I came across the following problem about closures:
If $A$ is a bounded nonempty subset of $\mathbb{R}$, prove that $\sup A \in \overline{A}$ and $\inf A \in \overline{A}$.
Proof. By hypothesis, $A$ satisfies the least upper bound property (and the greatest lower bound property). So $\sup A$ and $\inf A$ exist. Now we know that $\overline{A} = \{x \in \mathbb{R}: a_n \to x \ \text{for some sequence} \ (a_n) \ \text{in} \ A \}$
Moreover, $x \in \overline{A} \Longleftrightarrow (\forall \epsilon >0) \ \exists a \in A \ni |x-a| < \epsilon$. If $\sup A$ and $\inf A$ are in $A$, then we are done. So suppose they are not. Let $x_1 = \sup A$ and $x_2 = \inf A$. By definition, for every $\epsilon >0$, there exists $x_A \in A$ such that $x_A \leq x_2 +\epsilon$. Likewise, there exists $x_B \in A$ such that $x_1- \epsilon \leq x_B$. So we get $x_A-x_2 \leq \epsilon$ and $x_1-x_B \leq \epsilon$
It seems like we would be done if we put absolute value signs on those quantities. Because then we can approximate the supremum and infimum as closely as we like with members of $A$.
How would I get the absolute value signs on the above two quantities?