Showing that the floor is an integer

Question

Say you define $\lfloor x \rfloor := \sup\{y \in \mathbb{Z}~|~y \leq x\}$. How do you show that $\lfloor x \rfloor$ is an integer, i.e. that the supremum is a member of the set $\{y \in \mathbb{Z}~|~y \leq x\}$?

Answer 1

To prove this, you must use some property of the real number system, as there are other ordered fields for which this supremum may not exist.

Let $S=\{y\in \mathbb Z\mid y\leq x\}$.

Now if integers $a<b$ and $b\in S$ then certainly $a\in S$, so either $S=\emptyset$, $S=\mathbb Z$, or $$S=\{z\in \mathbb Z\mid z\leq n\}$$ for some $n$. (This could be proved by induction as Qiaochu Yuan suggests in the comments.)

The Archimedean property of the reals guarantees $S$ is nonempty and not all of $\mathbb Z$, and therefore has a maximum element $n$. But a maximum element of a subset is always the supremum, since $n$ is clearly an upper bound and if $r<n$ then $r$ is not an upper bound for $S$.

Remark

You could in fact show directly that the existence of a supremum for $S$, in any ordered field, is already enough to guarantee $S$ cannot be $\mathbb Z$ or $\emptyset$, without further appealing to the Archimedean property, but this proof would be quite similar to the proof of the Archimedean property itself from the LUB property.