I'm just starting with some of these concepts, and I need someone to check my work. This is an excercise from Baby Rudin, chapter one.
Question: Let $A$ be a subset of $\mathbb{R}$, bounded from below. Prove:
$$\text{inf} \ A = -\text{sup} (-A)$$
where $-A = \{-x \ | \ x \in A\}$.
My attempt:
Since $A$ is bounded, it has an infimum. So it holds that for all $x \in A, x \geq \alpha = \text{inf} \ A$. Thus, for all $-x \in -A$, it holds that $-x \leq -\alpha$. So $-\alpha$ is an upper bound of $-A$. Since $-A$ is bounded from above, it has a supremum.
Let $-\beta$ be any upper bound of $-A$. Show then, that $-\beta \geq -\alpha$, so that $-a$ is the supremum, since it's the lowest of the upper bounds.
Since $-\beta$ is an upper bound, for all $-x \in -A, \ -x \leq -\beta$. Then, for all $x \in A, \ x \geq \beta$, so $\beta$ is also a lower bound for $A$. But since $\alpha$ was the $\it{greatest}$ lower bound we have $\beta \leq \alpha$, thus we must have $-\beta \geq -\alpha$. Since $-\beta$ was an upper bound for $-A$, and it must be greater than $-\alpha$, then $\text{sup}(-A) = -\alpha$.
Then, of course: $\alpha = \text{inf}(A) = -\text{sup}(-A) $.
Holds water? Are there shortuts and good-to-know things when doing these kinds of problems?
Thanks in advance