Finite Extremums get arbitrarily close
# Statement 1
Suppose $X \subset \mathbb{R}$ and $X$ has an Upper Bound. Then for any $\epsilon > 0$, there exists $y \in X$ so that $\sup X - y < \epsilon$.
# Proof
Because $X$ has an Upper Bound, the Completeness of the Real Numbers tells us that $\sup X < \infty$. Suppose there was some $\epsilon > 0$ so that $\forall y \in X$, $\sup X - y \geq \epsilon$. Then we have that $\forall y \in X$, $\sup X - \epsilon \geq y$, and thus $\sup X - \epsilon$ is an Upper Bound for $X$. But since $\sup X - \epsilon < \sup X$, this violates the definition of Supremum, giving us a contradiction. $\blacksquare$
# Statement 2
Suppose $X \subset \mathbb{R}$ and $X$ has a Lower Bound. Then for any $\epsilon > 0$, there exists $y \in X$ so that $y - \inf X < \epsilon$.
# Proof
If $X$ has a Lower Bound then $-X = {-x \in \mathbb{R} : x \in X}$ has an Upper Bound. By Statement 1, we have that for any $\epsilon > 0$, $\exists y \in X$, $$\begin{align*} \epsilon &> \sup (-X) - (-y)\\ &= -\inf X + y\\ &= y - \inf X \end{align*}$$ establishing the result., since Supremum of Negative is Negative of Infimum. $\blacksquare$