Prove the least upper bound of a set is in the closure of that set

Question

If $S \subset \mathbb{R}$ and $x$ is the least upper bound of $S$, prove $x$ is in the closure of $S$.

I think this means I either have to show that $x \in S$ or $x$ is a limit point of $S$, so I have to do $2$ different cases, but in the solution it says it suffices to show that there exists a $y$ such that $x-r < y < x+r$, $r \in \mathbb{R}$. Why is this true?

Answer 1

You have two cases. If $x \in S$, then you're done. So let's not worry about that case. Now suppose that $x \not \in S$.

Since $x$ is the least upper bound of $S$, for every $\epsilon > 0$, there must be an element $y$ in $S$ such that $y \in (x - \epsilon, x)$. Otherwise, $x - \epsilon$ is a smaller upper bound. This is how I would rephrase that aspect of the solution.

Since this must be true for all $\epsilon > 0$, you can find a sequence of $y$ in $S$ which have limit $x$, and thus $x$ is in the closure.

Answer 2

This has a more general formulation.

Let $\langle X,\le\rangle$ be a totally ordered set equipped with the order topology.

Proposition: IF a nonempty subset $A$ of $X$ has least upper bound $\alpha \in X$
THEN
$\;\;\alpha$ is in the closure of $A$.

Proof:
We can assume without harm that at least one point in $X$ is different than $\alpha$.

An open ray containing $\alpha$ of the form $(-\infty,\beta)$ will automatically include all of $A$. So let any open ray $(\beta,+\infty)$ be given containing $\alpha$. We have to show that this open ray intersects $A$.

Assume to the contrary that it has an empty intersection with $A$. If $\beta \lt a$ for $a \in A$, then $a$ would be in $(\beta,\alpha) \subset (\beta,+\infty)$; so $\beta$ must be an upper bound of $A$. But $\beta \lt \alpha$. This is a contradiction.