real analysis

If $A$ is bounded from above, there exists $m \in \mathbb{R}$ such that $x \leq M$ for $x \in A$. Let $y \in A^\downarrow$. Then there exists $x \in A$ such that $y \leq x$, but $x \leq M$ so $y \leq M$, so $A^\downarrow$ is bounded from above.

Let $\sup A = s$. Then $s = \min\{x \in \mathbb{R} \mid x > y \; \forall y \in A\}$. By the same argument of part a), $\sup A^\downarrow \leq s$. But clearly $A \subseteq A^\downarrow$, so $\sup A \leq \sup A^\downarrow$, so they must be equal.

By definition, $A^\downarrow$ is an interval. Let $x \in A$, and let $\epsilon > 0$. Then there exists $y \in A^\downarrow \cap \mathbb{Q}$ such that $x-\epsilon \leq y \leq x$, because between any two reals there is a rational (this is the fact about rationals we use). But this means $\sup A^\downarrow \cap \mathbb{Q} \geq s-\epsilon$ for all $\epsilon > 0$, so $\sup A^\downarrow \cap \mathbb{Q} \geq s$. But since $\sup A^\downarrow = s$ and $(A^\downarrow \cap \mathbb{Q}) \subseteq A^\downarrow$, then $\sup A^\downarrow \cap \mathbb{Q} \leq s$ so they must be equal.