Supremum of Bounded Above Set of Reals is in Closure

Theorem
Let $\R$ be the real number line under the Euclidean metric.

Let $H \subseteq \R$ be a bounded above subset of $\R$ such that $H \ne \varnothing$.

Let $u = \sup \left({H}\right)$ be the supremum of $H$.

Then:
 * $u \in \operatorname{cl}\left({H}\right)$

where $\operatorname{cl}\left({H}\right)$ denotes the closure of $H$ in $\R$.

Proof
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

Let $B_\epsilon \left({u}\right)$ be the open $\epsilon$-ball of $u$ in $\R$.

From Distance from Subset of Real Numbers:
 * $d \left({u, H}\right) = 0$

Thus by definition of distance from subset:


 * $\exists x \in H: d \left({u, x}\right) < \epsilon$

Thus $x \in B_\epsilon \left({u}\right)$.

As $x \in H$ and $x \in B_\epsilon \left({u}\right)$, from the definition of intersection:
 * $x \in H \cap B_\epsilon \left({u}\right)$

The result follows from Condition for Point being in Closure.

Also see

 * Infimum of Bounded Below Set of Reals is in Closure