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?
Best Answer
You are extremely close.
$|x_A-x_2|\leq \varepsilon$ is equivalent to $-\varepsilon\leq x_A-x_2\leq \varepsilon$, so you'd be done with the inf part if you could show that $-\varepsilon\leq x_A-x_2$, or in other words, $x_2\leq x_A+\varepsilon$. Can you see why that is true?
Similarly, show that $-\varepsilon\leq x_1-x_B$, or, $x_B\leq x_1+\varepsilon$.
(A pedantic point about notation: Matching $A$ with $2$ and $B$ with $1$ may catch a reader off guard. An even more pedantic point: The "$A$" used as a subscript is the same symbol used to denote the original set, which might also be best avoided.)
Generally speaking, if you are trying to prove something about sups, you will often need to use both key components of their definitions.
The analogous statements hold for infs.
Once you have the essentials down, it's also worth noting that your assumption that the inf and sup are not in the set is superfluous. That is, whether or not $\sup A$ and $\inf A$ are in $A$, the existence of $x_A$ and $x_B$ as you stated is guaranteed by the definition of inf and sup.