 2019 at 19:25
• Is the union of cuts on the interval (0,1) equal to 1? How do I show this? Commented Dec 31, 2019 at 20:44

The short answer is: Yes. If A does have an upper bound, it will definitely have a least upper bound in the subset, and the least upper bound is the union of all cuts in A. The union we denote as $$\gamma$$.
More comments: the $$\gamma$$, the least upper bound of A, is also a cut in R. For proof of this, u can check appendix A of Principles of Mathematical Analysis