I was looking at the proof for the theorem "If a nonempty set S has an upper bound, then S has a least upper bound L." This is the first few lines of the proof:
Since $S \neq \emptyset $, there exists $s \in S$. Select any number $a_1 \lt s$ so that $a_1$ is too small to be an upper bound for S.
I understand that $S$ contains some element $s$ but how do we know there is also an element $a_1$? What if the set was a set with just one element, namely $s$? As the proof goes on further, it seems like there are actually infinitely many elements in the set $S$.
It is obvious that if $\max S$ exists then $\max S =\sup S.$ So to prove that any non-empty $S$ with an upper bound has a $\sup,$ it suffices to prove it for the case of a non-empty $S$ with no largest member. And a non-empty set of numbers with no largest member must be an infinite set.
There is a variety of ways to define $\mathbb R,$ but they all produce the same structure. Proving that any non-empty subset of $\mathbb R$ with an upper bound has a $\sup$ may depend on which def'n of $\mathbb R$ you choose ( unless you first prove its equivalence to another def'n.)