I recently read that a real number $r$ is any subset of the set $\mathbb{Q}$ of rational numbers such that it satisfies the following.
1) $r$ is nonempty
2) $r\neq\mathbb{Q}$
3) $r$ is closed downwards
4) $r$ contains no greatest element
My question is how can a single real number $r$ be a set of rational numbers?
Best Answer
To elaborate perhaps a bit on what Y. Forman wrote, it's more that we think of the real numbers as the Dedekind-complete ordered field - they satisfy the field axioms (just like $\mathbb{Q}$, for example) and they have the least-upper-bound property (that any set $S \subset \mathbb{R}$ such that $\forall s \in S, s \le b$ for some $b \in \mathbb{R}$ means there is a least $\lambda \in \mathbb{R}$ such that $\forall s \in S, s \le \lambda$).
What the Dedekind cuts are is a specific construction of a Dedekind-complete ordered field. There are actually different constructions! But, they are all the same in an important sense: they are isomorphic.
Proving this stuff is an interesting exercise - but the takeaway is, you don't need to think of the reals as being Dedekind cuts. Rather, we're proving that the concept of Dedekind-complete ordered field makes sense, by showing that it can be constructed - e.g., as Dedekind cuts - and then regardless of how it's constructed, we use the properties of "Dedekind-complete and an ordered field" to deduce results.